NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Exercise 3.1: In this exercise, students are introduced to trigonometric ratios of acute angles and their applications in solving problems related to heights and distances. The exercise covers basic concepts such as the definition of trigonometric ratios, Pythagoras theorem, and the concept of complementary angles.

Exercise 3.2: This exercise focuses on the evaluation of trigonometric ratios of special angles such as 0, 30, 45, 60, and 90 degrees. The exercise also includes the derivation of trigonometric ratios of 30, 45, and 60 degrees using the concept of the unit circle.

Exercise 3.3: This exercise covers the trigonometric ratios of angles between 0 and 90 degrees. The exercise includes problems on finding the values of trigonometric ratios using various techniques such as the use of Pythagoras theorem, the double-angle formula, and the half-angle formula.

Miscellaneous Exercise: This exercise includes a variety of problems that cover different topics such as the use of trigonometric ratios in solving real-life problems, the use of trigonometric identities to simplify expressions, and the solution of trigonometric equations. This exercise provides an opportunity for students to apply the concepts learned in the chapter to solve more complex problems.

Access NCERT Solution for Class 11 Maths Chapter 3 - Trigonometric Functions

Exercise 3.1

1. Find the radian measures corresponding to the following degree measures:

(i) $\text{2}{\text{5}}^{\text{o}}$

Ans: We know that $\text{18}{\text{0}}^{\text{o}}\text{= }\pi$ radian

Therefore  $1^{\circ }={\displaystyle \frac{\pi }{180}}$ radian 

hence, 

$\text{2}{\text{5}}^{\text{o}}\text{=}{\displaystyle \frac{\pi }{\text{180}}}\times \text{ 25}$ radian  

      $\text{=}{\displaystyle \frac{\text{5 }\pi }{\text{36}}}$radian

(ii) $\text{-4}{\text{7}}^{\text{o}}\text{30 }{}^{\prime }$

Ans: Here we have,

  $\text{-4}{\text{7}}^{\text{o}}\text{30 }{}^{\prime }\text{ =-47}{{\displaystyle \frac{\text{1}}{\text{2}}}}^{\text{o}}$   

           $\text{=-}{\displaystyle \frac{\text{95}}{\text{2}}}$ degree

Since we know that, $\text{18}{\text{0}}^{\text{o}}\text{= }\pi$ radian

Therefore  ${\text{1}}^{\text{o}}\text{=}{\displaystyle \frac{\pi }{\text{180}}}$ radian 

Hence,

$\text{-}{\displaystyle \frac{\text{95}}{\text{2}}}$ degree$\text{=}{\displaystyle \frac{\pi }{\text{180}}}\times \left({\displaystyle \frac{\text{-95}}{\text{2}}}\right)$ radian

                  $\text{=}\left({\displaystyle \frac{\text{-19}}{\text{36 }\times \text{ 2}}}\right)\pi$  radian

                  $\text{=}{\displaystyle \frac{\text{-19}}{\text{72}}}\pi$radian

Therefore,

$\text{-4}{\text{7}}^{\text{o}}\text{30 }{}^{\prime }\text{ =-}{\displaystyle \frac{\text{19}}{\text{72}}}\pi$ radian

(iii) $\text{24}{\text{0}}^{\text{o}}$

Ans: We know that,

$\text{18}{\text{0}}^{\text{o}}\text{= }\pi$ radian

Therefore  ${\text{1}}^{\text{o}}\text{=}{\displaystyle \frac{\pi }{\text{180}}}$ radian 

Hence,

$\text{24}{\text{0}}^{\text{o}}\text{=}{\displaystyle \frac{\pi }{\text{180}}}\times \text{ 240}$ radian

       $\text{=}{\displaystyle \frac{\text{4}}{\text{3}}}\pi$radian

(iv) $\text{52}{\text{0}}^{\text{o}}$

Ans: We know that,

$\text{18}{\text{0}}^{\text{o}}\text{= }\pi$ radian

Therefore  ${\text{1}}^{\text{o}}\text{=}{\displaystyle \frac{\pi }{\text{180}}}$ radian 

Hence,

$\text{52}{\text{0}}^{\text{o}}\text{=}{\displaystyle \frac{\pi }{\text{180}}}\times \text{ 520}$ radian

       $\text{=}{\displaystyle \frac{\text{26 }\pi }{\text{9}}}$radian

2. Find the degree measures corresponding to the following radian measures

(Use$\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}$ ) 

(i) ${\displaystyle \frac{\text{11}}{\text{16}}}$

Ans: We know that,

 $\pi$ radian$\text{=18}{\text{0}}^{\text{o}}$

Therefore  $\text{1 radian =}{{\displaystyle \frac{\text{180}}{\pi }}}^{\text{o}}$ 

Hence,

${\displaystyle \frac{\text{11}}{\text{16}}}$ radian$\text{=}{\displaystyle \frac{\text{180}}{\pi }}\times {\displaystyle \frac{\text{11}}{\text{16}}}$ degree

              $\text{=}{\displaystyle \frac{\text{45 }\times \text{ 11}}{\pi \times \text{ 4}}}$degree

              $$\text{=}{\displaystyle \frac{\text{45 }\times \text{ 11 }\times \text{ 7}}{\text{22 }\times \text{ 4}}}$$ degree

              $\text{=}{\displaystyle \frac{\text{315}}{\text{8}}}$degree

Further computing,

${\displaystyle \frac{\text{11}}{\text{16}}}$ radian$\text{=39}{\displaystyle \frac{\text{3}}{\text{8}}}$ degree

                $\text{=3}{\text{9}}^{\text{o}}\text{+}{\displaystyle \frac{\text{3 }\times \text{ 60}}{\text{8}}}$ minutes

Since ${\text{1}}^{\text{o}}\text{=60 }{}^{\prime }$

    ${\displaystyle \frac{\text{11}}{\text{16}}}$ radian $\text{=3}{\text{9}}^{\text{o}}\text{+22 }{}^{\prime }\text{ +}{\displaystyle \frac{\text{1}}{\text{2}}}$minutes

Since  $\text{1 }{}^{\prime }\text{ =60''}$ 

${\displaystyle \frac{\text{11}}{\text{16}}}$ radian$\text{ =3}{\text{9}}^{\text{o}}\text{22 }{}^{\prime }\text{ 30''}$

(ii) $\text{-4}$

Ans: We know that,

$\pi$ radian$\text{=18}{\text{0}}^{\text{o}}$

Therefore  $\text{1 radian =}{{\displaystyle \frac{\text{180}}{\pi }}}^{\text{o}}$ 

Hence,

$\text{-4}$ radian$\text{=}{\displaystyle \frac{\text{180}}{\pi }}\times \left(\text{-4}\right)$ degree

                $\text{=}{\displaystyle \frac{\text{180 }\times \text{ 7}\left(\text{-4}\right)}{\text{22}}}$degree

                $\text{=}{\displaystyle \frac{\text{-2520}}{\text{11}}}$ degree

                $\text{=-229}{\displaystyle \frac{\text{1}}{\text{11}}}$degree

Since ${\text{1}}^{\text{o}}\text{=60 }{}^{\prime }$

We have,

$\text{-4}$ radian$\text{=-22}{\text{9}}^{\text{o}}\text{+}{\displaystyle \frac{\text{1 }\times \text{ 60}}{\text{11}}}$ minutes                        

               $\text{=-22}{\text{9}}^{\text{o}}\text{+5 }{}^{\prime }\text{ +}{\displaystyle \frac{\text{5}}{\text{11}}}$ minutes

Since $\text{1 }{}^{\prime }\text{ =60 }{}^{\prime }{}^{\prime }$

$\text{-4}$ radian$\text{=-22}{\text{9}}^{\text{o}}\text{5 }{}^{\prime }\text{ 27''}$

(iii) ${\displaystyle \frac{\text{5 }\pi }{\text{3}}}$

Ans: We know that,

$\pi$ radian$\text{=18}{\text{0}}^{\text{o}}$

Therefore  $\text{1 radian =}{{\displaystyle \frac{\text{180}}{\pi }}}^{\text{o}}$ 

Hence,

${\displaystyle \frac{\text{5 }\pi }{\text{3}}}$ radian$\text{=}{\displaystyle \frac{\text{180}}{\pi }}\times {\displaystyle \frac{\text{5 }\pi }{\text{3}}}$ degree

                 $\text{=30}{\text{0}}^{\text{o}}$

(iv)${\displaystyle \frac{\text{7 }\pi }{\text{6}}}$

Ans: We know that,

 $\pi$ radian$\text{=18}{\text{0}}^{\text{o}}$

Therefore  $\text{1 radian =}{{\displaystyle \frac{\text{180}}{\pi }}}^{\text{o}}$ 

Hence,

${\displaystyle \frac{\text{7 }\pi }{\text{6}}}$ radian$\text{=}{\displaystyle \frac{\text{180}}{\pi }}\times {\displaystyle \frac{\text{7 }\pi }{\text{6}}}$

                 $\text{=21}{\text{0}}^{\text{o}}$

3. A wheel makes $\text{360}$ revolutions in one minute. Through how many radians does it turn in one second?

Ans: Number of revolutions the wheel makes in $\text{1}$ minute$\text{=360}$

 Number of revolutions the wheel make in $\text{1}$ second$\text{=}{\displaystyle \frac{\text{360}}{\text{60}}}$

                                                                                   $\text{=6}$

In one complete revolution, the wheel turns an angle of $$\text{2 }\pi$$ radian.

Hence, it will turn an angle of $\text{6 }\times \text{ 2 }\pi \text{ =12 }\pi$ radian, in $\text{6}$ complete revolutions.

Therefore, the wheel turns an angle of $\text{12 }\pi$ radian in one second.

4. Find the degree measure of the angle subtended at the centre of a circle of radius $\text{100}$cm by an arc of length  $\text{22}$  cm.

(Use$\pi \text{ =}{\displaystyle \frac{\text{22}}{\text{7}}}$ ) 

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\theta$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\theta \text{ =}{\displaystyle \frac{\text{l}}{\text{r}}}$                                ……(1)

Therefore, 

Substituting $\text{r=100cm}$ ,$$\text{l=22cm}$$ in the formula (1) , we have,

$\theta \text{ =}{\displaystyle \frac{\text{22}}{\text{100}}}$ radian

Since $\text{1 radian=}{\displaystyle \frac{\text{180}}{\pi }}$ 

Therefore,

 $\theta \text{ =}{\displaystyle \frac{\text{180}}{\pi }}\times {\displaystyle \frac{\text{22}}{\text{100}}}$ degree

     $\text{=}{\displaystyle \frac{\text{180 }\times \text{ 7 }\times \text{ 22}}{\text{22 }\times \text{ 100}}}$degree

     $\text{=}{\displaystyle \frac{\text{63}}{\text{5}}}$ degree

    $\text{=12}{\displaystyle \frac{\text{3}}{\text{5}}}$degree

Since  ${\text{1}}^{\text{o}}\text{=60 }{}^{\prime }$ , we have,

$\theta \text{ =1}{\text{2}}^{\text{o}}\text{36 }{}^{\prime }$

Hence , the required angle is $\text{1}{\text{2}}^{\text{o}}\text{36 }{}^{\prime }$.

5. In a circle of diameter $\text{40}$ cm, the length of a chord is $\text{20}$ cm. Find the length of minor arc of the chord.

Ans: Given that, diameter of the circle$=40$ cm

 Hence Radius $\left(r\right)$ of the circle$={\displaystyle \frac{40}{2}}cm$

                                                   $=20cm$

Let $\text{AB}$ be a chord  of the circle whose length is $20$ cm.

In $\mathrm{\Delta }\text{OAB,}$

$\text{OA=OB}$ 

     $=$ Radius of the circle

     $\text{=20cm}$

Now also, $\text{AB=20cm}$

Therefore, $\mathrm{\Delta }\text{OAB}$ is an equilateral triangle.

$$\therefore \theta \text{ =6}{\text{0}}^{\text{o}}$$

     $\text{=}{\displaystyle \frac{\pi }{\text{3}}}$ radian

We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\theta$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\theta \text{ =}{\displaystyle \frac{\text{l}}{\text{r}}}$                                ……(1)

Substituting $\theta \text{ =}{\displaystyle \frac{\pi }{\text{3}}}$  in the formula (1),

       ${\displaystyle \frac{\pi }{\text{3}}}\text{=}{\displaystyle \frac{\text{arc AB}}{\text{20}}}$

$\text{arc AB=}{\displaystyle \frac{\text{20 }\pi }{\text{3}}}\text{cm}$

Therefore, the length of the minor arc of the chord is ${\displaystyle \frac{\text{20 }\pi }{\text{3}}}\text{cm}$ .

6. If in two circles, arcs of the same length subtend angles $\text{6}{\text{0}}^{\text{o}}$ and $\text{7}{\text{5}}^{\text{o}}$ at the centre, find the ratio of their radii.

Ans: Let the radii of the two circles be ${\text{r}}_{\text{1}}$ and ${\text{r}}_{\text{2}}$ . Let an arc of length ${\text{l}}_{\text{1}}$ subtends an angle of $\text{6}{\text{0}}^{\text{o}}$ at the centre of the circle of radius ${\text{r}}_{\text{1}}$ , whereas let an arc of length ${\text{l}}_{\text{2}}$ subtends an angle of $\text{7}{\text{5}}^{\text{o}}$ at the centre of the circle of radius ${\text{r}}_{\text{2}}$ .

Now, we have,

$\text{6}{\text{0}}^{\text{o}}\text{=}{\displaystyle \frac{\pi }{\text{3}}}$ radian  and 

${75}^{\circ }={\displaystyle \frac{5\pi }{12}}$ radian

We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\theta$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$$\theta \text{ =}{\displaystyle \frac{\text{l}}{\text{r}}}$$  

$\text{l=r }\theta$          

Hence we obtain,                   

$\text{l=}{\displaystyle \frac{{\text{r}}_{\text{1}}\pi }{\text{3}}}$ 

and  $\text{l=}{\displaystyle \frac{{\text{r}}_{\text{2}}\text{5 }\pi }{\text{12}}}$

according to the ${\text{l}}_{\text{1}}\text{=}{\text{l}}_{\text{2}}$ 

thus we have,

${\displaystyle \frac{{\text{r}}_{\text{1}}\pi }{\text{3 }\pi }}\text{=}{\displaystyle \frac{{\text{r}}_{\text{2}}\text{5 }\pi }{\text{12}}}$

  ${\text{r}}_{\text{1}}\text{=}{\displaystyle \frac{{\text{r}}_{\text{2}}\text{5}}{\text{4}}}$

 ${\displaystyle \frac{{\text{r}}_{\text{1}}}{{\text{r}}_{\text{2}}}}\text{=}{\displaystyle \frac{\text{5}}{\text{4}}}$

Hence , the ratio of the radii is $\text{5:4}$ .

7. Find the angle in radian through which a pendulum swings if its length is $\text{75}$ cm and the tip describes an arc of length.

(i)  $\text{10}$ cm

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\theta$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\theta \text{ =}{\displaystyle \frac{\text{l}}{\text{r}}}$  

 Given that $\text{r=75cm}$

And here, $\text{l=10cm}$

Hence substituting the values in the formula,

$\theta \text{ =}{\displaystyle \frac{\text{10}}{\text{75}}}$ radian

  $\text{=}{\displaystyle \frac{\text{2}}{\text{15}}}$radian

(ii) $\text{15}$ cm

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\theta$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\theta \text{ =}{\displaystyle \frac{\text{l}}{\text{r}}}$  

 Given that $\text{r=75cm}$

And here, $\text{l=15cm}$  

Hence substituting the values in the formula,

$\theta \text{ =}{\displaystyle \frac{\text{15}}{\text{75}}}$ radian

       $\text{=}{\displaystyle \frac{\text{1}}{\text{5}}}$radian

(iii) $\text{21}$ cm

Ans: We know that,

 in a circle of radius $\text{r}$ unit, if  an angle $\theta$ radian at the centre is subtended by an arc of length $\text{l}$ unit then

$\theta \text{ =}{\displaystyle \frac{\text{l}}{\text{r}}}$  

And here, $\text{l=21cm}$

Hence substituting the values in the formula,

$\theta \text{ =}{\displaystyle \frac{\text{21}}{\text{75}}}$ radian

       $\text{=}{\displaystyle \frac{\text{7}}{\text{25}}}$radian

Exercise 3.2

1. Find the values of the other five trigonometric functions if  $\text{cos x=-}{\displaystyle \frac{\text{1}}{\text{2}}}$ , $x$ lies in the third quadrant.

Ans: Here given that,  $\text{cos x=-}{\displaystyle \frac{\text{1}}{\text{2}}}$

Therefore we have,

$\text{sec x=}{\displaystyle \frac{\text{1}}{\text{cos x}}}$

          $\text{=}{\displaystyle \frac{\text{1}}{\left(\text{-}{\displaystyle \frac{\text{1}}{\text{2}}}\right)}}$

          $\text{=-2}$

Now we know that,$\text{si}{\text{n}}^{\text{2}}\text{x+co}{\text{s}}^{\text{2}}\text{x=1}$

Therefore we have, $\text{si}{\text{n}}^{\text{2}}\text{x=1-co}{\text{s}}^{\text{2}}\text{x}$

Substituting  $\text{cos x=-}{\displaystyle \frac{\text{1}}{\text{2}}}$ in the formula, we obtain,

$\text{si}{\text{n}}^{\text{2}}\text{x=1-}{\left(\text{-}{\displaystyle \frac{\text{1}}{\text{2}}}\right)}^{\text{2}}$

$\text{si}{\text{n}}^{\text{2}}\text{x=1-}{\displaystyle \frac{\text{1}}{\text{4}}}$

            $\text{=}{\displaystyle \frac{\text{3}}{\text{4}}}$

   $\text{sin x= }\pm {\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}$

Since $\text{x}$ lies in the ${\text{3}}^{\text{rd}}$quadrant, the value of $\sin x$ will be negative.

$\text{sin x=-}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}$

Therefore, $\text{cosec x=}{\displaystyle \frac{\text{1}}{\text{sin x}}}$ 

                            $\text{=}{\displaystyle \frac{\text{1}}{\left(\text{-}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)}}$ 

                            $\text{=-}{\displaystyle \frac{\text{2}}{\sqrt{\text{3}}}}$ 

Hence ,

 $\text{tan x=}{\displaystyle \frac{\text{sin x}}{\text{cos x}}}$ 

        $\text{=}{\displaystyle \frac{\left(\text{-}{\displaystyle \frac{\sqrt{\text{3}}}{\text{2}}}\right)}{\left(\text{-}{\displaystyle \frac{\text{1}}{\text{2}}}\right)}}$

        $\text{=}\sqrt{\text{3}}$

And 

$\text{cot x=}{\displaystyle \frac{\text{1}}{\text{tan x}}}$

       $\text{=}{\displaystyle \frac{\text{1}}{\sqrt{\text{3}}}}$

2. Find the values of other five trigonometric functions if $\text{sin x=}{\displaystyle \frac{\text{3}}{\text{5}}}$  , $\text{x}$ lies in second quadrant.

Ans:

Here given that,  $\text{sin x=}{\displaystyle \frac{\text{3}}{\text{5}}}$

Therefore we have,

$\text{cosec x=}{\displaystyle \frac{\text{1}}{\text{sin x}}}$

          $={\displaystyle \frac{1}{\left({\displaystyle \frac{3}{5}}\right)}}$

           $={\displaystyle \frac{5}{3}}$

Now we know that , ${\sin }^{2}x+{\cos }^{2}x=1$

Therefore we have, $\text{co}{\text{s}}^{\text{2}}\text{x=1-si}{\text{n}}^{\text{2}}\text{x}$

Substituting  $\sin x={\displaystyle \frac{3}{5}}$  in the formula, we obtain,

$\text{co}{\text{s}}^{\text{2}}\text{x=1-}{\left({\displaystyle \frac{\text{3}}{\text{5}}}\right)}^{\text{2}}$

$\text{co}{\text{s}}^{\text{2}}\text{x=1-}{\displaystyle \frac{\text{9}}{\text{25}}}$

         $\text{=}{\displaystyle \frac{\text{16}}{\text{25}}}$ 

   $\text{cos x= }\pm {\displaystyle \frac{\text{4}}{\text{5}}}$

Since $x$ lies in the $2^{nd}$quadrant, the value of $\cos x$ will be negative.

$\text{cos x=-}{\displaystyle \frac{\text{4}}{\text{5}}}$

Therefore, $secx={\displaystyle \frac{1}{\cos x}}$ 

                          $\text{=}{\displaystyle \frac{\text{1}}{\left(\text{-}{\displaystyle \frac{\text{4}}{\text{5}}}\right)}}$ 

                          $\text{=-}{\displaystyle \frac{\text{5}}{\text{4}}}$ 

Hence ,

 $\tan x={\displaystyle \frac{\sin x}{\cos x}}$ 

        $\text{=}{\displaystyle \frac{\left({\displaystyle \frac{\text{3}}{\text{5}}}\right)}{\left(\text{-}{\displaystyle \frac{\text{4}}{\text{5}}}\right)}}$

        $\text{=-}{\displaystyle \frac{\text{3}}{\text{4}}}$

And 

$\cot x={\displaystyle \frac{1}{\tan x}}$

        $\text{=-}{\displaystyle \frac{\text{4}}{\text{3}}}$

3. Find the values of other five trigonometric functions if   $\text{cot x=}{\displaystyle \frac{\text{3}}{\text{4}}}$ , $\text{x}$ lies in third quadrant.

Ans: Here given that,  $\cot x={\displaystyle \frac{3}{4}}$

Therefore we have,

$\tan x={\displaystyle \frac{1}{\cot x}}$

       $={\displaystyle \frac{1}{\left({\displaystyle \frac{3}{4}}\right)}}$

       $={\displaystyle \frac{4}{3}}$

Now we know that , $$\text{se}{\text{c}}^{\text{2}}\text{x-ta}{\text{n}}^{\text{2}}\text{x=1}$$

Therefore we have, $\text{se}{\text{c}}^{\text{2}}\text{x=1+ta}{\text{n}}^{\text{2}}\text{x}$

Substituting  $\text{tan x=}{\displaystyle \frac{\text{4}}{\text{3}}}$  in the formula, we obtain,

${\sec }^{2}x=1+{\left({\displaystyle \frac{4}{3}}\right)}^2$

$\text{se}{\text{c}}^{\text{2}}\text{x=1+}{\displaystyle \frac{\text{16}}{\text{9}}}$

             $={\displaystyle \frac{25}{9}}$ 

   $\text{sec x= }\pm {\displaystyle \frac{\text{5}}{\text{3}}}$

Since $x$ lies in the $3^{rd}$quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}{\displaystyle \frac{\text{5}}{\text{3}}}$

Therefore, $\cos x={\displaystyle \frac{1}{\sec x}}$ 

                         $\text{=}{\displaystyle \frac{\text{1}}{\left(\text{-}{\displaystyle \frac{\text{5}}{\text{3}}}\right)}}$ 

                          $\text{=-}{\displaystyle \frac{\text{3}}{\text{5}}}$ 

Now  , $\text{tan x=}{\displaystyle \frac{\text{sin x}}{\text{cos x}}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}{\displaystyle \frac{\text{4}}{\text{3}}}\times \left(\text{-}{\displaystyle \frac{\text{3}}{\text{5}}}\right)$   

                                    $$\text{=}\left(\text{-}{\displaystyle \frac{\text{4}}{\text{5}}}\right)$$ 

 And 

$\text{cosec x=}{\displaystyle \frac{\text{1}}{\text{sin x}}}$

          $\text{=-}{\displaystyle \frac{\text{5}}{\text{4}}}$

4. Find the values of other five trigonometric functions if $\text{sec x=}{\displaystyle \frac{\text{13}}{\text{5}}}$  , $\text{x}$ lies in fourth quadrant.

Ans: Here given that,  $\sec x={\displaystyle \frac{13}{5}}$

Therefore we have,

$\cos x={\displaystyle \frac{1}{\sec x}}$

        $={\displaystyle \frac{1}{\left({\displaystyle \frac{13}{5}}\right)}}$

        $={\displaystyle \frac{5}{13}}$

Now we know that , $\text{se}{\text{c}}^{\text{2}}\text{x-ta}{\text{n}}^{\text{2}}\text{x=1}$

Therefore we have, $\text{ta}{\text{n}}^{\text{2}}\text{x=se}{\text{c}}^{\text{2}}\text{x-1}$

Substituting  $$\text{sec x=}{\displaystyle \frac{\text{13}}{\text{5}}}$$  in the formula, we obtain,

$$\text{ta}{\text{n}}^{\text{2}}\text{x=}{\left({\displaystyle \frac{\text{13}}{\text{5}}}\right)}^{\text{2}}\text{-1}$$

$\text{ta}{\text{n}}^{\text{2}}\text{x=}{\displaystyle \frac{\text{169}}{\text{25}}}\text{-1}$

         $\text{=}{\displaystyle \frac{\text{144}}{\text{25}}}$ 

   $$\text{tanx= }\pm {\displaystyle \frac{\text{12}}{\text{5}}}$$

Since $x$ lies in the $4^{th}$ quadrant, the value of $\tan x$ will be negative.

$\text{tan x=-}{\displaystyle \frac{\text{12}}{\text{5}}}$

Therefore, $$\text{cot x=}{\displaystyle \frac{\text{1}}{\text{tan x}}}$$ 

                          $\text{=-}{\displaystyle \frac{\text{5}}{\text{12}}}$ 

Now  , $\text{tan x=}{\displaystyle \frac{\text{sin x}}{\text{cos x}}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}{\displaystyle \frac{\text{5}}{\text{13}}}\times \left(\text{-}{\displaystyle \frac{\text{12}}{\text{5}}}\right)$   

                                  $\text{=}\left(\text{-}{\displaystyle \frac{\text{12}}{\text{13}}}\right)$ 

 And 

$\text{cosec x=}{\displaystyle \frac{\text{1}}{\text{sin x}}}$

            $\text{=-}{\displaystyle \frac{\text{13}}{\text{12}}}$

5. Find the values of other five trigonometric functions if  $\text{tan x=-}{\displaystyle \frac{\text{5}}{\text{12}}}$ , $\text{x}$ lies in second quadrant.

Ans: Here given that,  $\text{tan x=-}{\displaystyle \frac{\text{5}}{\text{12}}}$

Therefore we have,

$\text{cot x=}{\displaystyle \frac{\text{1}}{\text{tan x}}}$

       $\text{=}{\displaystyle \frac{\text{1}}{\left(\text{-}{\displaystyle \frac{\text{5}}{\text{12}}}\right)}}$

       $\text{=-}{\displaystyle \frac{\text{12}}{\text{5}}}$

Now we know that , $\text{se}{\text{c}}^{\text{2}}\text{x-ta}{\text{n}}^{\text{2}}\text{x=1}$

Therefore we have, $\text{se}{\text{c}}^{\text{2}}\text{x=1+ta}{\text{n}}^{\text{2}}\text{x}$

Substituting  $\text{tan x=-}{\displaystyle \frac{\text{5}}{\text{12}}}$  in the formula, we obtain,

$\text{se}{\text{c}}^{\text{2}}\text{x=1+}{\left(\text{-}{\displaystyle \frac{\text{5}}{\text{12}}}\right)}^{\text{2}}$

$\text{se}{\text{c}}^{\text{2}}\text{x=1+}{\displaystyle \frac{\text{25}}{\text{144}}}$

             $={\displaystyle \frac{169}{144}}$ 

   $\text{sec x= }\pm {\displaystyle \frac{\text{13}}{\text{12}}}$

Since $x$ lies in the $2^{nd}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}{\displaystyle \frac{\text{13}}{\text{12}}}$

Therefore, $\text{cos x=}{\displaystyle \frac{\text{1}}{\text{sec x}}}$ 

                          $\text{=-}{\displaystyle \frac{\text{12}}{\text{13}}}$ 

Now  , $\text{tan x=}{\displaystyle \frac{\text{sin x}}{\text{cos x}}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

 Hence we have, $\text{sin x=}\left(\text{-}{\displaystyle \frac{\text{5}}{\text{12}}}\right)\times \left(\text{-}{\displaystyle \frac{\text{12}}{\text{13}}}\right)$   

                                    $=\left({\displaystyle \frac{5}{13}}\right)$ 

 And 

$\text{cosec x=}{\displaystyle \frac{\text{1}}{\text{sin x}}}$

           $\text{=}{\displaystyle \frac{\text{13}}{\text{5}}}$

6. Find the value of the trigonometric function $\text{sin76}{\text{5}}^{\text{o}}$ .

Ans: We know that the values of $\sin x$ repeat after an interval of $2\pi$ or ${360}^{\circ }$ .

Therefore we can write,

$\text{sin76}{\text{5}}^{\text{o}}\text{=sin}(\text{2 }\times \text{ 36}{\text{0}}^{\text{o}}\text{+4}{\text{5}}^{\text{o}})$

            $\text{=sin4}{\text{5}}^{\text{o}}$

            $\text{=}{\displaystyle \frac{\text{1}}{\sqrt{\text{2}}}}\text{.}$

7. Find the value of the trigonometric function $\text{cosec}\left(\text{-141}{\text{0}}^{\text{o}}\right)$  

Ans: We  know that the values of $\text{cosec x}$ repeat after an interval of $\text{2 }\pi$ or ${360}^{\circ }$ .

Therefore we can write,

           $\text{cosec}\left(\text{-141}{\text{0}}^{\text{o}}\right)\text{=cosec}(\text{-141}{\text{0}}^{\text{o}}\text{+4 }\times \text{ 36}{\text{0}}^{\text{o}})$

                               $\text{=cosec}\left(\text{-141}{\text{0}}^{\text{o}}\text{+144}{\text{0}}^{\text{o}}\right)$

                               $\text{=cosec3}{\text{0}}^{\text{o}}$

                               $=2$ 

8. Find the value of the trigonometric function  $\text{tan}{\displaystyle \frac{\text{19 }\pi }{\text{3}}}$ .

Ans: We know that the values of $\text{tan x}$ repeat after an interval of $\pi$ or $${180}^{\circ }$$.

Therefore we can write,

$\text{tan}{\displaystyle \frac{\text{19 }\pi }{\text{3}}}\text{=tan6}{\displaystyle \frac{\text{1}}{\text{3}}}\pi$

            $\text{=tan}\left(\text{6 }\pi \text{ +}{\displaystyle \frac{\pi }{\text{3}}}\right)$

            $$\text{=tan}{\displaystyle \frac{\pi }{\text{3}}}$$

            $\text{=}\sqrt{\text{3}}$      

9. Find the value of the trigonometric function $\text{sin}\left(\text{-}{\displaystyle \frac{\text{11 }\pi }{\text{3}}}\right)$ 

Ans: We know that the values of $\text{sin x}$ repeat after an interval of $\text{2 }\pi$ or ${360}^{\circ }$ .

Therefore we can write,

$\text{sin}\left(\text{-}{\displaystyle \frac{\text{11 }\pi }{\text{3}}}\right)\text{=sin}(\text{-}{\displaystyle \frac{\text{11 }\pi }{\text{3}}}\text{+2 }\times \text{ 2 }\pi )$ 

                   $\text{=sin}\left({\displaystyle \frac{\pi }{\text{3}}}\right)$

                    $={\displaystyle \frac{\sqrt{3}}{2}}$

10. Find the value of the trigonometric function $\text{cot}\left(\text{-}{\displaystyle \frac{\text{15 }\pi }{\text{4}}}\right)$ 

Ans: We  know that the values of $\text{cot x}$ repeat after an interval of $\pi$ or $${180}^{\circ }$$.

Therefore we can write,

$\text{cot}\left(\text{-}{\displaystyle \frac{\text{15 }\pi }{\text{4}}}\right)\text{=cot}\left(\text{-}{\displaystyle \frac{\text{15 }\pi }{\text{4}}}\text{+4 }\pi \right)$ 

                   $\text{=cot}{\displaystyle \frac{\pi }{\text{4}}}$

              $=1$

Exercise 3.3

1. Prove that $\text{si}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}\text{+co}{\text{s}}^{\text{2}}{\displaystyle \frac{\pi }{\text{3}}}\text{-ta}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{4}}}\text{=-}{\displaystyle \frac{\text{1}}{\text{2}}}$

Ans: Substituting the values of  $$\text{sin}{\displaystyle \frac{\pi }{\text{6}}}\text{,cos}{\displaystyle \frac{\pi }{\text{6}}}\text{,tan}{\displaystyle \frac{\pi }{\text{4}}}$$ on left hand side,

$\text{si}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}\text{+co}{\text{s}}^{\text{2}}{\displaystyle \frac{\pi }{\text{3}}}\text{-ta}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{4}}}\text{=}{\left({\displaystyle \frac{\text{1}}{\text{2}}}\right)}^{\text{2}}\text{+}{\left({\displaystyle \frac{\text{1}}{\text{2}}}\right)}^{\text{2}}\text{-}{\left(\text{1}\right)}^{\text{2}}$

                                  $$\text{=}{\displaystyle \frac{\text{1}}{\text{4}}}\text{+}{\displaystyle \frac{\text{1}}{\text{4}}}\text{-1}$$

                                  $=-{\displaystyle \frac{1}{2}}$

                                  $=$ R.H.S.

Hence proved.

2. Prove that $\text{2si}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}\text{+cose}{\text{c}}^{\text{2}}{\displaystyle \frac{\text{7 }\pi }{\text{6}}}\text{co}{\text{s}}^{\text{2}}{\displaystyle \frac{\pi }{\text{3}}}\text{=}{\displaystyle \frac{\text{3}}{\text{2}}}$

Ans: Substituting the values of  $\text{sin}{\displaystyle \frac{\pi }{\text{6}}}\text{,cosec}{\displaystyle \frac{\text{7 }\pi }{\text{6}}}\text{,cos}{\displaystyle \frac{\pi }{\text{3}}}$ on left hand side,

L.H.S.$\text{=2si}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}\text{+cose}{\text{c}}^{\text{2}}{\displaystyle \frac{\text{7 }\pi }{\text{6}}}\text{co}{\text{s}}^{\text{2}}{\displaystyle \frac{\pi }{\text{3}}}$

           $\text{=2}{\left({\displaystyle \frac{\text{1}}{\text{2}}}\right)}^{\text{2}}\text{+cose}{\text{c}}^{\text{2}}\left(\pi \text{ +}{\displaystyle \frac{\pi }{\text{6}}}\right){\left({\displaystyle \frac{\text{1}}{\text{2}}}\right)}^{\text{2}}$

           $\text{=2 }\times {\displaystyle \frac{\text{1}}{\text{4}}}\text{+}{\left(\text{-cosec}{\displaystyle \frac{\pi }{\text{6}}}\right)}^{\text{2}}\left({\displaystyle \frac{\text{1}}{\text{4}}}\right)$

          $\text{=}{\displaystyle \frac{\text{1}}{\text{2}}}\text{+}{\left(\text{-2}\right)}^{\text{2}}\left({\displaystyle \frac{\text{1}}{\text{4}}}\right)$

Since $\text{cosec x}$ repeat its value after an interval of $\text{2 }\pi$ , 

we have, $\text{cosec}{\displaystyle \frac{\text{7 }\pi }{\text{6}}}\text{=-cosec}{\displaystyle \frac{\pi }{\text{6}}}$  

L.H.S  $={\displaystyle \frac{1}{2}}+{\displaystyle \frac{4}{4}}$

           $={\displaystyle \frac{3}{2}}$ 

          $=$ R.H.S.

Hence proved.

3. Prove that $\text{co}{\text{t}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}\text{+cosec}{\displaystyle \frac{\text{5 }\pi }{\text{6}}}\text{+3ta}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}\text{=6}$

Ans: Substituting the values of  $\text{cot}{\displaystyle \frac{\pi }{\text{6}}}\text{,cosec}{\displaystyle \frac{\text{5 }\pi }{\text{6}}}\text{,tan}{\displaystyle \frac{\pi }{\text{6}}}$ on left hand side,

L.H.S.$\text{=co}{\text{t}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}\text{+cosec}{\displaystyle \frac{\text{5 }\pi }{\text{6}}}\text{+3ta}{\text{n}}^{\text{2}}{\displaystyle \frac{\pi }{\text{6}}}$

           $\text{=}{\left(\sqrt{\text{3}}\right)}^{\text{2}}\text{+cosec}\left(\pi \text{ -}{\displaystyle \frac{\pi }{\text{6}}}\right)\text{+3}{\left({\displaystyle \frac{\text{1}}{\sqrt{\text{3}}}}\right)}^{\text{2}}$

         $\text{=3+cosec}{\displaystyle \frac{\pi }{\text{6}}}\text{+3 }\times {\displaystyle \frac{\text{1}}{\text{3}}}$

Since $\text{cosec x}$ repeat its value after an interval of $\text{2 }\pi$ , 

we have, $\text{cosec}{\displaystyle \frac{\text{5 }\pi }{\text{6}}}\text{=cosec}{\displaystyle \frac{\pi }{\text{6}}}$  

L.H.S $=3+2+1$ 

          $=1$ 

          $=$ R.H.S.

Hence proved.

4. Prove that $\text{2si}{\text{n}}^{\text{2}}{\displaystyle \frac{\text{3 }\pi }{\text{4}}}\text{+2co}{\text{s}}^{\text{2}}{\displaystyle \frac{\pi }{\text{4}}}\text{+2se}{\text{c}}^{\text{2}}{\displaystyle \frac{\pi }{\text{3}}}\text{=10}$

Ans:

Substituting the values of  $\text{sin}{\displaystyle \frac{\text{3 }\pi }{\text{4}}}\text{,cos}{\displaystyle \frac{\pi }{\text{4}}}\text{,sec}{\displaystyle \frac{\pi }{\text{3}}}$ on left hand side,

L.H.S.$\text{=2si}{\text{n}}^{\text{2}}{\displaystyle \frac{\text{3 }\pi }{\text{4}}}\text{+2co}{\text{s}}^{\text{2}}{\displaystyle \frac{\pi }{\text{4}}}\text{+2se}{\text{c}}^{\text{2}}{\displaystyle \frac{\pi }{\text{3}}}$

           $\text{=2}{\left\{\text{sin}\left(\pi \text{ -}{\displaystyle \frac{\pi }{\text{4}}}\right)\right\}}^{\text{2}}\text{+2}{\left({\displaystyle \frac{\text{1}}{\sqrt{\text{2}}}}\right)}^{\text{2}}\text{+2}{\left(\text{2}\right)}^{\text{2}}$

           $\text{=2}{\left\{\text{sin}{\displaystyle \frac{\pi }{\text{4}}}\right\}}^{\text{2}}\text{+2 }\times {\displaystyle \frac{\text{1}}{\text{2}}}\text{+8}$

Since $\text{sin x}$ repeat its value after an interval of $$\text{2 }\pi$$ , 

we have, $$\text{sin}{\displaystyle \frac{\text{3 }\pi }{\text{4}}}\text{=sin}{\displaystyle \frac{\pi }{\text{4}}}$$  

 L.H.S  $=1+1+8$ 

            $=10$

            $=$ R.H.S.

Hence proved.

5. Find the value of :

(i) $\text{sin7}{\text{5}}^{\text{o}}$

Ans: We have,

$\text{sin7}{\text{5}}^{\text{o}}\text{=sin(4}{\text{5}}^{\text{o}}\text{+3}{\text{0}}^{\text{o}}\text{)}$

          $\text{=sin4}{\text{5}}^{\text{o}}\text{cos3}{\text{0}}^{\text{o}}\text{+cos4}{\text{5}}^{\text{o}}\text{sin3}{\text{0}}^{\text{o}}$

Since we know that, $\text{sin}\left(\text{x+y}\right)\text{=sin x cos y+cos x sin y}$ 

Therefore we have,

$Sin{75}^o={\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{\sqrt{3}}{2}}+{\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{1}{2}}$

$Sin{75}^o={\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}}$

(ii) $\text{tan1}{\text{5}}^{\text{o}}$

Ans: We have,

$$\text{tan1}{\text{5}}^{\text{o}}\text{=tan}\left(\text{4}{\text{5}}^{\text{o}}\text{-3}{\text{0}}^{\text{o}}\right)$$

        $\text{=}{\displaystyle \frac{\text{tan4}{\text{5}}^{\text{o}}\text{-tan3}{\text{0}}^{\text{o}}}{\text{1+tan4}{\text{5}}^{\text{o}}\text{tan3}{\text{0}}^{\text{o}}}}$

Since we know, $\text{tan}\left(\text{x-y}\right)\text{=}{\displaystyle \frac{\text{tan x-tan y}}{\text{1+tan x tan y}}}$

Therefore we have,

$\text{tan1}{\text{5}}^{\text{o}}\text{=}{\displaystyle \frac{\text{1-}{\displaystyle \frac{\text{1}}{\sqrt{\text{3}}}}}{\text{1+1}\left({\displaystyle \frac{\text{1}}{\sqrt{\text{3}}}}\right)}}$

           $\text{=}{\displaystyle \frac{{\displaystyle \frac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}}}}{{\displaystyle \frac{\sqrt{\text{3}}\text{+1}}{\sqrt{\text{3}}}}}}$

         $\text{=}{\displaystyle \frac{\sqrt{\text{3}}\text{-1}}{\sqrt{\text{3}}\text{+1}}}$

         $\text{=}{\displaystyle \frac{{\left(\sqrt{\text{3}}\text{-1}\right)}^{\text{2}}}{\left(\sqrt{\text{3}}\text{+1}\right)\left(\sqrt{\text{3}}\text{-1}\right)}}$ 

Further computing we have,

$\text{tan1}{\text{5}}^{\text{o}}\text{=}{\displaystyle \frac{\text{3+1-2}\sqrt{\text{3}}}{{\left(\sqrt{\text{3}}\right)}^{\text{2}}\text{-}{\left(\text{1}\right)}^{\text{2}}}}$ 

           $$\text{=}{\displaystyle \frac{\text{4-2}\sqrt{\text{3}}}{\text{3-1}}}$$

           $$\text{=2-}\sqrt{\text{3}}$$

6. Prove that $\text{cos}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-x}\right)\text{cos}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-y}\right)\text{-sin}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-x}\right)\text{sin}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-y}\right)\text{=sin}\left(\text{x+y}\right)$

Ans: We know that, $\text{cos}\left(\text{x+y}\right)\text{=cos xcos y-sin xsin y}$ 

$\text{cos}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-x}\right)\text{cos}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-y}\right)\text{-sin}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-x}\right)\text{sin}\left({\displaystyle \frac{\pi }{\text{4}}}\text{-y}\right)\text{=cos}\left[{\displaystyle \frac{\pi }{\text{4}}}\text{-x+}{\displaystyle \frac{\pi }{\text{4}}}\text{-y}\right]$

$\text{=cos}\left[{\displaystyle \frac{\pi }{\text{2}}}\text{-}\left(\text{x+y}\right)\right]$