NCERT Solutions for Exercise 3.2 Class 11 Maths Chapter 3 - Trigonometric Functions

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.2

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

Answer:

Solution
$\cos x=-\frac{1}{2}$
$\because \sec x=\frac{1}{\cos x}=\frac{1}{-\frac{1}{2}}=-2$
x lies in III quadrants. Therefore sec x is negative

$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x=1 \\ {\sin }^{2}x=1-{\cos }^{2}x \\ {\sin }^{2}x=1-{(-\frac{1}{2})}^2 \\ {\sin }^{2}x=1-\frac{1}{4}=\frac{3}{4} \\ \sin x=\sqrt{\frac{3}{4}}=\pm \frac{\sqrt{3}}{2} \end{gathered}$$
x lies in III quadrants. Therefore sin x is negative
$\therefore \sin x=-\frac{\sqrt{3}}{2}$

$\because cosecx=\frac{1}{\sin x}=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}$

x lies in III quadrants. Therefore cosec x is negative

$\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3}$
x lies in III quadrants. Therefore tan x is positive

$\cot x=\frac{1}{\tan x}=\frac{1}{\sqrt{3}}$
x lies in III quadrants. Therefore cot x is positive

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

Answer:

Solution

$\sin x=\frac{3}{5}$

$cosecx=\frac{1}{\sin x}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$
x lies in the second quadrant. Therefore cosec x is positive

$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x=1 \\ {\cos }^{2}x=1-{\sin }^{2}x \\ {\cos }^{2}x=1-{\left(\frac{3}{5}\right)}^2 \\ {\cos }^{2}x=1-\frac{9}{25}=\frac{16}{25} \\ \cos x=\sqrt{\frac{16}{25}}=\pm \frac{4}{5} \end{gathered}$$
x lies in the second quadrant. Therefore cos x is negative
$\therefore \cos x=-\frac{4}{5}$

$\sec x=\frac{1}{\cos x}=\frac{1}{-\frac{4}{5}}=-\frac{5}{4}$
x lies in the second quadrant. Therefore sec x is negative

$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$
x lies in the second quadrant. Therefore tan x is negative

$\cot x=\frac{1}{\tan x}=\frac{1}{-\frac{3}{4}}=-\frac{4}{3}$
x lies in the second quadrant. Therefore cot x is negative

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

Answer:

Solution

$\cot x=\frac{3}{4}$

$\tan x=\frac{1}{\cot x}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
$$\begin{gathered} 1+{\tan }^{2}x={\sec }^{2}x \\ 1+\frac{4^2}{3^2}={\sec }^{2}x \\ 1+\frac{16}{9}={\sec }^{2}x \\ \frac{25}{9}={\sec }^{2}x \\ \sec x=\sqrt{\frac{25}{9}}=\pm \frac{5}{3} \end{gathered}$$
x lies in x lies in third quadrant. therefore sec x is negative
$\sec x=-\frac{5}{3}$

$\cos x=\frac{1}{\sec x}=\frac{1}{-\frac{5}{3}}=-\frac{3}{5}$
$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x=1 \\ {\sin }^{2}x=1-{\cos }^{2}x \\ {\sin }^{2}x=1-{(-\frac{3}{5})}^2 \\ {\sin }^{2}x=1-\frac{9}{25} \\ {\sin }^{2}x=\frac{16}{25} \\ \sin x=\sqrt{\frac{16}{25}}=\pm \frac{4}{5} \end{gathered}$$
x lies in x lies in third quadrant. Therefore sin x is negative
$\sin x=-\frac{4}{5}$
$cosecx=\frac{1}{\csc }=\frac{1}{-\frac{4}{5}}=-\frac{5}{4}$

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

Answer:

Solution
$\sec x=\frac{13}{5}$
$\cos x=\frac{1}{\sec x}=\frac{1}{\frac{13}{5}}=\frac{5}{13}$
$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x=1 \\ {\sin }^{2}x=1-{\cos }^{2}x \\ {\sin }^{2}x=1-\frac{5}{13} \\ {\sin }^{2}x=1-\frac{25}{169}=\frac{144}{169} \\ \sin x=\sqrt{\frac{144}{169}}=\pm \frac{12}{13} \end{gathered}$$
lies in fourth quadrant. Therefore sin x is negative
$\sin x=-\frac{12}{13}$
$\csc x=\frac{1}{\sin x}=\frac{1}{-\frac{12}{13}}=-\frac{13}{12}$
$\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{12}{13}}{\frac{5}{13}}=-\frac{12}{5}$
$\cot x=\frac{1}{\tan x}=\frac{1}{-\frac{12}{5}}=-\frac{5}{12}$

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

Answer:
$\tan x=-\frac{5}{12}$
$\cot x=\frac{1}{\tan x}=\frac{1}{-\frac{5}{12}}=-\frac{12}{5}$
$$\begin{gathered} 1+{\tan }^{2}x={\sec }^{2}x \\ 1+{(-\frac{5}{12})}^2={\sec }^{2}x \\ 1+\frac{25}{144}={\sec }^{2}x \\ \frac{169}{144}={\sec }^{2}x \\ \sec x=\sqrt{\frac{169}{144}}=\pm \frac{13}{12} \end{gathered}$$
x lies in second quadrant. Therefore the value of sec x is negative
$\sec x=-\frac{13}{12}$
$\cos x=\frac{1}{\sec x}=\frac{1}{-\frac{13}{12}}=-\frac{12}{13}$
$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x=1 \\ {\sin }^{2}x=1-{\cos }^{2}x \\ {\sin }^{2}x=1-{(-\frac{12}{13})}^2 \\ {\sin }^{2}x=1-\frac{144}{169} \\ {\sin }^{2}x=\frac{25}{169} \\ \sin x=\sqrt{\frac{25}{169}}=\pm \frac{5}{13} \end{gathered}$$
x lies in the second quadrant. Therefore the value of sin x is positive
$\sin x=\frac{5}{13}$
$\csc =\frac{1}{\sin x}=\frac{1}{\frac{5}{13}}=\frac{13}{5}$

Question:6 Find the values of the trigonometric functions $\sin {765}^{\circ }$

Answer:
We know that values of sin x repeat after an interval of $2\pi or360degree$

$$\begin{gathered} \sin {765}^{\circ }=\sin (2\times {360}^{\circ }+{45}^{\circ })=\sin {45}^{\circ } \\ sin{45}^{\circ }=\frac{1}{\sqrt{2}} \end{gathered}$$

Question:7 Find the values of the trigonometric functions $cosec(-{1410}^{\circ })$

Answer:

We know that value of cosec x repeats after an interval of $2\pi or{360}^{\circ }$
$$\begin{gathered} cosec(-{1410}^{\circ })=cosec(-{1410}^{\circ }+{360}^{\circ }\times 4) \\ cosec{30}^{\circ }=2 \end{gathered}$$

or

$$\begin{gathered} cosec(-{1410}^{\circ })=-cosec({1410}^{\circ }) \\ =-cosec(4\times {360}^{\circ }-{30}^{\circ })=-cosec(-{30}^{\circ })=2 \end{gathered}$$

Question:8 Find the values of the trigonometric functions $\tan \frac{19\pi }{3}$

Answer:

We know that tan x repeats after an interval of $\pi$ or 180 degree
$\tan (\frac{19\pi }{3})=\tan (6\pi +\frac{\pi }{3})=\tan \frac{\pi }{3}=\tan {60}^{\circ }=\sqrt{3}$

Question:9 Find the values of the trigonometric functions $\sin (-\frac{11\pi }{3})$

Answer:

We know that sin x repeats after an interval of $2\pi or{360}^{\circ }$
$\sin (-\frac{11\pi }{3})=\sin (-4\pi +\frac{\pi }{3})=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$

Question:10 Find the values of the trigonometric functions $\cot (-\frac{15\pi }{4})$

Answer:

We know that cot x repeats after an interval of $\pi or{180}^{\circ }$
$\cot (-\frac{15\pi }{4})=\cot (-4\pi +\frac{\pi }{4})=\cot \left(\frac{\pi }{4}\right)=1$