Ex 3.2 Trigonomety ncert maths solution class 11

07/02/2023 by Team Gmath

EXERCISE 3.2(Trigonometric Function)

Find the values of other five trigonometric functions in Exercises 1 to 5.(Ex 3.2 Trigonomety ncert maths solution class 11)

Question 1: $\cos x = -\frac{1}{2}$ , x lies in third quadrant.

Solution: Given, $\cos x = -\frac{1}{2}$

Since, $\sec x = \frac{1}{\cos x}$

$\Rightarrow \sec x = \frac{1}{-1/2}$

$\Rightarrow \sec x = -2$

We know that

$\sin^2 x+\cos^2x=1$

$\Rightarrow \sin^2 x = 1-\cos^2x$

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

$\Rightarrow \sin^2x = 1-\frac{1}{4}=\frac{3}{4}$

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

Since x lies in third quadrant

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

We know that

$\operatorname{cosec x}=\frac{1}{\sin x}$

$\Rightarrow \operatorname{cosec x}= \frac{1}{-\sqrt{3}/2}=-\frac{2}{\sqrt{3}}$

Since, $\tan x = \frac{\sin x}{\cos x}$

$\Rightarrow \tan x = \frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}$

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

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

Question 2: $\sin x = \frac{3}{5}$ , x lies in second quadrant

Solution: Given, $\sin x = \frac{3}{5}$

Since, $\operatorname{cosec x} =\frac{1}{\sin x}= \frac{1}{3/5}$

$\Rightarrow \operatorname{cosec x}=\frac{5}{3}$

We know that

$\cos^2 x = 1-\sin^2 x$

$\Rightarrow \cos^2 x = 1-(\frac{3}{5})^2 = 1-\frac{9}{25}$

$\Rightarrow \cos^2 x = \frac{16}{25}$

$\Rightarrow \cos x = =\pm \frac{4}{5}$

X lies in second quadrant

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

Since, $\sec x =\frac{1}{\cos x}=\frac{1}{-4/5}$

$\Rightarrow \sec x = -\frac{5}{4}$

Again we know that

$\tan x = \frac{\sin x}{\cos x}= \frac{3/5}{-4/5}$

$\Rightarrow \tan x = -\frac{3}{4}$

And $\cot x = \frac{1}{\tan x}=\frac{1}{-3/4}$

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

Question 3: $\cot x = \frac{3}{4}$ , x lies in third quadrant.

Solution: Given $\cot x =\frac{3}{4}$

Since, $\tan x = \frac{1}{\cot x}$

$\Rightarrow \tan x = \frac{1}{3/4}=\frac{4}{3}$

We know that

$\sec^2x = 1+\tan^2x$

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

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

$\Rightarrow \sec x = \pm \frac{5}{3}$

x lies in third quadrant

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

Now, $\cos x = \frac{1}{\sec x}$

$\Rightarrow \cos x = \frac{1}{-5/3}=-\frac{3}{5}$

Again we know that

$\sin^2x = 1-\cos^2x$

$\Rightarrow \sin^2x = 1-(-\frac{3}{5})^2$

$= 1-\frac{9}{25}$

$\Rightarrow \sin^2x= \frac{16}{25}$

$\Rightarrow \sin x = \pm \frac{4}{5}$

x lies in third quadrant

$\sin x = -\frac{4}{5}$

Since, $\operatorname{cosec x}=\frac{1}{\sin x}$

$= \frac{1}{-4/5}= -\frac{5}{4}$

Question 4: $\sec x = \frac{13}{5}$ , x lies in fourth quadrant

Solution: Given, $\sec x = \frac{13}{5}$

Sinc, $\cos x =\frac{1}{\sec x}$

$\Rightarrow \cos x = \frac{1}{13/5}=\frac{5}{13}$

We know that

$\sin^2 x = 1-\cos^2 x$

$\Rightarrow \sin^2x = 1-(\frac{5}{13})^2$

$= 1-\frac{25}{169}=\frac{144}{169}$

$\Rightarrow \sin x= \pm\frac{12}{13}$

x lies in fourth quadrant

$\sin x = -\frac{12}{13}$

$\operatorname{cosec x}=\frac{1}{\sin x}=-\frac{13}{12}$

Since, $\tan x =\frac{\sin x}{\cos x} = \frac{-12/13}{5/13}$

$\Rightarrow \tan x = -\frac{12}{5}$

Since, $\cot x = \frac{1}{\tan x} = -\frac{5}{12}$

Question 5: $\tan x = -\frac{5}{12}$ , x lies in second quadrant.

Solution: Given $\tan x = -\frac{5}{12}$

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

We know that

$\sec^2x = 1+\tan^2x$

$\Rightarrow \sec^2 x = 1+(-\frac{5}{12})^2$

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

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

x lies in second quadrant

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

Now, $\cos x = \frac{1}{\sec x}=-\frac{12}{13}$

Again we know that

$\sin^2x = 1-\cos^2x$

$\Rightarrow \sin^2x = 1-(-\frac{12}{13})^2$

$= 1-\frac{144}{169}= \frac{25}{169}$

$\Rightarrow \sin x=\pm\frac{5}{13}$

x lies in second quadrant

$\therefore \sin x = \frac{5}{13}$

$\operatorname{cosec x} = \frac{1}{\sin x} = \frac{13}{5}$

Find the values of the trigonometric functions in Exercises 6 to 10.

Question 6: sin 765°

Solution: The values of sin x repeat after an interval of 2π or 360°

So we get

$\sin 765^{\circ}=\sin(45+2\times 360)$

$= \sin 45^{\circ}$

$=\frac{1}{\sqrt{2}}$

question 7: $\operatorname{cosec}(-1410)$

Solution: The values of sin x repeat after an interval of 2π or 360°

$\operatorname{cosec}(-1410)=\operatorname{cosec}(-1410+4\times 360)$

$=\operatorname{cosec}(30)$

$=2$

Question 8: $\tan(\frac{19\pi}{3})$

Solution: The values of sin x repeat after an interval of 2π or 360°

$\tan(\frac{19\pi}{3})=\tan(\frac{19\pi}{3}-3\times 2\pi)$

$=\tan(\frac{19\pi}{3}-6\pi)$

$= \tan(\frac{\pi}{3})$

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

Question 9: $\sin(-\frac{11\pi}{3})$

Solution: The values of sin x repeat after an interval of 2π or 360°

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

$= \sin(\frac{\pi}{3})$

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

Question 10: $\cot(-\frac{15\pi}{4})$

Solution: The values of sin x repeat after an interval of 2π or 360°

$\cot(-\frac{15\pi}{4})=\cot(-\frac{15\pi}{4}+2\times 2\pi)$

$=\cot(-\frac{15\pi}{4}+4\pi)$

$=\cot(\frac{\pi}{4})$

$=1$

Ex 3.1 Trigonometric function ncert solution class 11

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