Class 11 Maths Chapter 3 – Trigonometric Functions Exercise 3.2

Q1. Find the values of other five trigonometric functions: $\cos x=-\frac12$ , $x$ lies in third quadrant.

Sol. Given, $\cos x=-\frac12$ .

\sec x=\frac{1}{\cos x}=\frac{1}{-\dfrac12}=-2

We know that

\textcolor{#1B5E20}{ \begin{array}{l} \displaystyle \sin^2x+\cos^2x=1 \\[0.4em] \displaystyle \sin^2x=1-\cos^2x \end{array} }

Substituting the value of $\cos x$ ,

\displaystyle \sin^2x=1-\left(-\frac12\right)^2

\displaystyle =1-\frac14=\frac34

\displaystyle \sin x=\pm\frac{\sqrt3}{2}

Since $x$ lies in the third quadrant, $\sin x$ is negative.

\begin{array}{l} \displaystyle \sin x=-\frac{\sqrt3}{2} \\[0.8em] \displaystyle \text{cosec}\,x=\frac{1}{\sin x}=\frac{1}{-\dfrac{\sqrt3}{2}}=-\frac{2}{\sqrt3} \end{array}

Now,

\begin{array}{l} \displaystyle \tan x=\frac{\sin x}{\cos x}=\frac{-\dfrac{\sqrt3}{2}}{-\dfrac12}=\sqrt3 \\[1.2em] \displaystyle \cot x=\frac{1}{\tan x}=\frac{1}{\sqrt3} \end{array}

∴ $\sin x=-\frac{\sqrt3}{2}$ , $\text{cosec}\,x=-\frac{2}{\sqrt3}$ , $\sec x=-2$ , $\tan x=\sqrt3$ and $\cot x=\frac{1}{\sqrt3}$ .

Q2. Find the values of other five trigonometric functions: $\sin x=\frac35$ , $x$ lies in second quadrant.

Sol. Given, $\sin x=\frac35$ .

\text{cosec}\,x=\frac{1}{\sin x}=\frac{1}{\dfrac35}=\frac53

We know that

\textcolor{#1B5E20}{ \begin{array}{l} \displaystyle \sin^2x+\cos^2x=1 \\[0.4em] \displaystyle \cos^2x=1-\sin^2x \end{array} }

Substituting the value of $\sin x$ ,

\displaystyle \cos^2x=1-\left(\frac35\right)^2

\displaystyle =1-\frac{9}{25}=\frac{16}{25}

\displaystyle \cos x=\pm\frac45

Since $x$ lies in the second quadrant, $\cos x$ is negative.

\begin{array}{l} \displaystyle \cos x=-\frac45 \\[0.8em] \displaystyle \sec x=\frac{1}{\cos x}=\frac{1}{-\dfrac45}=-\frac54 \end{array}

Now,

\begin{array}{l} \displaystyle \tan x=\frac{\sin x}{\cos x}=\frac{\dfrac35}{-\dfrac45}=-\frac34 \\[2em] \displaystyle \cot x=\frac{1}{\tan x}=\frac{1}{-\dfrac34}=-\frac43 \end{array}

∴ $\cos x=-\frac45$ , $\text{cosec}\,x=\frac53$ , $\sec x=-\frac54$ , $\tan x=-\frac34$ and $\cot x=-\frac43$ .

Q3. Find the values of other five trigonometric functions: $\cot x=\frac34$ , $x$ lies in third quadrant.

Sol. Given, $\cot x=\frac34$ .

\tan x=\frac{1}{\cot x}=\frac{1}{\dfrac34}=\frac43

We know that

\textcolor{#1B5E20}{ \sec^2x=1+\tan^2x }

Substituting the value of $\tan x$ ,

\displaystyle \sec^2x=1+\left(\frac43\right)^2

\displaystyle =1+\frac{16}{9}=\frac{25}{9}

\displaystyle \sec x=\pm\frac53

Since $x$ lies in the third quadrant, $\sec x$ is negative.

\begin{array}{l} \displaystyle \sec x=-\frac53 \\[0.8em] \displaystyle \cos x=\frac{1}{\sec x}=\frac{1}{-\dfrac53}=-\frac35 \end{array}

Now,

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

\displaystyle \sin x=\tan x\cos x=\frac43\times\left(-\frac35\right)=-\frac45

\displaystyle \text{cosec}\,x=\frac{1}{\sin x}=\frac{1}{-\dfrac45}=-\frac54

∴ $\sin x=-\frac45$ , $\cos x=-\frac35$ , $\tan x=\frac43$ , $\text{cosec}\,x=-\frac54$ and $\sec x=-\frac53$ .

Q4. Find the values of other five trigonometric functions: $\sec x=\frac{13}{5}$ , $x$ lies in fourth quadrant.

Sol. Given, $\sec x=\frac{13}{5}$

\cos x=\frac{1}{\sec x}=\frac{1}{\dfrac{13}{5}}=\frac{5}{13}

We know that

\textcolor{#1B5E20}{ \begin{array}{l} \displaystyle \sin^2x+\cos^2x=1 \\[0.4em] \displaystyle \sin^2x=1-\cos^2x \end{array} }

Substituting the value of $\cos x$ ,

\begin{array}{l} \displaystyle \sin^2x=1-\left(\frac{5}{13}\right)^2 \\[1.6em] \hspace{2.4em} \displaystyle =1-\frac{25}{169}=\frac{144}{169} \\[0.8em] \displaystyle \sin x=\pm\frac{12}{13} \end{array}

Since $x$ lies in the fourth quadrant, $\sin x$ is negative.

\begin{array}{l} \displaystyle \sin x=-\frac{12}{13} \\[0.8em] \displaystyle \text{cosec}\,x=\frac{1}{\sin x}=\frac{1}{-\dfrac{12}{13}}=-\frac{13}{12} \end{array}

Now,

\begin{array}{l} \displaystyle \tan x=\frac{\sin x}{\cos x}=\frac{-\dfrac{12}{13}}{\dfrac{5}{13}}=-\frac{12}{5} \\[2em] \displaystyle \cot x=\frac{1}{\tan x}=\frac{1}{-\dfrac{12}{5}}=-\frac{5}{12} \end{array}

∴ $\sin x=-\frac{12}{13}$ , $\cos x=\frac{5}{13}$ , $\tan x=-\frac{12}{5}$ , $\text{cosec}\,x=-\frac{13}{12}$ and $\cot x=-\frac{5}{12}$ .

Q5. Find the values of other five trigonometric functions: $\tan x=-\frac{5}{12}$ , $x$ lies in second quadrant.

Sol. Given, $\tan x=-\frac{5}{12}$ .

\cot x=\frac{1}{\tan x}=\frac{1}{-\dfrac{5}{12}}=-\frac{12}{5}

We know that

\textcolor{#1B5E20}{ \sec^2x=1+\tan^2x }

Substituting the value of $\tan x$ ,

\displaystyle \sec^2x=1+\left(-\frac{5}{12}\right)^2

\displaystyle =1+\frac{25}{144}=\frac{169}{144}

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

Since $x$ lies in the second quadrant, $\sec x$ is negative.

\begin{array}{l} \displaystyle \sec x=-\frac{13}{12} \\[0.8em] \displaystyle \cos x=\frac{1}{\sec x}=\frac{1}{-\dfrac{13}{12}}=-\frac{12}{13} \end{array}

Now,

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

\displaystyle \sin x=\tan x\cos x=\left(-\frac{5}{12}\right)\times\left(-\frac{12}{13}\right)=\frac{5}{13}

\displaystyle \text{cosec}\,x=\frac{1}{\sin x}=\frac{1}{\dfrac{5}{13}}=\frac{13}{5}

∴ $\sin x=\frac{5}{13}$ , $\cos x=-\frac{12}{13}$ , $\text{cosec}\,x=\frac{13}{5}$ , $\sec x=-\frac{13}{12}$ and $\cot x=-\frac{12}{5}$ .

Q6. Find the value of the trigonometric function $\sin765^\circ$ .

Sol. We know that the values of $\sin x$ repeat after an interval of $360^\circ$ .

Writing $765^\circ$ as a multiple of $360^\circ$ plus an angle,

765^\circ=2\times360^\circ+45^\circ

Therefore,

\begin{array}{l} \displaystyle \sin765^\circ=\sin(2\times360^\circ+45^\circ) \\[0.4em] \hspace{3em} \displaystyle =\sin45^\circ=\frac{1}{\sqrt2} \end{array}

Q7. Find the value of the trigonometric function $\text{cosec}(-1410^\circ)$ .

Sol. We know that the values of $\text{cosec}\,x$ repeat after an interval of $360^\circ$ .

Writing $-1410^\circ$ as a multiple of $360^\circ$ plus an angle,

-1410^\circ=-4\times360^\circ+30^\circ

Therefore,

\begin{array}{l} \displaystyle \text{cosec}(-1410^\circ)=\text{cosec}(-4\times360^\circ+30^\circ) \\[0.8em] \displaystyle =\text{cosec}\,30^\circ=2 \end{array}

∴ $\text{cosec}(-1410^\circ)=2$ .

Q8. Find the value of the trigonometric function $\tan\frac{19\pi}{3}$ .

Sol. We know that the values of $\tan x$ repeat after an interval of $\pi$ .

Writing $\frac{19\pi}{3}$ as a multiple of $\pi$ plus an angle,

\frac{19\pi}{3}=6\pi+\frac{\pi}{3}

Therefore,

\begin{array}{l} \displaystyle \tan\frac{19\pi}{3}=\tan\left(6\pi+\frac{\pi}{3}\right) \\[0.8em] \hspace{3.2em} \displaystyle =\tan\frac{\pi}{3}=\sqrt3 \end{array}

Q9. Find the value of the trigonometric function $\sin\left(-\frac{11\pi}{3}\right)$ .

Sol. We know that the values of $\sin x$ repeat after an interval of $2\pi$ .

Writing $-\frac{11\pi}{3}$ as a multiple of $2\pi$ plus an angle,

-\frac{11\pi}{3}=-4\pi+\frac{\pi}{3}

Therefore,

\displaystyle \sin\left(-\frac{11\pi}{3}\right)=\sin\left(-4\pi+\frac{\pi}{3}\right)

\displaystyle =\sin\frac{\pi}{3}=\frac{\sqrt3}{2}

∴ $\sin\left(-\frac{11\pi}{3}\right)=\frac{\sqrt3}{2}$ .

Q10. Find the value of the trigonometric function $\cot\left(-\frac{15\pi}{4}\right)$ .

Sol. We know that the values of $\cot x$ repeat after an interval of $\pi$ .

Writing $-\frac{15\pi}{4}$ as a multiple of $\pi$ plus an angle,

-\frac{15\pi}{4}=-4\pi+\frac{\pi}{4}

Therefore,

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

\displaystyle =\cot\frac{\pi}{4}=1

∴ $\cot\left(-\frac{15\pi}{4}\right)=1$ .