Answer
437.7k+ views
Hint: Here we will substitute the value of x and y on the left hand side of the equation and prove it equal to the right hand side using various trigonometric identities:-
\[
1 + {\tan ^2}\theta = {\sec ^2}\theta \\
\Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\
\]
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Complete step-by-step answer:
The given equation is:-
\[{{\text{x}}^2} - {y^2} = {a^2} - {b^2}\]
Let us consider the left hand side of the given equation:-
\[LHS = {{\text{x}}^2} - {y^2}\]
Now it is given that:-
\[x = a\sec \theta + b\tan \theta \] and \[y = a\tan \theta + b\sec \theta \]
Hence putting in the respective values of x and y in LHS we get:-
\[LHS = {\left( {a\sec \theta + b\tan \theta } \right)^2} - {\left( {a\tan \theta + b\sec \theta } \right)^2}\]
Now applying the following identity on both he terms:-
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
We get:-
\[LHS = {\left( {a\sec \theta } \right)^2} + {\left( {b\tan \theta } \right)^2} + 2\left( {a\sec \theta } \right)\left( {b\tan \theta } \right) - \left[ {{{\left( {a\tan \theta } \right)}^2} + {{\left( {b\sec \theta } \right)}^2} + 2\left( {b\sec \theta } \right)\left( {a\tan \theta } \right)} \right]\]
Now simplifying it further we get:-
\[
LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - \left[ {{a^2}{{\tan }^2}\theta + {b^2}{{\sec }^2}\theta + 2ab\sec \theta \tan \theta } \right] \\
\Rightarrow LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta - 2ab\sec \theta \tan \theta \\
\]
Now cancelling the required terms we get:-
\[LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta \]
Now taking \[{a^2}\] and \[{b^2}\] common we get:-
\[LHS = {a^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) - {b^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\]
Again taking common we get:-
\[LHS = \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {{a^2} - {b^2}} \right)\]
Now applying the following identity:-
\[
1 + {\tan ^2}\theta = {\sec ^2}\theta \\
\Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\
\]
We get:-
\[
\Rightarrow LHS = \left( {{a^2} - {b^2}} \right) \\
{\text{ }} = RHS \\
\]
Therefore,
\[LHS = RHS\]
Hence proved.
Note: The student may make mistakes while applying the identity, so the identity should be first simplified in the required form and then apply it.
The student can also use the following identity on the initial stage:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Hence,
\[{{\text{x}}^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\]
Then put in the values of x and y and proceed further to get the desired answer.
\[
1 + {\tan ^2}\theta = {\sec ^2}\theta \\
\Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\
\]
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Complete step-by-step answer:
The given equation is:-
\[{{\text{x}}^2} - {y^2} = {a^2} - {b^2}\]
Let us consider the left hand side of the given equation:-
\[LHS = {{\text{x}}^2} - {y^2}\]
Now it is given that:-
\[x = a\sec \theta + b\tan \theta \] and \[y = a\tan \theta + b\sec \theta \]
Hence putting in the respective values of x and y in LHS we get:-
\[LHS = {\left( {a\sec \theta + b\tan \theta } \right)^2} - {\left( {a\tan \theta + b\sec \theta } \right)^2}\]
Now applying the following identity on both he terms:-
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
We get:-
\[LHS = {\left( {a\sec \theta } \right)^2} + {\left( {b\tan \theta } \right)^2} + 2\left( {a\sec \theta } \right)\left( {b\tan \theta } \right) - \left[ {{{\left( {a\tan \theta } \right)}^2} + {{\left( {b\sec \theta } \right)}^2} + 2\left( {b\sec \theta } \right)\left( {a\tan \theta } \right)} \right]\]
Now simplifying it further we get:-
\[
LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - \left[ {{a^2}{{\tan }^2}\theta + {b^2}{{\sec }^2}\theta + 2ab\sec \theta \tan \theta } \right] \\
\Rightarrow LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta - 2ab\sec \theta \tan \theta \\
\]
Now cancelling the required terms we get:-
\[LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta \]
Now taking \[{a^2}\] and \[{b^2}\] common we get:-
\[LHS = {a^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) - {b^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\]
Again taking common we get:-
\[LHS = \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {{a^2} - {b^2}} \right)\]
Now applying the following identity:-
\[
1 + {\tan ^2}\theta = {\sec ^2}\theta \\
\Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\
\]
We get:-
\[
\Rightarrow LHS = \left( {{a^2} - {b^2}} \right) \\
{\text{ }} = RHS \\
\]
Therefore,
\[LHS = RHS\]
Hence proved.
Note: The student may make mistakes while applying the identity, so the identity should be first simplified in the required form and then apply it.
The student can also use the following identity on the initial stage:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Hence,
\[{{\text{x}}^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\]
Then put in the values of x and y and proceed further to get the desired answer.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)