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ICSE Class 10 Mathematics Chapter 21 - Trigonometric Identities Selina Solutions

ICSE Class 10 Mathematics Chapter 21 - Trigonometric Identities Selina Solutions

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ICSE Class 10 Mathematics Chapter 21 Selina Concise Solutions - Free PDF Download

Updated ICSE Class 10 Mathematics Chapter 21 - Trigonometric Identities Selina Solutions are provided by Vedantu in a step by step method. Selina is the most famous publisher of ICSE textbooks. Studying these solutions by Selina Concise Mathematics Class 10 Solutions which are explained and solved by our subject matter experts will help you in preparing for ICSE exams. Concise Mathematics Class 10 ICSE Solutions can be easily downloaded in the given PDF format. These solutions for Class 10 ICSE will help you to score good marks in ICSE Exams 2024-25.

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Access ICSE Selina Solutions for Grade 10 Mathematics Chapter 21 - Trigonometric Identities

Exercise - 21(A)

1. Prove the following identity:

$\frac{\sec A - 1}{\sec A + 1} = \frac{1 - \cos A}{1 + \cos A}$

Ans: Taking LHS -

$LHS = \frac{sec A - 1}{sec A + 1}$

$= \frac{\frac{1}{\cos A} - 1}{\frac{1}{\cos A} + 1}$ $[We know that : \cos A = \frac{1}{\sec A}]$

$= \frac{\frac{1 - \cos A}{\cos A}}{\frac{1 + \cos A}{\cos A}}$

$= \frac{1 - \cos A}{1 + \cos A} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

2. Prove the following identity:

$\frac{1 + \sin A}{1 - \sin A} = \frac{cosec A + 1}{cosec A - 1}$

Ans: Taking RHS -

$RHS = \frac{cosec A + 1}{cosec A - 1}$

$= \frac{\frac{1}{\sin A} + 1}{\frac{1}{\sin A} - 1}$ $[We know that : cosec A = \frac{1}{\sin A}]$

$= \frac{\frac{1 + \sin A}{\sin A}}{\frac{1 - \sin A}{\sin A}}$

$= \frac{1 + \sin A}{1 - \sin A} = LHS$

$\Rightarrow LHS = RHS$

Hence proved.

3. Prove the following identity:

$\frac{1}{\tan A + \cot A} = \cos A \sin A$

Ans: Taking LHS -

$LHS = \frac{1}{\tan A + \cot A}$

$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$ $[We know that : \tan A = \frac{\sin A}{\cos A}, \cot A = \frac{\cos A}{\sin A}]$

$= \frac{1}{\frac{si n^{2} A + co s^{2} A}{\sin A \cos A}}$

$= \frac{\sin A \cos A}{\frac{si n^{2} A + co s^{2} A}{1}}$

$= \sin A \cos A$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \cos A \sin A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

4. Prove the following identity:

$\tan A - \cot A = \frac{1 - 2 co s^{2} A}{\sin A \cos A}$

Ans: Taking LHS -

$LHS = \tan A - \cot A$

$= \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A}$ $[We know that : \tan A = \frac{\sin A}{\cos A}, \cot A = \frac{\cos A}{\sin A}]$

$= \frac{si n^{2} A - co s^{2} A}{\cos A \sin A}$

$= \frac{1 - co s^{2} A - co s^{2} A}{\cos A \sin A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{1 - 2 co s^{2} A}{\sin A \cos A} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

5. Prove the following identity:

$si n^{4} A - co s^{4} A = 2 si n^{2} A - 1$

Ans: Taking LHS -

$LHS = si n^{4} A - co s^{4} A$

$= (si n^{2} A - co s^{2} A) (si n^{2} A + co s^{2} A)$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= (si n^{2} A - co s^{2} A)$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= si n^{2} A - (1 - si n^{2} A)$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 2 si n^{2} A - 1 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

6. Prove the following identity:

${(1 - \tan A)}^{2} + {(1 + \tan A)}^{2} = 2 se c^{2} A$

Ans: Taking LHS -

$LHS = {(1 - \tan A)}^{2} + {(1 + \tan A)}^{2}$

$= 1 + ta n^{2} A - 2 \tan A + 1 + ta n^{2} A + 2 \tan A$ $[{(a \pm b)}^{2} = a^{2} + b^{2} \pm 2 ab]$

$= 2 + 2 ta n^{2} A$

$= 2 (1 + ta n^{2} A)$

$= 2 se c^{2} A = RHS$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$\Rightarrow LHS = RHS$

Hence proved.

7. Prove the following identity:

$cose c^{4} A - cose c^{2} A = co t^{4} A + co t^{2} A$

Ans: Taking LHS -

$LHS = cose c^{4} A - cose c^{2} A$

Taking $cose c^{2} A$ common,

$= cose c^{2} A (cose c^{2} A - 1)$

$= cose c^{2} A \times co t^{2} A$ $[We know that : 1 + co t^{2} A = cose c^{2} A]$

$= (1 + co t^{2} A) \times co t^{2} A$ $[We know that : 1 + co t^{2} A = cose c^{2} A]$

$= co t^{4} A + co t^{2} A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

8. Prove the following identity:

$\sec A (1 - \sin A) (\sec A + \tan A) = 1$

Ans: Taking LHS -

$LHS = \sec A (1 - \sin A) (\sec A + \tan A)$

$= \frac{1}{\cos A} (1 - \sin A) (\frac{1}{\cos A} + \frac{\sin A}{\cos A})$ $[We know : \tan A = \frac{\sin A}{\cos A}, \sec A = \frac{1}{\cos A}]$

$= \frac{(1 - \sin A)}{\cos A} (\frac{1 + \sin A}{\cos A})$

$= \frac{(1^{2} - si n^{2} A)}{co s^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{(1 - si n^{2} A)}{co s^{2} A}$

$= \frac{co s^{2} A}{co s^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 1 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

9. Prove the following identity:

$cosec A (1 + \cos A) (cosec A - \cot A) = 1$

Ans: Taking LHS -

$LHS = cosec A (1 + \cos A) (cosec A - \cot A)$

$= \frac{1}{\sin A} (1 + \cos A) (\frac{1}{\sin A} - \frac{\cos A}{\sin A})$ $[\cot A = \frac{\cos A}{\sin A}, cosec A = \frac{1}{\sin A}]$

$= \frac{(1 + \cos A)}{\sin A} (\frac{1 - \cos A}{\sin A})$

$= \frac{(1^{2} - co s^{2} A)}{si n^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{(1 - co s^{2} A)}{si n^{2} A}$

$= \frac{si n^{2} A}{si n^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 1 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

10. Prove the following identity:

$se c^{2} A + cose c^{2} A = se c^{2} A . cose c^{2} A$

Ans: Taking LHS -

$LHS = \frac{1}{co s^{2} A} + \frac{1}{si n^{2} A}$ $[We know that : \sec A = \frac{1}{\cos A}, cosec A = \frac{1}{\sin A}]$

$= \frac{si n^{2} A + co s^{2} A}{si n^{2} A \times co s^{2} A}$

$= \frac{1}{si n^{2} A \times co s^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= se c^{2} A . cose c^{2} A = RHS$ $[We know that : \sec A = \frac{1}{\cos A}, cosec A = \frac{1}{\sin A}]$

$\Rightarrow LHS = RHS$

Hence proved.

11. Prove the following identity:

$\frac{(1 + ta n^{2} A) \cot A}{cose c^{2} A} = \tan A$

Ans: Taking LHS -

$LHS = \frac{(1 + ta n^{2} A) \cot A}{cose c^{2} A}$

$= \frac{(se c^{2} A) \cot A}{cose c^{2} A}$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$= \frac{\frac{1}{co s^{2} A} \times \frac{\cos A}{\sin A}}{\frac{1}{si n^{2} A}}$ $[We know that : \sec A = \frac{1}{\cos A}, cosec A = \frac{1}{\sin A}]$

$= \frac{\frac{1}{\cos A} \times \frac{1}{\sin A}}{\frac{1}{si n^{2} A}}$

$= \frac{\sin A}{\cos A}$ $[We know that : \tan A = \frac{\sin A}{\cos A}]$

$= \tan A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

12. Prove the following identity:

$ta n^{2} A - si n^{2} A = ta n^{2} A . si n^{2} A$

Ans: Taking LHS -

$LHS = \frac{si n^{2} A}{co s^{2} A} - \frac{si n^{2} A}{1}$ $[We know that : \tan A = \frac{\sin A}{\cos A}]$

$= \frac{si n^{2} A - co s^{2} A \times si n^{2} A}{co s^{2} A}$

$= \frac{si n^{2} A (1 - co s^{2} A)}{co s^{2} A}$

$= ta n^{2} A . (si n^{2} A) = RHS$ $[We know that : si n^{2} A + co s^{2} A = 1, \tan A = \frac{\sin A}{\cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

13. Prove the following identity:

$co t^{2} A - co s^{2} A = co s^{2} A . co t^{2} A$

Ans: Taking LHS -

$LHS = \frac{co s^{2} A}{si n^{2} A} - \frac{co s^{2} A}{1}$ $[We know that : \cot A = \frac{\cos A}{\sin A}]$

$= \frac{co s^{2} A - co s^{2} A \times si n^{2} A}{si n^{2} A}$

$= \frac{co s^{2} A (1 - si n^{2} A)}{si n^{2} A}$

$= co t^{2} A . (co s^{2} A) = RHS$ $[We know that : si n^{2} A + co s^{2} A = 1, \cot A = \frac{\cos A}{\sin A}]$

$\Rightarrow LHS = RHS$

Hence proved.

14. Prove the following identity:

$(cosec A + \sin A) (cosec A - \sin A) = co t^{2} A + co s^{2} A$

Ans: Taking LHS -

$LHS = (cosec A + \sin A) (cosec A - \sin A)$

$= (cose c^{2} A - si n^{2} A)$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$[We know that : si n^{2} A + co s^{2} A = 1 & 1 + co t^{2} A = cose c^{2} A]$

$= 1 + co t^{2} A - (1 - co s^{2} A)$

$= co t^{2} A + co s^{2} A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

15. Prove the following identity:

$(\sec A - \cos A) (\sec A + \cos A) = ta n^{2} A + si n^{2} A$

Ans: Taking LHS -

$LHS = (\sec A - \cos A) (\sec A + \cos A)$

$= (se c^{2} A - co s^{2} A)$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$[We know that : si n^{2} A + co s^{2} A = 1 & 1 + ta n^{2} A = se c^{2} A]$

$= 1 + ta n^{2} A - (1 - si n^{2} A)$

$= ta n^{2} A + si n^{2} A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

16. Prove the following identity:

${(\cos A + \sin A)}^{2} + {(\cos A - \sin A)}^{2} = 2$

Ans: Taking LHS -

$LHS = {(\cos A + \sin A)}^{2} {(\cos A - \sin A)}^{2}$

$= co s^{2} A + si n^{2} A + 2 \cos A . \sin A + co s^{2} A + si n^{2} A - 2 \cos A . \sin A$

$[{(a \pm b)}^{2} = a^{2} + b^{2} \pm 2 ab]$

$= 2 (co s^{2} A + si n^{2} A)$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 2 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

17. Prove the following identity:

$(cosec A - \sin A) (\sec A - \cos A) (\tan A + \cot A) = 1$

Ans: Taking LHS -

$LHS = (cosec A - \sin A) (\sec A - \cos A) (\tan A + \cot A)$

$[We know that : \cot A = \frac{\cos A}{\sin A}, \tan A = \frac{\sin A}{\cos A}, cosec A = \frac{1}{\sin A}, \sec = \frac{1}{\cos A}]$

$= (\frac{1}{\sin A} - \sin A) (\frac{1}{\cos A} - \cos A) (\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})$

$= (\frac{1 - si n^{2} A}{\sin A}) (\frac{1 - co s^{2} A}{\cos A}) (\frac{si n^{2} A + co s^{2} A}{\cos A . \sin A})$

$= (\frac{co s^{2} A}{\sin A}) (\frac{si n^{2} A}{\cos A}) (\frac{1}{\cos A . \sin A})$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 1 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

18. Prove the following identity:

$\frac{1}{\sec A + \tan A} = \sec A - \tan A$

Ans: Taking LHS -

$LHS = \frac{1}{\sec A + \tan A}$

$= \frac{se c^{2} A - ta n^{2} A}{\sec A + \tan A}$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$= \frac{(\sec A - \tan A) (\sec A + \tan A)}{\sec A + \tan A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \sec A - \tan A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

19. Prove the following identity:

$cosec A + \cot A = \frac{1}{cosec A - \cot A}$

Ans: Taking RHS -

$RHS = \frac{1}{cosec A - \cot A}$

$= \frac{cose c^{2} A - co t^{2} A}{cosec A - \cot A}$ $[We know that : 1 + co t^{2} A = cose c^{2} A]$

$= \frac{(cosec A - \cot A) (cosec A + \cot A)}{cosec A - \cot A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= cosec A + \cot A = LHS$

$\Rightarrow LHS = RHS$

Hence proved.

20. Prove the following identity:

$\frac{\sec A - \tan A}{\sec A + \tan A} = 1 - 2 \sec A . \tan A + 2 ta n^{2} A$

Ans: Taking LHS -

$LHS = \frac{\sec A - \tan A}{\sec A + \tan A}$

$= \frac{\sec A - \tan A}{\sec A + \tan A} \times \frac{\sec A - \tan A}{\sec A - \tan A}$

$= \frac{{(\sec A - \tan A)}^{2}}{se c^{2} A - ta n^{2} A}$

$[We know that : se c^{2} A - ta n^{2} A = 1 & {(a + b)}^{2} = a^{2} + b^{2} + 2 ab]$

$= se c^{2} A + ta n^{2} A - 2 \sec A . \tan A$

$[We know that : se c^{2} A = 1 + ta n^{2} A]$

$= 1 + ta n^{2} A + ta n^{2} A - 2 \sec A . \tan A$

$= 1 - 2 \sec A . \tan A + 2 ta n^{2} A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

21. Prove the following identity:

${(\sin A + cosec A)}^{2} + {(\cos A + \sec A)}^{2} = 7 + ta n^{2} A + co t^{2} A$

Ans: Taking LHS -

$LHS = {(\sin A + cosec A)}^{2} {(\cos A + \sec A)}^{2}$

$= si n^{2} A + cose c^{2} A + 2 cosec A . \sin A + co s^{2} A + se c^{2} A + 2 \cos A . \sec A$

$[{(a + b)}^{2} = a^{2} + b^{2} + 2 ab]$

$= (co s^{2} A + si n^{2} A) + (cose c^{2} A + se c^{2} A) + 2 + 2$ $[We know that : cosec A . \sin A = 1 & \cos A . \sec A = 1]$

$= 2 + 2 + 1 + (1 + co t^{2} A) + (1 + ta n^{2} A)$

$[We know that : se c^{2} A = 1 + ta n^{2} A & 1 + co t^{2} A = cose c^{2} A]$

$= 7 + ta n^{2} A + co t^{2} A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

22. Prove the following identity:

$se c^{2} A . cose c^{2} A = ta n^{2} A + co t^{2} A + 2$

Ans: Taking LHS -

$LHS = se c^{2} A . cose c^{2} A$

$= \frac{1}{co s^{2} A} \times \frac{1}{si n^{2} A}$

$= \frac{si n^{2} A + co s^{2} A}{co s^{2} A . si n^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{si n^{2} A}{co s^{2} A . si n^{2} A} + \frac{co s^{2} A}{co s^{2} A . si n^{2} A}$

$= \frac{1}{co s^{2} A} + \frac{1}{si n^{2} A}$

$= se c^{2} A + cose c^{2} A$ $[We know that : cosec A = \frac{1}{\sin A}, \sec A = \frac{1}{\cos A}]$

$= 1 + ta n^{2} A + 1 + co t^{2} A$ $[se c^{2} A = 1 + ta n^{2} A & 1 + co t^{2} A = cose c^{2} A]$

$= ta n^{2} A + co t^{2} A + 2 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

23. Prove the following identity:

$\frac{1}{1 + \cos A} + \frac{1}{1 - \cos A} = 2 cose c^{2} A$

Ans: Taking LHS -

$LHS = \frac{1}{1 + \cos A} + \frac{1}{1 - \cos A}$

$= \frac{1 - \cos A + 1 + \cos A}{(1 - \cos A) (1 + \cos A)}$

$= \frac{2}{1^{2} - co s^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2}{1 - co s^{2} A}$

$= \frac{2}{si n^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 2 cose c^{2} A = RHS$ $[We know that : cosec A = \frac{1}{\sin A}]$

$\Rightarrow LHS = RHS$

Hence proved.

24. Prove the following identity:

$\frac{1}{1 - \sin A} + \frac{1}{1 + \sin A} = 2 se c^{2} A$

Ans: Taking LHS -

$LHS = \frac{1}{1 - \sin A} + \frac{1}{1 + \sin A}$

$= \frac{1 + \sin A + 1 - \sin A}{(1 - \sin A) (1 + \sin A)}$

$= \frac{2}{1^{2} - si n^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2}{1 - si n^{2} A}$

$= \frac{2}{co s^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 2 se c^{2} A = RHS$ $[We know that : \sec A = \frac{1}{\cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

25. Prove the following identity:

$\frac{cosec A}{cosec A - 1} + \frac{cosec A}{cosec A + 1} = 2 se c^{2} A$

Ans: Taking LHS -

$LHS = \frac{cosec A}{cosec A - 1} + \frac{cosec A}{cosec A + 1}$

$= cosec A \times [\frac{1 + cosec A + cosec A - 1}{(cosec - 1) (cosec + 1)}]$

$= \frac{2 cose c^{2} A}{cose c^{2} A - 1^{2}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2 cose c^{2} A}{cose c^{2} A - 1}$

$= \frac{2 cose c^{2} A}{co t^{2} A}$ $[We know that : 1 + co t^{2} A = cose c^{2} A]$

$= 2 \times \frac{\frac{1}{si n^{2} A}}{\frac{co s^{2} A}{si n^{2} A}}$

$= 2 \times \frac{1}{co s^{2} A}$

$= 2 se c^{2} A = RHS$ $[We know that : \sec A = \frac{1}{\cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

26. Prove the following identity:

$\frac{\sec A}{\sec A + 1} + \frac{\sec A}{\sec A - 1} = 2 cose c^{2} A$

Ans: Taking LHS -

$LHS = \frac{\sec A}{\sec A + 1} + \frac{\sec A}{\sec A - 1}$

$= \sec A \times [\frac{\sec A - 1 + \sec A + 1}{(\sec + 1) (\sec - 1)}]$

$= \frac{2 se c^{2} A}{se c^{2} A - 1^{2}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2 se c^{2} A}{se c^{2} A - 1^{}}$

$= \frac{2 se c^{2} A}{ta n^{2} A}$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$= 2 \times \frac{\frac{1}{co s^{2} A}}{\frac{si n^{2} A}{co s^{2} A}}$

$= 2 \times \frac{1}{si n^{2} A}$

$= 2 cose c^{2} A = RHS$ $[We know that : cosec A = \frac{1}{\sin A}]$

$\Rightarrow LHS = RHS$

Hence proved.

27. Prove the following identity:

$\frac{1 + \cos A}{1 - \cos A} = \frac{ta n^{2} A}{{(\sec A - 1)}^{2}}$

Ans: Taking LHS -

$LHS = \frac{1 + \cos A}{1 - \cos A}$

Divide N & D by cos A,

$= \frac{\frac{1 + \cos A}{\cos A}}{\frac{1 - \cos A}{\cos A}}$

$= \frac{\frac{1}{\cos A} + 1}{\frac{1}{\cos A} - 1^{}}$

$= \frac{\sec A + 1}{\sec A - 1}$ $[We know that : \sec A = \frac{1}{\cos A}]$

$= \frac{\sec A + 1}{\sec A - 1} \times \frac{\sec A - 1}{\sec A - 1}$

$= \frac{se c^{2} A - 1^{2}}{{(\sec A - 1)}^{2}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{se c^{2} A - 1}{{(\sec A - 1)}^{2}}$

$= \frac{ta n^{2} A}{{(\sec A - 1)}^{2}} = RHS$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$\Rightarrow LHS = RHS$

Hence proved.

28. Prove the following identity:

$\frac{co t^{2} A}{{(cosec A + 1)}^{2}} = \frac{1 - \sin A}{1 + \sin A}$

Ans: Taking RHS -

$RHS = \frac{1 - \sin A}{1 + \sin A}$

Divide N & D by sin A,

$= \frac{\frac{1 - \sin A}{\sin A}}{\frac{1 + \sin A}{\sin A}}$

$= \frac{\frac{1}{\sin A} - 1}{\frac{1}{\sin A} + 1^{}}$

$= \frac{cosec A - 1}{cosec A + 1}$ $[We know that : cosec A = \frac{1}{\sin A}]$

$= \frac{cosec A - 1}{cosec A + 1} \times \frac{cosec A + 1}{cosec A + 1}$

$= \frac{cose c^{2} A - 1^{2}}{{(cosec A + 1)}^{2}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{cose c^{2} A - 1}{{(cosec A + 1)}^{2}}$

$= \frac{co t^{2} A}{{(cosec A + 1)}^{2}} = LHS$ $[We know that : 1 + co t^{2} A = cose c^{2} A]$

$\Rightarrow LHS = RHS$

Hence proved.

29. Prove the following identity:

$\frac{1 + \sin A}{\cos A} + \frac{\cos A}{1 + \sin A} = 2 \sec A$

Ans: Taking LHS -

$LHS = \frac{1 + \sin A}{\cos A} + \frac{\cos A}{1 + \sin A}$

$= \frac{{(1 + \sin A)}^{2} + co s^{2} A}{\cos A (1 + \sin A)}$

$= \frac{1 + si n^{2} A + 2 \sin A + co s^{2} A}{\cos A (1 + \sin A)}$ $[{(a + b)}^{2} = a^{2} + b^{2} + 2 ab]$

$= \frac{1 + 1 + 2 \sin A}{\cos A (1 + \sin A)}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{2 (1 + \sin A)}{\cos A (1 + \sin A)}$

$= \frac{2}{\cos A}$

$= 2 secA = RHS$ $[We know that : \sec A = \frac{1}{\cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

30. Prove the following identity:

$\frac{1 - \sin A}{1 + \sin A} = {(\sec A - \tan A)}^{2}$

Ans: Taking LHS -

$LHS = \frac{1 - \sin A}{1 + \sin A}$

Divide N & D by cos A,

$= \frac{\frac{1 - \sin A}{\cos A}}{\frac{1 + \sin A}{\cos A}}$

$= \frac{\frac{1}{\cos A} - \frac{\sin A}{\cos A}}{\frac{1}{\cos A} + \frac{\sin A}{\cos A}}$

$= \frac{\sec A - \tan A}{\sec A + \tan A}$ $[We know that : \sec A = \frac{1}{\cos A}, \tan A = \frac{\sin A}{\cos A}]$

$= \frac{\sec A - \tan A}{\sec A + \tan A} \times \frac{\sec A - \tan A}{\sec A - \tan A}$

$= \frac{{(\sec A - \tan A)}^{2}}{se c^{2} A - ta n^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{{(\sec A - \tan A)}^{2}}{1}$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$= {(\sec A - \tan A)}^{2} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

31. Prove the following identity:

${(\cot A - cosec A)}^{2} = \frac{1 - \cos A}{1 + \cos A}$

Ans: Taking RHS -

Divide N & D by sin A,

$= \frac{\frac{1 - \cos A}{\sin A}}{\frac{1 + \cos A}{\sin A}}$

$= \frac{\frac{1}{\sin A} - \frac{\cos A}{\sin A}}{\frac{1}{\sin A} + \frac{\cos A}{\sin A}}$

$= \frac{cosec A - \cot A}{cosec A + \cot A}$ $[We know that : cosec A = \frac{1}{\sin A}, \cot A = \frac{\cos A}{\sin A}]$

$= \frac{cosec A - \cot A}{cosec A + \cot A} \times \frac{cosec A - \cot A}{cosec A - \cot A}$

$= \frac{{(cosec A - \cot A)}^{2}}{cose c^{2} A - co t^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{{[- (\cot A - cosec A)]}^{2}}{1}$ $[We know that : 1 + co t^{2} A = cose c^{2} A]$

$= {(\cot A - cosec A)}^{2} = LHS$

$\Rightarrow LHS = RHS$

Hence proved.

32. Prove the following identity:

$\frac{cosec A - 1}{cosec A + 1} = {(\frac{\cos A}{1 + \sin A})}^{2}$

Ans: Taking LHS -

$LHS = \frac{\frac{1}{\sin A} - 1}{\frac{1}{\sin A} + 1}$ $[We know that : cosec A = \frac{1}{\sin A}]$

$= \frac{\frac{1 - \sin A}{\sin A}}{\frac{1 + \sin A}{\sin A}}$

$= \frac{1 - \sin A}{1 + \sin A}$

$= \frac{1 - \sin A}{1 + \sin A} \times \frac{1 + \sin A}{1 + \sin A}$

$= \frac{1^{2} - si n^{2} A}{{(1 + \sin A)}^{2}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{1 - si n^{2} A}{{(1 + \sin A)}^{2}}$

$= \frac{co s^{2} A}{{(1 + \sin A)}^{2}}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= {(\frac{\cos A}{1 + \sin A})}^{2} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

33. Prove the following identity:

$ta n^{2} A - ta n^{2} B = \frac{si n^{2} A - si n^{2} B}{co s^{2} A . co s^{2} B}$

Ans: Taking LHS -

$LHS = ta n^{2} A - ta n^{2} B$

$= \frac{si n^{2} A}{co s^{2} A} - \frac{si n^{2} B}{co s^{2} B}$ $[We know that : \tan A = \frac{\sin A}{\cos A}]$

$= \frac{si n^{2} A . co s^{2} B - si n^{2} B . co s^{2} A}{co s^{2} A . co s^{2} B}$

$= \frac{si n^{2} A (1 - si n^{2} B) - si n^{2} B (1 - si n^{2} A)}{co s^{2} A . co s^{2} B}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{si n^{2} A - si n^{2} A . si n^{2} B - si n^{2} B + si n^{2} A . si n^{2} B}{co s^{2} A . co s^{2} B}$

$= \frac{si n^{2} A - si n^{2} B}{co s^{2} A . co s^{2} B} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

34. Prove the following identity:

$\frac{\sin θ - 2 si n^{3} θ}{2 co s^{3} θ - \cos θ} = \tan θ$

Ans: Taking LHS -

$LHS = \frac{\sin θ - 2 si n^{3} θ}{2 co s^{3} θ - \cos θ}$

$= \frac{\sin θ}{\cos θ} [\frac{1 - 2 si n^{2} θ}{2 co s^{2} θ - 1}]$

$= \frac{\sin θ}{\cos θ} [\frac{si n^{2} θ + co s^{2} θ - 2 si n^{2} θ}{2 co s^{2} θ - (si n^{2} θ + co s^{2} θ)}]$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \tan θ [\frac{co s^{2} θ - si n^{2} θ}{co s^{2} θ - si n^{2} θ}]$

$= \tan θ = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

35. Prove the following identity:

$\frac{\sin A}{1 + \cos A} = cosec A - \cot A$

Ans: Taking LHS -

$LHS = \frac{\sin A}{1 + \cos A}$

$= \frac{\sin A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}$

$= \frac{\sin A (1 - \cos A)}{1^{2} - co s^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{\sin A - \sin A . \cos A}{1 - co s^{2} A}$

$= \frac{\sin A}{1 - co s^{2} A} - \frac{\sin A . \cos A}{1 - co s^{2} A}$

$= \frac{\sin A}{si n^{2} A} - \frac{\sin A . \cos A}{si n^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{1}{\sin A} - \frac{\cos A}{\sin A}$

$= cosec A - \cot A = RHS$ $[We know that : cosec A = \frac{1}{\sin A}, \cot A = \frac{\cos A}{\sin A}]$

$\Rightarrow LHS = RHS$

Hence proved.

36. Prove the following identity:

$\frac{\cos A}{1 - \sin A} = \sec A + \tan A$

Ans: Taking LHS -

$LHS = \frac{\cos A}{1 - \sin A}$

$= \frac{\cos A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}$

$= \frac{\cos A (1 + \sin A)}{1^{2} - si n^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{\cos A + \sin A . \cos A}{1 - si n^{2} A}$

$= \frac{\cos A}{1 - si n^{2} A} + \frac{\sin A . \cos A}{1 - si n^{2} A}$

$= \frac{\cos A}{co s^{2} A} + \frac{\sin A . \cos A}{co s^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{1}{\cos A} + \frac{\sin A}{\cos A}$

$= \sec A + \tan A = RHS$ $[We know that : \sec A = \frac{1}{\cos A}, \tan A = \frac{\sin A}{\cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

37. Prove the following identity:

$\frac{\sin A . \tan A}{1 - \cos A} = 1 + \sec A$

Ans: Taking LHS -

$LHS = \frac{\sin A . \tan A}{1 - \cos A}$

$= \frac{\sin A . \frac{\sin A}{\cos A}}{1 - \cos A}$ $[We know that : \tan A = \frac{\sin A}{\cos A}]$

$= \frac{\frac{si n^{2} A}{\cos A}}{1 - \cos A}$

$= \frac{si n^{2} A}{\cos A (1 - \cos A)}$

$= \frac{1 - co s^{2} A}{\cos A (1 - \cos A)}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{(1 - \cos A) (1 + \cos A)}{\cos A (1 - \cos A)}$

$= \frac{(1 + \cos A)}{\cos A}$

$= \frac{1}{\cos A} + \frac{\cos A}{\cos A}$ $[We know that : \sec A = \frac{1}{\cos A}]$

$= 1 + \sec A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

38. Prove the following identity:

$(1 + \cot A - cosec A) (1 + \tan A + \sec A) = 2$

Ans: Taking LHS -

$LHS = (1 + \cot A - cosec A) (1 + \tan A + \sec A)$

$= (1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A}) (1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A})$

$[We know that : \sec A = \frac{1}{\cos A}, \tan A = \frac{\sin A}{\cos A}, cosec A = \frac{1}{\sin A}, \cot = \frac{\cos A}{\sin A}]$

$= (\frac{sin A + cos A - 1}{sin A}) (\frac{cos A + sin A + 1}{cos A})$

$= \frac{cos A (sin A + cos A - 1) + sin A (sin A + cos A - 1) + sin A + cos A - 1}{sin A . cos A}$

$= \frac{\cos A \sin A + \cos^{2} A - \cos A + \sin^{2} A + \sin A \cos A - \sin A + \sin A + \cos A - 1}{\sin A . \cos A}$

$= \frac{cos A.sin A - cos A + 1 + sin A.cos A - sin A + sin A + cos A - 1}{sin A . cos A}$

$[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{2 cos A . sin A}{sin A . cos A}$

$= 2 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

39. Prove the following identity:

$\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$

Ans: Taking LHS -

$LHS = \sqrt{\frac{1 + \sin A}{1 - \sin A}}$

$= \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}}$

$= \sqrt{\frac{{(1 + \sin A)}^{2}}{1^{2} - si n^{2} A}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \sqrt{\frac{{(1 + \sin A)}^{2}}{co s^{2} A}}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{1 + \sin A}{\cos A}$

$= \frac{1}{\cos A} + \frac{\sin A}{\cos A}$ $[We know that : \sec A = \frac{1}{\cos A}, \tan A = \frac{\sin A}{\cos A}]$

$= \sec A + \tan A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

40. Prove the following identity:

$\sqrt{\frac{1 - \cos A}{1 + \cos A}} = cosec A - \cot A$

Ans: Taking LHS -

$LHS = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$= \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}}$

$= \sqrt{\frac{{(1 - \cos A)}^{2}}{1^{2} - co s^{2} A}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \sqrt{\frac{{(1 - \cos A)}^{2}}{si n^{2} A}}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{1 - \cos A}{\sin A}$

$= \frac{1}{\sin A} - \frac{\cos A}{\sin A}$ $[We know that : cosec A = \frac{1}{\sin A}, \cot A = \frac{\cos A}{\sin A}]$

$= cosec A - \cot A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

41. Prove the following identity:

$\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A}$

Ans: Taking LHS -

$LHS = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$= \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A}}$

$= \sqrt{\frac{1^{2} - co s^{2} A}{{(1 + \cos A)}^{2}}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \sqrt{\frac{si n^{2} A}{{(1 + \cos A)}^{2}}}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{\sin A}{1 + \cos A} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

42. Prove the following identity:

$\sqrt{\frac{1 - \sin A}{1 + \sin A}} = \frac{\cos A}{1 + \sin A}$

Ans: Taking LHS -

$LHS = \sqrt{\frac{1 - \sin A}{1 + \sin A}}$

$= \sqrt{\frac{1 - \sin A}{1 + \sin A} \times \frac{1 + \sin A}{1 + \sin A}}$

$= \sqrt{\frac{1^{2} - si n^{2} A}{{(1 + \sin A)}^{2}}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \sqrt{\frac{co s^{2} A}{{(1 + \sin A)}^{2}}}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{\cos A}{1 + \sin A} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

43. Prove the following identity:

$1 - \frac{co s^{2} A}{1 + \sin A} = \sin A$

Ans: Taking LHS -

$LHS = 1 - \frac{co s^{2} A}{1 + \sin A}$

$= 1 - \frac{1^{2} - si n^{2} A}{1 + \sin A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 1 - [\frac{(1 - \sin A) (1 + \sin A)}{1 + \sin A}]$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= 1 - [1 - \sin A]$

$= \sin A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

44. Prove the following identity:

$\frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A} = \frac{2 \sin A}{1 - 2 co s^{2} A}$

Ans: Taking LHS -

$LHS = \frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A}$

$= \frac{\sin A - \cos A + \sin A + \cos A}{(\sin A + \cos A) (\sin A - \cos A)}$

$= \frac{2 \sin A}{(\sin^{2} A - co s^{2} A)}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2 \sin A}{(1 - co s^{2} A - co s^{2} A)}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{2 \sin A}{1 - 2 co s^{2} A} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

45. Prove the following identity:

$\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 si n^{2} A - 1}$

Ans: Taking LHS -

$LHS = \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A}$

$= \frac{{(\sin A + \cos A)}^{2} + {(\sin A - \cos A)}^{2}}{(\sin A + \cos A) (\sin A - \cos A)}$

$= \frac{si n^{2} A + co s^{2} A + 2 \sin A . \cos A + si n^{2} A + co s^{2} A - 2 \sin A . \cos A}{(\sin^{2} A - co s^{2} A)}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2 (si n^{2} A + co s^{2} A)}{(si n^{2} A - (1 - si n^{2} A)}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{2}{2 si n^{2} A - 1} = RHS$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$\Rightarrow LHS = RHS$

Hence proved.

46. Prove the following identity:

$\frac{\cot A + cosec A - 1}{\cot A - cosec A + 1} = \frac{1 + \cos A}{\sin A}$

Ans: Taking LHS -

$LHS = \frac{\cot A + cosec A - 1}{\cot A - cosec A + 1}$

$= [\frac{\cot A + cosec A - (cose c^{2} A - co t^{2} A)}{\cot A - cosec A + 1}]$ $[We know that : 1 + co t^{2} A = cose c^{2} A]$

$= [\frac{\cot A + cosec A - (cosec A - cotA) (\cot A + cosec A)}{\cot A - cosec A + 1}]$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= [\frac{(\cot A + cosec A) (1 - cosec A + cotA)}{\cot A - cosec A + 1}]$

$= \cot A + cosec A$

$= \frac{\cos A}{\sin A} + \frac{1}{\sin A}$ $[We know that : cosec A = \frac{1}{\sin A}, \cot A = \frac{\cos A}{\sin A}]$

$= \frac{1 + \cos A}{\sin A} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

47. Prove the following identity:

$\frac{1 + \sin A}{cosec A - \cot A} - \frac{1 - \sin A}{cosec A + \cot A} = 2 (1 + \cot A)$

Ans: Taking LHS -

$LHS = \frac{1 + \sin A}{cosec A - \cot A} - \frac{1 - \sin A}{cosec A + \cot A}$

$= \frac{(1 + \sin A) (cosec A + \cot A) - (1 - \sin A) (cosec A - \cot A)}{(cosec A - \cot A) (cosec A + \cot A)}$

$= \frac{cosec A + \cot A + \sin A cosec A + \sin A \cot A - cosec A + \cot A + \sin A cosec A - \sin A \cot A}{(cosec A - \cot A) (cosec A + \cot A)}$ $= \frac{2 (\cot A + \sin A cosec A)}{(cose c^{2} A - co t^{2} A)}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2 (\cot A + 1)}{1}$ $[We know that : 1 + co t^{2} A = cose c^{2} A & cosec A = \frac{1}{\sin A}]$

$= 2 (1 + \cot A) = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

48. Prove the following identity:

$\frac{\cos θ . \cot θ}{1 + \sin θ} = cosec θ - 1$

Ans: Taking LHS -

$LHS = \frac{\cos θ . \cot θ}{1 + \sin θ}$

$= \frac{\cos θ . \frac{\cos θ}{\sin θ}}{1 + \sin θ}$ $[We know that : \cot A = \frac{\cos A}{\sin A}]$

$= \frac{\frac{co s^{2} θ}{\sin θ}}{1 + \sin θ}$

$= \frac{\frac{co s^{2} θ}{\sin θ}}{1 + \sin θ} \times \frac{1 - \sin θ}{1 - \sin θ}$

$= \frac{\frac{co s^{2} θ}{\sin θ} \times (1 - \sin θ)}{1^{2} - si n^{2} θ}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{\frac{co s^{2} θ}{\sin θ} \times (1 - \sin θ)}{co s^{2} θ}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{1 - \sin θ}{\sin θ}$

$= cosec θ - 1 = RHS$ $[We know that : cosec A = \frac{1}{\sin A}]$

$\Rightarrow LHS = RHS$

Hence proved.

49. Prove the following identity:

$\frac{\sin θ . \tan θ}{1 - \cos θ} = \sec θ + 1$

Ans: Taking LHS -

$LHS = \frac{\sin θ . \tan θ}{1 - \cos θ}$

$= \frac{\sin θ . \frac{\sin θ}{\cos θ}}{1 - \cos θ}$ $[We know that : \tan A = \frac{\sin A}{\cos A}]$

$= \frac{\frac{si n^{2} θ}{\cos θ}}{1 - \cos θ}$

$= \frac{\frac{si n^{2} θ}{\cos θ}}{1 - \cos θ} \times \frac{1 + \cos θ}{1 + \cos θ}$

$= \frac{\frac{si n^{2} θ}{\cos θ} \times (1 + \cos θ)}{1^{2} - co s^{2} θ}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{\frac{si n^{2} θ}{\cos θ} \times (1 + \cos θ)}{si n^{2} θ}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{1 + \cos θ}{\cos θ}$

$= \sec θ + 1 = RHS$ $[We know that : \sec A = \frac{1}{\cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

Exercise - 21(B)

1. Prove that:

i) $\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A$

Ans: Taking LHS -

$LHS = \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A}$

$= \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}$ $[We know that : \tan A = \frac{\sin A}{\cos A}, \cot A = \frac{\cos A}{\sin A}]$

$= \frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin A}{\frac{\sin A - \cos A}{\sin A}}$

$= \frac{co s^{2} A}{\cos A - \sin A} + \frac{si n^{2} A}{\sin A - \cos A}$

$= \frac{co s^{2} A}{\cos A - \sin A} - \frac{si n^{2} A}{\cos A - \sin A}$

$= \frac{co s^{2} A - si n^{2} A}{\cos A - \sin A}$

$= \frac{(\cos A - \sin A) (\cos A + \sin A)}{\cos A - \sin A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \cos A + \sin A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

ii) $\frac{co s^{3} A + si n^{3} A}{\cos A + \sin A} + \frac{co s^{3} A - si n^{3} A}{\cos A - \sin A} = 2$

Ans: Taking LHS -

$LHS = \frac{co s^{3} A + si n^{3} A}{\cos A + \sin A} + \frac{co s^{3} A - si n^{3} A}{\cos A - \sin A}$

$= \frac{(co s^{3} A + si n^{3} A) (\cos A - \sin A) + (co s^{3} A - si n^{3} A) (\cos A + \sin A)}{(\cos A + \sin A) (\cos A - \sin A)}$

$= \frac{(co s^{3} A + si n^{3} A) (\cos A - \sin A) + (co s^{3} A - si n^{3} A) (\cos A + \sin A)}{co s^{2} A - si n^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{(co s^{4} A + si n^{3} A . \cos A - co s^{3} A . \sin A - si n^{4} A) + (co s^{4} A - si n^{3} A . \cos A + co s^{3} A . \sin A - si n^{4} A)}{co s^{2} A - si n^{2} A}$ $= \frac{2 (co s^{4} A - si n^{4} A)}{co s^{2} A - si n^{2} A}$

$= \frac{2 (co s^{2} A - si n^{2} A) (co s^{2} A + si n^{2} A)}{co s^{2} A - si n^{2} A}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= 2 (co s^{2} A + si n^{2} A)$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 2 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

iii) $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = \sec A . cosec A + 1$

Ans: Taking LHS -

$LHS = \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$

$= \frac{\tan A}{1 - \frac{1}{\tan A}} + \frac{\frac{1}{\tan A}}{1 - \tan A}$ $[We know that : \cot A = \frac{1}{\tan A}]$

$= \frac{ta n^{2} A}{\tan A - 1} + \frac{1}{\tan A (1 - \tan A)}$

$= \frac{ta n^{2} A}{\tan A - 1} - \frac{1}{\tan A (\tan A - 1)}$

$= \frac{ta n^{3} A - 1}{\tan A (\tan A - 1)}$

$= \frac{(\tan A - 1) (ta n^{2} A + 1 + \tan A)}{\tan A (\tan A - 1)}$ $[a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + ab)]$

$= \frac{ta n^{2} A + 1 + \tan A}{\tan A}$

$= \frac{se c^{2} A + \tan A}{\tan A}$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$= \frac{se c^{2} A}{\tan A} + 1$

$= \frac{\frac{1}{co s^{2} A}}{\frac{\sin A}{\cos A}} + 1$ $[We know that : \sec A = \frac{1}{\cos A}, \tan A = \frac{\sin A}{\cos A}]$

$= \frac{1}{\sin A . \cos A} + 1$

$= cosec A . \sec A + 1 = RHS$ $[We know that : cosec A = \frac{1}{\sin A}]$

$\Rightarrow LHS = RHS$

Hence proved.

iv) ${(\tan A + \frac{1}{\cos A})}^{2} + {(\tan A - \frac{1}{\cos A})}^{2} = 2 (\frac{1 + si n^{2} A}{1 - si n^{2} A})$

Ans: Taking LHS -

$LHS = {(\tan A + \frac{1}{\cos A})}^{2} + {(\tan A - \frac{1}{\cos A})}^{2}$

$= {(\frac{\sin A}{\cos A} + \frac{1}{\cos A})}^{2} + {(\frac{\sin A}{\cos A} - \frac{1}{\cos A})}^{2}$ $[We know that : \tan A = \frac{\sin A}{\cos A}]$

$= {(\frac{\sin A + 1}{\cos A})}^{2} + {(\frac{\sin A - 1}{\cos A})}^{2}$

$= \frac{{(\sin A + 1)}^{2} + {(\sin A - 1)}^{2}}{co s^{2} A}$

$= \frac{si n^{2} A + 1 + 2 \sin A + si n^{2} A + 1 - 2 \sin A}{co s^{2} A}$ $[{(a \pm b)}^{2} = a^{2} + b^{2} \pm 2 ab]$

$= \frac{2 (1 + si n^{2} A)}{1 - si n^{2} A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 2 (\frac{1 + si n^{2} A}{1 - si n^{2} A}) = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

v) $2 si n^{2} A + co s^{4} A = 1 + si n^{4} A$

Ans: Taking LHS -

$LHS = 2 si n^{2} A + co s^{4} A$

$= 2 si n^{2} A + {(1 - si n^{2} A)}^{2}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= 2 si n^{2} A + 1 + si n^{4} A - 2 si n^{2} A$ $[{(a - b)}^{2} = a^{2} + b^{2} - 2 ab]$

$= 1 + si n^{4} A = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

vi) $\frac{\sin A - \sin B}{\cos A + \cos B} + \frac{\cos A - \cos B}{\sin A + \sin B} = 0$

Ans: Taking LHS -

$LHS = \frac{\sin A - \sin B}{\cos A + \cos B} + \frac{\cos A - \cos B}{\sin A + \sin B}$

$= \frac{(\sin A - \sin B) (\sin A + \sin B) + (\cos A - \cos B) (\cos A + \cos B)}{(\cos A + \cos B) (\sin A + \sin B)}$

$= \frac{(si n^{2} A - si n^{2} B) + (co s^{2} A - co s^{2} B)}{(\cos A + \cos B) (\sin A + \sin B)}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{(si n^{2} A + co s^{2} A) - (si n^{2} B + co s^{2} B)}{(\cos A + \cos B) (\sin A + \sin B)}$

$= \frac{1 - 1}{(\cos A + \cos B) (\sin A + \sin B)}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \frac{0}{(\cos A + \cos B) (\sin A + \sin B)}$

$= 0 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

vii) $(cosec A - \sin A) (\sec A - \cos A) = \frac{1}{\tan A + \cot A}$

Ans: Taking LHS -

$LHS = (cosec A - \sin A) (\sec A - \cos A)$

$= (\frac{1}{\sin A} - \sin A) (\frac{1}{\cos A} - \cos A)$ $[cosec A = \frac{1}{\sin A}, \sec A = \frac{1}{\cos A}]$

$= (\frac{1 - si n^{2} A}{\sin A}) (\frac{1 - co s^{2} A}{\cos A})$

$= (\frac{co s^{2} A}{\sin A}) (\frac{si n^{2} A}{\cos A})$

$= \sin A . \cos A$

Taking RHS -

$RHS = \frac{1}{\tan A + \cot A}$

$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$ $[We know that : \cot A = \frac{\cos A}{\sin A}, \tan A = \frac{\sin A}{\cos A}]$

$= \frac{1}{\frac{si n^{2} A + co s^{2} A}{\cos A \times \sin A}}$

$= \frac{\cos A \times \sin A}{si n^{2} A + co s^{2} A}$

$= \sin A . \cos A$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$\Rightarrow LHS = RHS$

Hence proved.

viii) ${(1 + \tan A . \tan B)}^{2} + {(\tan A - \tan B)}^{2} = se c^{2} A . se c^{2} B$

Ans: Taking LHS -

$LHS = {(1 + \tan A . \tan B)}^{2} + {(\tan A - \tan B)}^{2}$

$= 1 + ta n^{2} A . ta n^{2} B + 2 \tan A . \tan B + ta n^{2} A + ta n^{2} B - 2 \tan A . \tan B$

$[{(a \pm b)}^{2} = a^{2} + b^{2} \pm 2 ab]$

$= 1 + ta n^{2} A + ta n^{2} A . ta n^{2} B + ta n^{2} B$

$= se c^{2} A + ta n^{2} B (ta n^{2} A + 1)$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$= se c^{2} A + ta n^{2} B (se c^{2} A)$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$= se c^{2} A (1 + ta n^{2} B)$

$= se c^{2} A . se c^{2} B = RHS$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$\Rightarrow LHS = RHS$

Hence proved.

ix) $\frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1} = cosec A + \sec A$

Ans: Taking LHS -

$LHS = \frac{1}{\cos A + \sin A - 1} + \frac{1}{\cos A + \sin A + 1}$

$= \frac{\cos A + \sin A - 1 + \cos A + \sin A + 1}{(\cos A + \sin A - 1) (\cos A + \sin A + 1)}$

$= \frac{2 (\cos A + \sin A)}{{(\cos A + \sin A)}^{2} - 1^{2}}$ $[a^{2} - b^{2} = (a - b) (a + b)]$

$= \frac{2 (\cos A + \sin A)}{si n^{2} A + co s^{2} A + 2 \sin A . \cos A - 1}$ $[{(a + b)}^{2} = a^{2} + b^{2} + 2 ab]$

$= \frac{2 (\cos A + \sin A)}{1 + 2 \sin A . \cos A - 1}$

$= \frac{(\cos A + \sin A)}{\sin A . \cos A}$

$= \frac{(\cos A)}{\sin A . \cos A} + \frac{(\sin A)}{\sin A . \cos A}$

$= \frac{1}{\sin A} + \frac{1}{\cos A}$

$= cosec A + \sec A = RHS$ $[We know that : cosec A = \frac{1}{\sin A}, \sec A = \frac{1}{\cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

2.If $x . \cos A + y . \sin A = m$ and $x . \sin A - y . \cos A = n$ , then prove that: $x^{2} + y^{2} = m^{2} + n^{2}$

Ans: Taking RHS -

$RHS = m^{2} + n^{2}$

$= {(x . cos A + y . sin A)}^{2} + {(x . sin A - y . cos A)}^{2}$

$= x^{2} . co s^{2} A + y^{2} si n^{2} A + 2 xy . cos A . sin A + x^{2} . si n^{2} A + y^{2} co s^{2} A - 2 xy . cos A . sin A$ $[{(a \pm b)}^{2} = a^{2} + b^{2} \pm 2 ab]$

$= x^{2} (co s^{2} A + si n^{2} A) + y^{2} (co s^{2} A + si n^{2} A)$

$= x^{2} + y^{2} = LHS$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$\Rightarrow LHS = RHS$

Hence proved.

3. If $m = a . \sec A + b . \tan A$ and $n = a . \tan A + b . \sec A$ , then prove that: $m^{2} - n^{2} = a^{2} - b^{2}$

Ans: Taking LHS -

$LHS = m^{2} - n^{2}$

$= {(a . \sec A + b . \tan A)}^{2} - {(a . \tan A + b . \sec A)}^{2}$

$= a^{2} . se c^{2} A + b^{2} ta n^{2} A + 2 ab . sec A . tan A - a^{2} . ta n^{2} A - b^{2} se c^{2} - 2 ab . sec A . tan A$

$[{(a + b)}^{2} = a^{2} + b^{2} + 2 ab]$

$= a^{2} (se c^{2} A - ta n^{2} A) - b^{2} (se c^{2} A - ta n^{2} A)$

$= a^{2} - b^{2} = RHS$ $[We know that : 1 + ta n^{2} A = se c^{2} A]$

$\Rightarrow LHS = RHS$

Hence proved.

4. If $x = r \sin A \cos B, y = r \sin A \sin B$ and $z = r \cos A$ , then prove that: $x^{2} + y^{2} + z^{2} = r^{2}$

Ans: Taking LHS -

$LHS = x^{2} + y^{2} + z^{2}$

$= {(r \sin A \cos B)}^{2} + {(r \sin A \sin B)}^{2} + {(r \cos A)}^{2}$

$= {(r \sin A)}^{2} (co s^{2} B + si n^{2} B) + {(r \cos A)}^{2}$

$[We know that : si n^{2} A + co s^{2} A = 1]$

$= {(r \sin A)}^{2} + {(r \cos A)}^{2}$

$= r^{2} (si n^{2} A + co s^{2} A)$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= r^{2} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

5. If $\sin A + \cos A = m$ and $\sec A + cosec A = n$ , show that: $n (m^{2} - 1) = 2 m$

Ans: Taking LHS -

$LHS = n (m^{2} - 1)$

$= (sec\;A + cosec A) [{(sin A + cos A)}^{2} - 1]$

$= (sec\;A + cosec A) [si n^{2} A + co s^{2} A + 2 cos A . sin A - 1]$

$[{(a + b)}^{2} = a^{2} + b^{2} + 2 ab]$

$= (sec\;A + cosec\;A) [1 + 2 cos A . sin A - 1]$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= (sec A + cosec A) [2 cos A . sin A]$

$= (sec A) [2 cos A . sin A] + cosec A [2 cos A . sin A]$

$= [2 sin A] + [2 cos A]$ $[We know : cos A . sec A = 1, cosec A . sin A = 1]$

$= 2 [sin A + cos A]$

$= 2 m = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

6. If $x = r \cos A \cos B, y = r \cos A \sin B$ and $z = r \sin A$ , show that: $x^{2} + y^{2} + z^{2} = r^{2}$

Ans: Taking LHS -

$LHS = x^{2} + y^{2} + z^{2}$

$= {(r \cos A \cos B)}^{2} + {(r \cos A \sin B)}^{2} + {(r \sin A)}^{2}$

$= {(r \cos A)}^{2} (co s^{2} B + si n^{2} B) + {(r \sin A)}^{2}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= {(r \cos A)}^{2} + {(r \sin A)}^{2}$

$= r^{2} (co s^{2} A + si n^{2} A)$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= r^{2} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

7. If $\frac{\cos A}{\cos B} = m$ and $\frac{\cos A}{\sin B} = n$ , show that: $(m^{2} + n^{2}) co s^{2} B = n^{2} .$

Ans: Taking LHS -

$LHS = (m^{2} + n^{2}) co s^{2} B$

$= [{(\frac{cos A}{cos B})}^{2} + {(\frac{cos A}{sin B})}^{2}] co s^{2} B$

$= [{(\frac{1}{cos B})}^{2} + {(\frac{1}{sin B})}^{2}] co s^{2} A . co s^{2} B$

$= [{(\frac{si n^{2} B + co s^{2} B}{co s^{2} B . si n^{2} B})}^{}] co s^{2} A . co s^{2} B$

$= [{(\frac{1}{co s^{2} B . si n^{2} B})}^{}] co s^{2} A . co s^{2} B$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= [{(\frac{1}{si n^{2} B})}^{}] co s^{2} A$

$= {(\frac{cos A}{sin\;B})}^{2} = n^{2} = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

Exercise - 21(C)

1. Show that:

i) $\tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ} = 1$

Ans: Taking LHS -

$LHS = tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ}$

$= tan {(90 - 80)}^{\circ} tan {(90 - 75)}^{\circ} tan 75^{\circ} tan 80^{\circ}$

$= cot 80^{\circ} cot 75^{\circ} tan 75^{\circ} tan 80^{\circ}$ $[We know that : tan (90^{\circ} - A) = cot A]$

$= 1 = RHS$ $[We know that : tan A . cot A = 1]$

$\Rightarrow LHS = RHS$

Hence proved.

ii) $\sin 42^{\circ} \sec 48^{\circ} + \cos 42^{\circ} cosec 48^{\circ} = 2$

Ans: Taking LHS -

$LHS = sin 42^{\circ} sec 48^{\circ} + cos 42^{\circ} cosec 48^{\circ}$

$= sin 42^{\circ} sec {(90 - 42)}^{\circ} + cos 42^{\circ} cosec (90 - 42)^{\circ}$

$= sin 42^{\circ} cosec 42^{\circ} + cos 42^{\circ} sec 42^{\circ}$

$[We know that : sec (90^{\circ} - A) = cosec A, cosec (90^{\circ} - A) = sec A]$

$= 1 + 1$ $[We know that : sin A . cosec A = 1, cos A . sec A = 1]$

$= 2 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

iii) $\frac{\sin 26^{\circ}}{\sec 64^{\circ}} + \frac{\cos 26^{\circ}}{cosec 64^{\circ}} = 1$

Ans: Taking LHS -

$LHS = \frac{sin 26^{\circ}}{sec 64^{\circ}} + \frac{cos 26^{\circ}}{cosec 64^{\circ}}$

$= \frac{sin 26^{\circ}}{sec {(90 - 26)}^{\circ}} + \frac{cos 26^{\circ}}{cosec {(90 - 26)}^{\circ}}$

$= \frac{sin 26^{\circ}}{cosec {(26)}^{\circ}} + \frac{cos 26^{\circ}}{sec {(26)}^{\circ}}$

$[We know that : sec (90^{\circ} - A) = cosec A, cosec (90^{\circ} - A) = sec A]$

$= si n^{2} 26^{\circ} + co s^{2} 26^{\circ}$ $[We know that : cosec A = \frac{1}{sin A}, sec A = \frac{1}{cos A}]$

$= 1 = RHS$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$\Rightarrow LHS = RHS$

Hence proved.

2. Express each of the following in terms of angles between $0^{\circ}$ and $45^{\circ}$ :

i) $\sin 59^{\circ} + \tan 63^{\circ}$

Ans: $\sin 59^{\circ} + \tan 63^{\circ} = \sin {(90 - 31)}^{\circ} + \tan {(90 - 27)}^{\circ}$

$[We know that : sin (90^{\circ} - A) = cos A, tan (90^{\circ} - A) = cot A]$

$\Rightarrow \sin 59^{\circ} + \tan 63^{\circ} = \cos 31^{\circ} + \cot 27^{\circ}$

ii) $cosec 68^{\circ} + \cot 72^{\circ}$

Ans: $cosec 68^{\circ} + \cot 72^{\circ} = cosec {(90 - 22)}^{\circ} + \cot {(90 - 18)}^{\circ}$

$[We know that : cosec (90^{\circ} - A) = sec A, cot (90^{\circ} - A) = tan A]$

$\Rightarrow cosec 68^{\circ} + \cot 72^{\circ} = \sec 22^{\circ} + \tan 18^{\circ}$

iii) $\cos 74^{\circ} + \sec 67^{\circ}$

Ans: $\cos 74^{\circ} + \sec 67^{\circ} = \cos {(90 - 16)}^{\circ} + \sec {(90 - 23)}^{\circ}$

$[We know that : cos (90^{\circ} - A) = sin A, sec (90^{\circ} - A) = cosec A]$

$\Rightarrow \cos 74^{\circ} + \sec 67^{\circ} = \sin 16^{\circ} + cosec 23^{\circ}$

3. Show that:

i) $\frac{\sin A}{\sin (90 - A)} + \frac{\cos A}{\cos (90 - A)} = \sec A . cosec A$

Ans: Taking LHS -

$LHS = \frac{sin A}{sin (90 - A)} + \frac{cos A}{cos (90 - A)}$

$= \frac{sin A}{cos A} + \frac{cos A}{sin A}$ $[cos (90^{\circ} - A) = sin A, sin (90^{\circ} - A) = cos A]$

$= \frac{si n^{2} A + co s^{2} A}{cos A . sin A}$

$= \frac{1}{cos A . sin A}$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$= \sec A . cosec A = RHS$ $[We know that : cosec A = \frac{1}{sin A}, sec A = \frac{1}{cos A}]$

$\Rightarrow LHS = RHS$

Hence proved.

ii) $\sin A \cos A - \frac{\sin A \cos (90 - A) \cos A}{\sec (90 - A)} - \frac{\sin A \sin (90 - A) \cos A}{cosec (90 - A)} = 0$

Ans: Taking LHS -

$LHS = sin A. cos A - \frac{sin A.cos (90 - A) cos A}{sec (90 - A)} - \frac{sin A sin (90 - A) cos A}{cosec (90 - A)}$

$= sin A.cos A - \frac{sin A sin A cos A}{cosec A} - \frac{sin A.cos A cos A}{sec A}$ $[We know that : cos (90^{\circ} - A) = sin A, sin (90^{\circ} - A) = cos A]$

$[We know that : cosec (90^{\circ} - A) = sec A, sec (90^{\circ} - A) = cosec A]$

$= sin A.cos A - si n^{3} A cos A - co s^{3} A sin A$

$[We know that : cosecA = \frac{1}{sinA}, sec A = \frac{1}{cosA}]$

$= sinA.cosA - sinA.cosA (si n^{2} A + co s^{2} A)$

$[We know that : si n^{2} A + co s^{2} A = 1]$

$= sinA.cosA - sinA.cosA (1)$

$= sinA.cosA - sinA.cosA$

$= 0 = RHS$

$\Rightarrow LHS = RHS$

Hence proved.

4. For triangle ABC, show that:

i) $\sin (\frac{A + B}{2}) = \cos \frac{C}{2}$

Ans: $\sin (\frac{A + B}{2})$

We know that for a $Δ A B C,$

$∠ A + ∠ B + ∠ C = 180^{\circ}$

$\Rightarrow \frac{∠ A + ∠ B + ∠ C}{2} = \frac{180^{\circ}}{2}$

$\Rightarrow \frac{∠ A + ∠ B + ∠ C}{2} = {90^{\circ}}^{}$

$\Rightarrow \frac{∠ A + ∠ B}{2} = 90^{\circ} - {\frac{∠ C}{2}}^{}$

$\sin (\frac{A + B}{2}) = \sin (90^{\circ} - \frac{C}{2})$

$\Rightarrow \sin (\frac{A + B}{2}) = \cos (\frac{C}{2})$ $[We know that : sin (90^{\circ} - A) = cos A]$

Hence proved.

ii) $\tan (\frac{B + C}{2}) = \cos \frac{A}{2}$

Ans: $\tan (\frac{B + C}{2})$

We know that for a $\Delta ABC,$

$∠ A + ∠ B + ∠ C = 180^{\circ}$

$\Rightarrow \frac{∠ A + ∠ B + ∠ C}{2} = \frac{180^{\circ}}{2}$

$\Rightarrow \frac{∠ A + ∠ B + ∠ C}{2}$ = 90o

$\Rightarrow \frac{∠ B + ∠ C}{2} =$ 90o - $\frac{∠ A}{2}$

$\tan (\frac{B + C}{2}) = \tan (90^{\circ} - \frac{A}{2})$

$\Rightarrow \tan (\frac{B + C}{2}) = \cot (\frac{A}{2})$ $[We know that : tan (90^{\circ} - A) = cot A]$

Hence proved.

5. Evaluate:

i) $3 \frac{\sin 72^{\circ}}{\cos 18^{\circ}} - \frac{\sec 32^{\circ}}{cosec 58^{\circ}}$

Ans: $3 \frac{\sin 72^{\circ}}{\cos 18^{\circ}} - \frac{\sec 32^{\circ}}{cosec 58^{\circ}}$

$= 3 \frac{\sin 72^{\circ}}{\cos {(90 - 72)}^{\circ}} - \frac{\sec 32^{\circ}}{cosec {(90 - 32)}^{\circ}}$

$[We know that : cos (90^{\circ} - A) = sinA, cosec (90^{\circ} - A) = secA]$

$= 3 \frac{\sin 72^{\circ}}{\sin 72^{\circ}} - \frac{\sec 32^{\circ}}{\sec 32^{\circ}}$

$= 3 \times 1 - 1$

$= 3 - 1$

$= 2$

ii) $3 \cos 80^{\circ} cosec 10^{\circ} + 2 \sin 59^{\circ} \sec 31^{\circ}$

Ans: $3 \cos 80^{\circ} cosec 10^{\circ} + 2 \sin 59^{\circ} \sec 31^{\circ}$

$= 3 \frac{\cos 80^{\circ}}{\sin 10^{\circ}} + 2 \frac{\sin 59^{\circ}}{\cos 31^{\circ}}$ $[We know that : cosecA = \frac{1}{sin A}, secA = \frac{1}{cosA}]$

$= 3 \frac{\cos {(90 - 10)}^{\circ}}{\sin 10^{\circ}} + 2 \frac{\sin {(90 - 31)}^{\circ}}{\cos 31^{\circ}}$ $[We know that : cos (90^{\circ} - A) = sinA, sin (90^{\circ} - A) = cosA]$

$= 3 \frac{\sin 10^{\circ}}{\sin 10^{\circ}} + 2 \frac{\cos 31^{\circ}}{\cos 31^{\circ}}$

$= 3 \times 1 + (2 \times 1)$

$= 3 + 2$

$= 5$

iii) $\frac{\sin 80^{\circ}}{\cos 10^{\circ}} + \sin 59^{\circ} \sec 31^{\circ}$

Ans: $\frac{\sin 80^{\circ}}{\cos 10^{\circ}} + \sin 59^{\circ} \sec 31^{\circ}$

$= \frac{\sin 80^{\circ}}{\cos {(90 - 80)}^{\circ}} + \frac{\sin 59^{\circ}}{\cos 31^{\circ}}$ $[We know that : sec A = \frac{1}{cos A}]$

$= \frac{\sin 80^{\circ}}{\sin 80^{\circ}} + \frac{\sin 59^{\circ}}{\cos {(90 - 59)}^{\circ}}$ $[We know that : cos (90^{\circ} - A) = sin A]$

$= 1 + \frac{\sin 59^{\circ}}{\sin 59^{\circ}}$ $[We know that : cos (90^{\circ} - A) = sin A]$

$= 1 + 1$

$= 2$

iv) $\tan (55^{\circ} - A) - \cot (35^{\circ} + A)$

Ans: $\tan (55^{\circ} - A) - \cot (35^{\circ} + A)$

$= \tan [90^{\circ} - (35^{\circ} + A)] - \cot (35^{\circ} + A)$

$= \cot (35^{\circ} + A) - \cot (35^{\circ} + A)$ $[We know that : tan (90^{\circ} - A) = cot A]$

$= 0$

v) $cosec (65^{\circ} + A) - \sec (25^{\circ} - A)$

Ans: $cosec (65^{\circ} + A) - \sec (25^{\circ} - A)$

$= cosec [90^{\circ} - (25^{\circ} - A)] - \sec (25^{\circ} - A)$

$= \sec (25^{\circ} - A) - \sec (25^{\circ} - A)$ $[We know that : cosec (90^{\circ} - A) = sec A]$

$= 0$

vi) $2 \frac{\tan 57^{\circ}}{\cot 33^{\circ}} - \frac{\cot 70^{\circ}}{\tan 20^{\circ}} - \sqrt{2} \cos 45^{\circ}$

Ans: $2 \frac{\tan 57^{\circ}}{\cot 33^{\circ}} - \frac{\cot 70^{\circ}}{\tan 20^{\circ}} - \sqrt{2} \cos 45^{\circ}$

$= 2 \frac{\tan 57^{\circ}}{\cot {(90 - 57)}^{\circ}} - \frac{\cot {(90 - 20)}^{\circ}}{\tan 20^{\circ}} - \sqrt{2} \times \frac{1}{\sqrt{2}}$ $[We know that : \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

$= 2 \frac{\tan 57^{\circ}}{\tan 57^{\circ}} - \frac{\tan 20^{\circ}}{\tan 20^{\circ}} - 1$ $[We know that : cot (90^{\circ} - A) = tan A]$

$= 2 \times 1 - 1 - 1$

$= 0$

vii) $\frac{co t^{2} 41^{\circ}}{ta n^{2} 49^{\circ}} - 2 \frac{si n^{2} 75^{\circ}}{co s^{2} 15^{\circ}}$

Ans: $\frac{co t^{2} 41^{\circ}}{ta n^{2} 49^{\circ}} - 2 \frac{si n^{2} 75^{\circ}}{co s^{2} 15^{\circ}}$

$= \frac{co t^{2} 41^{\circ}}{ta n^{2} {(90 - 41)}^{\circ}} - 2 \frac{si n^{2} 75^{\circ}}{co s^{2} {(90 - 75)}^{\circ}}$

$= \frac{co t^{2} 41^{\circ}}{co t^{2} 41^{\circ}} - 2 \frac{si n^{2} 75^{\circ}}{si n^{2} 75^{\circ}}$ $[We know that : tan (90^{\circ} - A) = cot A, cos (90^{\circ} - A) = sin A]$

$= 1 - (2 \times 1)$

$= - 1$

viii) $\frac{\cos 70^{\circ}}{\sin 20^{\circ}} + \frac{\cos 59^{\circ}}{\sin 31^{\circ}} - 8 si n^{2} 30^{\circ}$

Ans: $\frac{\cos 70^{\circ}}{\sin 20^{\circ}} + \frac{\cos 59^{\circ}}{\sin 31^{\circ}} - 8 si n^{2} 30^{\circ}$

$= \frac{\cos 70^{\circ}}{\sin {(90 - 70)}^{\circ}} + \frac{\cos 59^{\circ}}{\sin {(90 - 59)}^{\circ}} - 8 {(\frac{1}{2})}^{2}$ $[We know that : \sin 30^{\circ} = \frac{1}{2}]$

$= \frac{\cos 70^{\circ}}{\cos 70^{\circ}} + \frac{\cos 59^{\circ}}{\cos 59^{\circ}} - 8 \times \frac{1}{4}$ $[We know that : sin (90^{\circ} - A) = cos A]$

$= 1 + 1 - 2$

$= 0$

ix) $14 \sin 30^{\circ} + 6 \cos 60^{\circ} - 5 \tan 45^{\circ}$

Ans: $14 \sin 30^{\circ} + 6 \cos 60^{\circ} - 5 \tan 45^{\circ}$

$= (14 \times \frac{1}{2}) + (6 \times \frac{1}{2}) - (5 \times 1)$ $[We know that : \sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}, \tan 45^{\circ} = 1]$

$= 7 + 3 - 5$

$= 5$

6. A triangle ABC is right angled at B; find the value of $\frac{\sec A . cosec C - \tan A . \cot C}{\sin B} .$

Ans: $\frac{\sec A . cosec C - \tan A . \cot C}{\sin B}$

Since, triangle ABC is right angle triangle and right angled at B.

$\Rightarrow ∠ B = 90^{\circ}$

We know that $∠ A + ∠ B + ∠ C = 180^{\circ}$

$\Rightarrow ∠ A + ∠ C = 90^{\circ}$

$\frac{\sec A . cosec C - \tan A . \cot C}{\sin B} = \frac{\sec (90 - C) . cosec C - \tan (90 - C) . \cot C}{\sin 90^{\circ}}$

$\Rightarrow \frac{\sec A . cosec C - \tan A . \cot C}{\sin B} = \frac{cosec C . cosec C - \cot C . \cot C}{1}$

$[We know that : sec (90^{\circ} - A) = cosec A, tan (90^{\circ} - A) = cot A]$

$\Rightarrow \frac{\sec A . cosec C - \tan A . \cot C}{\sin B} = cose c^{2} C - co t^{2} C$

$\Rightarrow \frac{\sec A . cosec C - \tan A . \cot C}{\sin B} = 1$ $[We know that : cose c^{2} C - co t^{2} C = 1]$

7. Find ( in each case, given below) the value of $x$ if:

i) $\sin x = \sin 60^{\circ} . \cos 30^{\circ} - \cos 60^{\circ} . \sin 30^{\circ}$

Ans: $\sin x = \sin 60^{\circ} . \cos 30^{\circ} - \cos 60^{\circ} . \sin 30^{\circ}$

$[We know that : \sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}, \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$

$\Rightarrow \sin x = (\frac{\sqrt{3}}{2}) . (\frac{\sqrt{3}}{2}) - (\frac{1}{2}) . (\frac{1}{2})$

$\Rightarrow \sin x = (\frac{3}{4}) - (\frac{1}{4})$

$\Rightarrow \sin x = (\frac{1}{2})$

We know that $[sin 30^{\circ} = \frac{1}{2}]$

$\Rightarrow \sin x = sin 30^{\circ}$

$\Rightarrow x = 30^{\circ}$

ii) $\sin x = \sin 60^{\circ} . \cos 30^{\circ} + \cos 60^{\circ} . \sin 30^{\circ}$

Ans: $\sin x = \sin 60^{\circ} . \cos 30^{\circ} + \cos 60^{\circ} . \sin 30^{\circ}$

$[We know that : \sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}, \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$

$\Rightarrow \sin x = (\frac{\sqrt{3}}{2}) . (\frac{\sqrt{3}}{2}) + (\frac{1}{2}) . (\frac{1}{2})$

$\Rightarrow \sin x = (\frac{3}{4}) + (\frac{1}{4})$

$\Rightarrow \sin x = 1$

We know that $[sin 90^{\circ} = 1]$

$\Rightarrow \sin x = sin 90^{\circ}$

$\Rightarrow x = 90^{\circ}$

iii) $\cos x = \cos 60^{\circ} . \cos 30^{\circ} - \sin 60^{\circ} . \sin 30^{\circ}$

Ans: $\cos x = \cos 60^{\circ} . \cos 30^{\circ} - \sin 60^{\circ} . \sin 30^{\circ}$

$[We know that : \sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}, \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$

$\Rightarrow \cos x = (\frac{1}{2}) . (\frac{\sqrt{3}}{2}) - (\frac{\sqrt{3}}{2}) . (\frac{1}{2})$

$\Rightarrow \cos x = (\frac{\sqrt{3}}{4}) - (\frac{\sqrt{3}}{4})$

$\Rightarrow \cos x = 0$

We know that $[cos 90^{\circ} = 0]$

$\Rightarrow \cos x = co s 90^{\circ}$

$\Rightarrow x = 90^{\circ}$

iv) $\tan x = \frac{\tan 60^{\circ} - \tan 30^{\circ}}{1 + \tan 60^{\circ} . \tan 30^{\circ}}$

Ans: $\tan x = \frac{\tan 60^{\circ} - \tan 30^{\circ}}{1 + \tan 60^{\circ} . \tan 30^{\circ}}$

$[We know that : \tan 60^{\circ} = \sqrt{3}, \tan 30^{\circ} = \frac{1}{\sqrt{3}}]$

$\Rightarrow \tan x = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} . \frac{1}{\sqrt{3}}}$

$\Rightarrow \tan x = \frac{\frac{3 - 1}{\sqrt{3}}}{1 + 1}$

$\Rightarrow \tan x = \frac{\frac{2}{\sqrt{3}}}{2}$

$\Rightarrow \tan x = \frac{1}{\sqrt{3}}$

We know that $[\tan 30^{\circ} = \frac{1}{\sqrt{3}}]$

$\Rightarrow \tan x = \tan 30^{\circ}$

$\Rightarrow x = 30^{\circ}$

v) $\sin 2 x = 2 \sin 45^{\circ} . \cos 45^{\circ}$

Ans: $\sin 2 x = 2 \sin 45^{\circ} . \cos 45^{\circ}$

$[We know that : \sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

$\Rightarrow \sin 2 x = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

$\Rightarrow \sin 2 x = 2 \times \frac{1}{2}$

$\Rightarrow \sin 2 x = 1$

We know that $[sin 90^{\circ} = 1]$

$\Rightarrow \sin 2 x = sin 90^{\circ}$

$\Rightarrow 2 x = 90^{\circ}$

$\Rightarrow x = 45^{\circ}$

vi) $\sin 3 x = 2 \sin 30^{\circ} . \cos 30^{\circ}$

Ans: $\sin 3 x = 2 \sin 30^{\circ} . \cos 30^{\circ}$

$[We know that : \sin 30^{\circ} = \frac{1}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$

$\Rightarrow \sin 3 x = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}$

$\Rightarrow \sin 3 x = \frac{\sqrt{3}}{2}$

We know that $[sin 60^{\circ} = \frac{\sqrt{3}}{2}]$

$\Rightarrow \sin 3 x = sin 60^{\circ}$

$\Rightarrow 3 x = 60^{\circ}$

$\Rightarrow x = 20^{\circ}$

vii) $\cos (2 x - 6^{\circ}) = co s^{2} 30^{\circ} - co s^{2} 60^{\circ}$

Ans: $\cos (2 x - 6^{\circ}) = co s^{2} 30^{\circ} - co s^{2} 60^{\circ}$

$[We know that : co s 30^{\circ} = \frac{\sqrt{3}}{2}, \cos 60^{\circ} = \frac{1}{2}]$

$\Rightarrow \cos (2 x - 6^{\circ}) = {(\frac{\sqrt{3}}{2})}^{2} - {(\frac{1}{2})}^{2}$

$\Rightarrow \cos (2 x - 6^{\circ}) = (\frac{3}{4}) - (\frac{1}{4})$

$\Rightarrow \cos (2 x - 6^{\circ}) = (\frac{1}{2})$

We know that $[cos 60^{\circ} = \frac{1}{2}]$

$\Rightarrow \cos (2 x - 6^{\circ}) = cos 60^{\circ}$

$\Rightarrow 2 x - 6^{\circ} = 60^{\circ}$

$\Rightarrow 2 x = 66^{\circ}$

$\Rightarrow x = 33^{\circ}$

8. In each case, given below, find the value of angle A, where $0^{\circ} ⩽ A ⩽ 90^{\circ}$ .

i) $\sin (90^{\circ} - 3 A) . cosec 42^{\circ} = 1$

Ans: $\sin (90^{\circ} - 3 A) . cosec 42^{\circ} = 1$

[We know that: $\sin A . cosec A = 1$ ]

So, $\sin (90^{\circ} - 3 A) = \sin 42^{\circ}$

$\Rightarrow (90^{\circ} - 3 A) = 42^{\circ}$

$\Rightarrow 3 A = 90^{\circ} - 42^{\circ}$

$\Rightarrow 3 A = 48^{\circ}$

$\Rightarrow A = 16^{\circ}$

ii) $\cos (90^{\circ} - A) . \sec 77^{\circ} = 1$

Ans: $\cos (90^{\circ} - A) . \sec 77^{\circ} = 1$

[We know that: $\cos A . \sec A = 1$ ]

So, $\cos (90^{\circ} - A) = \cos 77^{\circ}$

$\Rightarrow (90^{\circ} - A) = 77^{\circ}$

$\Rightarrow A = 90^{\circ} - 77^{\circ}$

$\Rightarrow A = 13^{\circ}$

9. Prove that:

i) $\frac{\cos (90^{\circ} - θ) \cos θ}{\cot θ} = 1 - co s^{2} θ$

Ans: Taking LHS -

$LHS = \frac{\cos (90^{\circ} - θ) \cos θ}{\cot θ}$

$= \frac{\sin θ . \cos θ}{\frac{\cos θ}{\sin θ}}$ $[We know that : cos (90^{\circ} - A) = sin A, cot A = \frac{cos A}{sin A}]$

$= si n^{2} θ$

$= 1 - co s^{2} θ = RHS$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$\Rightarrow LHS = RHS$

Hence proved.

ii) $\frac{\sin θ . \sin (90^{\circ} - θ)}{\cot (90^{\circ} - θ)} = 1 - si n^{2} θ$

Ans: Taking LHS -

$LHS = \frac{\sin θ . \sin (90^{\circ} - θ)}{\cot (90^{\circ} - θ)}$

$= \frac{\sin θ . \cos θ}{\tan θ}$ $[We know : sin (90^{\circ} - A) = cosA, cot (90^{\circ} - A) = tan A]$

$= \frac{\sin θ . \cos θ}{\frac{\sin θ}{\cos θ}}$ $[We know that : tan A = \frac{sin A}{cos A}]$

$= co s^{2} θ$

$= 1 - si n^{2} θ = RHS$ $[We know that : si n^{2} A + co s^{2} A = 1]$

$\Rightarrow LHS = RHS$

Hence proved.

10. Evaluate:

$\frac{\sin 35^{\circ} \cos 55^{\circ} + \cos 35^{\circ} \sin 55^{\circ}}{cose c^{2} 10^{\circ} - ta n^{2} 80^{\circ}}$

Ans: $\frac{\sin 35^{\circ} \cos 55^{\circ} + \cos 35^{\circ} \sin 55^{\circ}}{cose c^{2} 10^{\circ} - ta n^{2} 80^{\circ}}$

$= \frac{\sin {(90 - 55)}^{\circ} \cos 55^{\circ} + \cos {(90 - 55)}^{\circ} \sin 55^{\circ}}{cose c^{2} 10^{\circ} - ta n^{2} {(90 - 10)}^{\circ}}$

$[∵ sin (90^{\circ} - A) = cos A, cos (90^{\circ} - A) = sin A, tan (90^{\circ} - A) = cot A]$

$= \frac{\cos 55^{\circ} . \cos 55^{\circ} + \sin 55^{\circ} . \sin 55^{\circ}}{cose c^{2} 10^{\circ} - co t^{2} 10^{\circ}}$

$= \frac{co s^{2} 55^{\circ} + si n^{2} 55^{\circ}}{cose c^{2} 10^{\circ} - co t^{2} 10^{\circ}}$

$[We know that : si n^{2} A + co s^{2} A = 1 & cose c^{2} A - co t^{2} A = 1]$

$= \frac{1}{1} = 1$

11. Evaluate:

$si n^{2} 34^{\circ} + si n^{2} 56^{\circ} + 2 \tan 18^{\circ} . \tan 72^{\circ} - co t^{2} 30^{\circ}$

Ans: $si n^{2} 34^{\circ} + si n^{2} 56^{\circ} + 2 \tan 18^{\circ} . \tan 72^{\circ} - co t^{2} 30^{\circ}$

$= si n^{2} {(90 - 56)}^{\circ} + si n^{2} 56^{\circ} + 2 \tan 18^{\circ} . \tan {(90 - 18)}^{\circ} - co t^{2} 30^{\circ}$

$[∵ sin (90^{\circ} - A) = cos A, tan (90^{\circ} - A) = cot A, cot 30^{\circ} = \sqrt{3}]$

$= co s^{2} 56^{\circ} + si n^{2} 56^{\circ} + 2 \tan 18^{\circ} . \cot 18^{\circ} - {(\sqrt{3})}^{2}$

$[We know that : si n^{2} A + co s^{2} A = 1 & \tan A . \cot A = 1]$

$= 1 + (2 \times 1) - 3$

$= 0$

12. Evaluate:

$cose c^{2} 57^{\circ} - ta n^{2} 33^{\circ} + \cos 44^{\circ} . cosec 46^{\circ} - \sqrt{2} \cos 45^{\circ} - ta n^{2} 60^{\circ}$

Ans: $cose c^{2} 57^{\circ} - ta n^{2} 33^{\circ} + \cos 44^{\circ} . cosec 46^{\circ} - \sqrt{2} \cos 45^{\circ} - ta n^{2} 60^{\circ}$

$[We know that : cos 45^{\circ} = \frac{1}{\sqrt{2}}, tan 60^{\circ} = \sqrt{3}]$

$= cose c^{2} 57^{\circ} - ta n^{2} {(90 - 57)}^{\circ} + \cos 44^{\circ} . cosec {(90 - 44)}^{\circ} - \sqrt{2} \times \frac{1}{\sqrt{2}} - {(\sqrt{3})}^{2}$

$[∵ cosec (90^{\circ} - A) = sec A, tan (90^{\circ} - A) = cot A]$

$= cose c^{2} 57^{\circ} - co t^{2} 57^{\circ} + \cos 44^{\circ} . \sec 44^{\circ} - 1 - 3$

$[We know that : cose c^{2} A - co t^{2} A = 1 & \cos A . \sec A = 1]$

$= 1 + 1 - 1 - 3$

$= - 2$

Exercise - 21(D)

1. Use table to find sine of:

i) $21^{\circ}$

Ans: $\sin 21^{\circ}$

According to the natural sine table,

$sin 21^{\circ} = 0.3584$

ii) $34^{\circ} 42^{'}$

Ans: $\sin 34^{\circ} 42^{'}$

According to the natural sine table,

$sin 34^{\circ} 42' = 0.5693$

iii) $47^{\circ} 32^{'}$

Ans: $\sin 47^{\circ} 32^{'}$

According to the natural sine table,

$sin 47^{\circ} 32' = sin (47^{\circ} 30' + 2')$

$\Rightarrow sin 47^{\circ} 32' = 0.7373 + 0.0004$

$\Rightarrow sin 47^{\circ} 32' = 0.7377$

iv) $62^{\circ} 57^{'}$

Ans: $\sin 62^{\circ} 57^{'}$

According to the natural sine table,

$sin 62^{\circ} 57' = sin (62^{\circ} 54' + 3')$

$\Rightarrow sin 62^{\circ} 57' = 0.8902 + 0.0004$

$\Rightarrow sin 62^{\circ} 57' = 0.8906$

v) $10^{\circ} 20^{'} + 20^{\circ} 45^{'}$

Ans: $\sin (10^{\circ} 20^{'} + 20^{\circ} 45^{'})$

According to the natural sine table,

$sin (10^{\circ} 20' + 20^{\circ} 45') = sin (30^{\circ} 65')$

$\Rightarrow sin (10^{\circ} 20' + 20^{\circ} 45') = sin (31^{\circ} 5')$

$\Rightarrow sin (10^{\circ} 20' + 20^{\circ} 45') = 0.5150 + 0.0012$

$\Rightarrow sin (10^{\circ} 20' + 20^{\circ} 45') = 0.5162$

2. Use table to find cosine of:

i) $2^{\circ} 4^{'}$

Ans: $\cos 2^{\circ} 4^{'}$