Exercise 2.1 (Revised) - Chapter 2 - Inverse Trigonometry - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 2 - Inverse Trigonometry

Ex 2.1 Question 1:

Find the principal values of the following:
1. $\sin ^{-1}\left(\frac{-1}{2}\right)$

Answer.

Let $\sin ^{-1}\left(\frac{-1}{2}\right)=y$
$
\begin{aligned}
& \Rightarrow \sin y=-\frac{1}{2} \\
& \Rightarrow \sin y=-\sin \frac{\pi}{6} \\
& \Rightarrow \sin y=\sin \left(-\frac{\pi}{6}\right)
\end{aligned}
$

Since, the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Therefore, Principal value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ is $-\frac{\pi}{6}$.
Ex 2.1 Question 2.

$\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer.

Let $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=y$

$
\begin{aligned}
& \Rightarrow \cos y=\frac{\sqrt{3}}{2} \\
& \Rightarrow \cos y=\cos \frac{\pi}{6} \\
& \Rightarrow y=\frac{\pi}{6}
\end{aligned}
$

Since, the principal value branch of $\cos ^{-1}$ is $[0, \pi]$.
Therefore, Principal value of $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is $\frac{\pi}{6}$.
Ex 2.1 Question 3.

$\operatorname{cosec}^{-1}(2)$

Answer.

Let $\operatorname{cosec}^{-1}(2)=y$
$\Rightarrow \operatorname{cosec} y=2$
$\Rightarrow \operatorname{cosec} y=\operatorname{cosec} \frac{\pi}{6}$
Since, the principal value branch of $\operatorname{cosec}^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$.
Therefore, Principal value of $\operatorname{cosec}^{-1}(2)$ is $\frac{\pi}{6}$.
Ex 2.1 Question 4.

$\tan ^{-1}(-\sqrt{3})$

Answer0.

Let $\tan ^{-1}(-\sqrt{3})=y$
$
\begin{aligned}
& \Rightarrow \tan y=-\sqrt{3} \\
& \Rightarrow \tan y=-\tan \frac{\pi}{3}
\end{aligned}
$

$
\Rightarrow \tan y=\tan \left(-\frac{\pi}{3}\right)
$

Since, the principal value branch of $\tan ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Therefore, Principal value of $\tan ^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$.
Ex 2.1 Question 5.

$\cos ^{-1}\left(\frac{-1}{2}\right)$

Answer.

Let $\cos ^{-1}\left(\frac{-1}{2}\right)=y$
$
\begin{aligned}
& \Rightarrow \cos y=-\frac{1}{2} \\
& \Rightarrow \cos y=-\cos \frac{\pi}{3} \\
& \Rightarrow \cos y=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}
\end{aligned}
$

Since, the principal value branch of $\cos ^{-1}$ is $[0, \pi]$.
Therefore, Principal value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ is $\frac{2 \pi}{3}$.

Ex 2.1 Question 6.

$\tan ^{-1}(-1)$

Answer.

Let $\tan ^{-1}(-1)=y$
$
\begin{aligned}
& \Rightarrow \tan y=-1 \\
& \Rightarrow \tan y=-\tan \frac{\pi}{4} \\
& \Rightarrow \tan y=\tan \left(-\frac{\pi}{4}\right)
\end{aligned}
$

Since, the principal value branch of $\tan ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Therefore, Principal value of $\tan ^{-1}(-1)$ is $-\frac{\pi}{4}$.
Ex 2.1 Question 7.

$\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer.

Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$
$
\begin{aligned}
& \Rightarrow \sec y=\frac{2}{\sqrt{3}} \\
& \Rightarrow \sec y=\sec \frac{\pi}{6}
\end{aligned}
$

Since, the principal value branch of $\sec ^{-1}$ is $[0, \pi]$.
Therefore, Principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$.

Ex 2.1 Question 8.

$\cot ^{-1}(\sqrt{3})$

Answer.

Let $\cot ^{-1}(\sqrt{3})=y$
$
\begin{aligned}
& \Rightarrow \cot y=\sqrt{3} \\
& \Rightarrow \cot y=\cot \frac{\pi}{6}
\end{aligned}
$

Since, the principal value branch of $\cot ^{-1}$ is $[0, \pi]$.
Therefore, Principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.

Ex 2.1 Question 9.

$\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$

Answer.

Let $\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=y$
$
\begin{aligned}
& \Rightarrow \cos y=-\frac{1}{\sqrt{2}} \\
& \Rightarrow \cos y=-\cos \frac{\pi}{4} \\
& \Rightarrow \cos y=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}
\end{aligned}
$

Since, the principal value branch of $\cos ^{-1}$ is $[0, \pi]$.
Therefore, Principal value of $\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$.

Ex 2.1 Question 10.

$\operatorname{cosec}^{-1}(-\sqrt{2})$

Answer.

Let $\operatorname{cosec}^{-1}(-\sqrt{2})=y$
$
\begin{aligned}
& \Rightarrow \operatorname{cosec} y=-\sqrt{2} \\
& \Rightarrow \operatorname{cosec} y=\operatorname{cosec} \frac{-\pi}{4}
\end{aligned}
$

Since, the principal value branch of $\operatorname{cosec}^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Therefore, Principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ is $\frac{-\pi}{4}$.

Find the value of the following:
Ex 2.1 Question 11.

$\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$

Answer.

$\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$
$
\begin{aligned}
& =\tan ^{-1} \tan \frac{\pi}{4}+\cos ^{-1}\left(-\cos \frac{\pi}{3}\right)+\sin ^{-1}\left(-\sin \frac{\pi}{6}\right) \\
& =\frac{\pi}{4}+\cos ^{-1}\left(\cos \left(\pi-\frac{\pi}{3}\right)\right)+\sin ^{-1}\left(\sin \left(-\frac{\pi}{6}\right)\right) \\
& =\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \\
& =\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}
\end{aligned}
$

Ex 2.1 Question 12.

$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$

Answer.

$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$
$
\begin{aligned}
& =\cos ^{-1} \cos \frac{\pi}{3}+2 \sin ^{-1} \sin \frac{\pi}{6} \\
& =\frac{\pi}{3}+2\left(\frac{\pi}{6}\right) \\
& =\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}
\end{aligned}
$
Ex 2.1 Question 13.

If $\sin ^{-1} x=y$, then:
A) $0 \leq y \leq \pi$

(B) $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
(C) 0

(D) $-\frac{\pi}{2}$

Answer.

By definition of principal value for $y=\sin ^{-1} x,-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$. Therefore, option (B) is correct.
Ex 2.1 Question 14.

$\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$ is equal to:
(A) $\pi$
(B) $-\frac{\pi}{3}$
(C) $\frac{\pi}{3}$
(D) $\frac{2 \pi}{3}$

Answer.

$\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$
$=\tan ^{-1} \tan \frac{\pi}{3}-\sec ^{-1}\left(-\sec \frac{\pi}{3}\right)$
$=\frac{\pi}{3}-\sec ^{-1} \sec \left(\pi-\frac{\pi}{3}\right)$
$=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}$
Therefore, option (B) is correct.