Exercise 3.2 (Revised) - Chapter 3 - Trigonometric Functions - Ncert Solutions class 11 - Maths

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Chapter 3: Trigonometric Functions | NCERT Solutions for Class 11 Maths

Find the values of other trigonometric functions in exercises 1 to 5.
Ex 3.2 Question 1.

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

Answer.

Given: $\cos x=-\frac{1}{2}$
$\because \sin ^2 \theta+\cos ^2 \theta=1$
$\Rightarrow \sin ^2 x+\left(-\frac{1}{2}\right)^2=1$
$\Rightarrow \sin ^2 x=1-\frac{1}{4}$
$\Rightarrow \sin ^2 x=\frac{3}{4}$
$\Rightarrow \sin x= \pm \frac{\sqrt{3}}{2}$
$\Rightarrow \sin x=-\frac{\sqrt{3}}{2}$ [ $x$ lies in third quadrant]
Now, $\operatorname{cosec} x=\frac{1}{\sin x}=-\frac{2}{\sqrt{3}}$

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

$
\begin{aligned}
& \tan x=\frac{\sin x}{\cos x}=\frac{-\sqrt{3} / 2}{-1 / 2}=\sqrt{3} \\
& \cot x=\frac{\cos x}{\sin x}=\frac{-1 / 2}{-\sqrt{3} / 2}=\frac{1}{\sqrt{3}}
\end{aligned}
$
Ex 3.2 Question 2.

$\sin x=\frac{3}{5}, x$ lies in second quadrant.

Answer.

Given: $\sin x=\frac{3}{5}$
$
\begin{aligned}
& \because \sin ^2 \theta+\cos ^2 \theta=1 \\
& \Rightarrow\left(\frac{3}{5}\right)^2+\cos ^2 x=1 \\
& \Rightarrow \cos ^2 x=1-\frac{9}{25} \\
& \Rightarrow \cos ^2 x=\frac{16}{25}
\end{aligned}
$

$
\Rightarrow \cos x= \pm \frac{4}{5}
$
$\Rightarrow \cos x=-\frac{4}{5}[x$ lies in second quadrant $]$
Now, $\operatorname{cosec} x=\frac{1}{\sin x}=\frac{5}{3}$
$
\sec x=\frac{1}{\cos x}=-\frac{5}{4}
$

$\begin{aligned}
& \tan x=\frac{\sin x}{\cos x}=\frac{3 / 5}{-4 / 5}=\frac{-3}{4} \\
& \cot x=\frac{\cos x}{\sin x}=\frac{-4 / 5}{3 / 5}=\frac{-4}{3}
\end{aligned}$

Ex 3.2 Question 3.

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

Answer.

Given: $\cot x=\frac{3}{4}$
$
\begin{aligned}
& \because \operatorname{cosec}^2 \theta-\cot ^2 \theta=1 \\
& \Rightarrow \operatorname{cosec}^2 x-\left(\frac{3}{4}\right)^2=1
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \operatorname{cosec}^2 x=1+\frac{9}{16} \\
& \Rightarrow \operatorname{cosec}^2 x=\frac{25}{16} \\
& \Rightarrow \operatorname{cosec} x= \pm \frac{5}{4}
\end{aligned}
$
$\Rightarrow \operatorname{cosec} x=-\frac{5}{4}[x$ lies in third quadrant $]$
Now, $\sin x=\frac{1}{\operatorname{cosec} x}=\frac{-4}{5}$

$\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\frac{16}{25}}=-\frac{3}{5}$

$
\begin{aligned}
& \tan x=\frac{1}{\cot x}=\frac{4}{3} \\
& \sec x=\frac{1}{\cos x}=\frac{-5}{3}
\end{aligned}
$
Ex 3.2 Question 4.

$\sec x=\frac{13}{5}, x$ lies in fourth quadrant.

Answer:

$
\begin{aligned}
& \text {Given: } \sec x=\frac{13}{5} \\
& \because \sec ^2 \theta-\tan ^2 \theta=1 \\
& \Rightarrow\left(\frac{13}{5}\right)^2-\tan ^2 x=1 \\
& \Rightarrow \tan ^2 x=\left(\frac{13}{5}\right)^2-1 \\
& \Rightarrow \tan ^2 x=\frac{169}{25}-1 \\
& \Rightarrow \tan ^2 x=\frac{144}{25}
\end{aligned}
$

$
\Rightarrow \tan x= \pm \frac{12}{5}
$
$\Rightarrow \tan x=\frac{-12}{5}[x$ lies in fourth quadrant $]$
Now $\cot x=\frac{1}{\tan x}=\frac{-5}{12}$
$
\cos x=\frac{1}{\sec x}=\frac{5}{13}
$

$
\begin{aligned}
& \sin x=-\sqrt{1-\cos ^2 \theta}=-\sqrt{1-\frac{144}{169}}=-\frac{12}{13} \\
& \operatorname{cosec} x=\frac{1}{\sin x}=\frac{-13}{12}
\end{aligned}
$
Ex 3.2 Question 5.

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

Answer.

Given: $\tan x=\frac{-5}{12}$
$
\begin{aligned}
& \therefore \cot x=\frac{1}{\tan x}=\frac{-12}{5} \\
& \because \sec ^2 \theta-\tan ^2 \theta=1 \\
& \Rightarrow \sec ^2 x-\left(\frac{-5}{12}\right)^2=1 \\
& \Rightarrow \sec ^2 x=1+\frac{25}{144} \\
& \Rightarrow \sec ^2 x=\frac{169}{144}
\end{aligned}
$

$\Rightarrow \sec x= \pm \frac{13}{12}$
$\Rightarrow \sec x=\frac{-13}{12}$ [ $x$ lies in second quadrant]
Now, $\cos x=\frac{1}{\sec x}=\frac{-12}{13}$
$\sin x=\sqrt{1-\cos ^2 x}=\sqrt{1-\frac{144}{169}}=\frac{5}{13}$

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

Find the values of the trigonometric functions in exercises 6 o 10 .
Ex 3.2 Question 6.

$\sin 765^{\circ}$

Answer.

Here $\sin 765^{\circ}=\sin \left(2 \times 360^{\circ}+45^{\circ}\right)=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
Ex 3.2 Question 7.

$\operatorname{cosec}(-1410)^{\circ}$

Answer.

Here $\operatorname{cosec}(-1410)^{\circ}=\operatorname{cosec}\left(-4 \times 360^{\circ}+30^{\circ}\right)=\operatorname{cosec} 30^{\circ}=2$
Ex 3.2 Question 8.

$\tan \frac{19 \pi}{3}$

Answer.
Here $\tan \frac{19 \pi}{3}=\tan \frac{19}{3} \times 180^{\circ}=\tan 1140^{\circ}=\tan \left(3 \times 360^{\circ}+60^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}$
Ex 3.2 Question 9.

$\sin \left(\frac{-11 \pi}{3}\right)$

Answer.
Here
$
\sin \left(\frac{-11 \pi}{3}\right)=\sin \left(\frac{-11 \times 180^{\circ}}{3}\right)=\sin \left(-660^{\circ}\right)=\sin \left(-2 \times 360^{\circ}+60^{\circ}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2}
$
Ex 3.2 Question 10.

$\cot \left(\frac{-15 \pi}{4}\right)$

Answer.
Here
$
\cot \left(\frac{-15 \pi}{4}\right)=\cot \left(\frac{-15 \times 180^{\circ}}{4}\right)=\cot \left(-675^{\circ}\right)=\cot \left(-2 \times 360^{\circ}+45^{\circ}\right)=\cot 45^{\circ}=1
$