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

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NCERT Class 12 Maths Solutions: Chapter 2 - Inverse Trigonometry

Example 1

Find the principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$.
Solution

Let $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=y$. Then, $\sin y=\frac{1}{\sqrt{2}}$.
We know that the range of the principal value branch of $\sin ^{-1}$ is $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ and $\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$. Therefore, principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is $\frac{\pi}{4}$

Example 2

Find the principal value of $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$
Solution

Let $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=y$. Then,
$
\cot y=\frac{-1}{\sqrt{3}}=-\cot \left(\frac{\pi}{3}\right)=\cot \left(\pi-\frac{\pi}{3}\right)=\cot \left(\frac{2 \pi}{3}\right)
$

We know that the range of principal value branch of $\cot ^{-1}$ is $(0, \pi)$ and $\cot \left(\frac{2 \pi}{3}\right)=\frac{-1}{\sqrt{3}}$. Hence, principal value of $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is $\frac{2 \pi}{3}$

Example 3

Show that
(i) $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)=2 \sin ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$
(ii) $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)=2 \cos ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$

Solution
(i) Let $x=\sin \theta$. Then $\sin ^{-1} x=\theta$. We have
$
\begin{aligned}
\sin ^{-1}\left(2 x \sqrt{1-x^2}\right) & =\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right) \\
& =\sin ^{-1}(2 \sin \theta \cos \theta)=\sin ^{-1}(\sin 2 \theta)=2 \theta \\
& =2 \sin ^{-1} x
\end{aligned}
$
(ii) Take $x=\cos \theta$, then proceeding as above, we get, $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)=2 \cos ^{-1} x$

Example 3

Show that
(i) $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)=2 \sin ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$
(ii) $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)=2 \cos ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$

Solution
(i) Let $x=\sin \theta$. Then $\sin ^{-1} x=\theta$. We have
$
\begin{aligned}
\sin ^{-1}\left(2 x \sqrt{1-x^2}\right) & =\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right) \\
& =\sin ^{-1}(2 \sin \theta \cos \theta)=\sin ^{-1}(\sin 2 \theta)=2 \theta \\
& =2 \sin ^{-1} x
\end{aligned}
$
(ii) Take $x=\cos \theta$, then proceeding as above, we get, $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)=2 \cos ^{-1} x$

Example 4

Express $\tan ^{-1} \frac{\cos x}{1-\sin x},-\frac{3 \pi}{2}<x<\frac{\pi}{2}$ in the simplest form.
Solution

We write
$
\begin{aligned}
\tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right) & =\tan ^{-1}\left[\frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right] \\
& =\tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}\right] \\
& =\tan ^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]=\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right] \\
& =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]=\frac{\pi}{4}+\frac{x}{2}
\end{aligned}
$

Example 5

Write $\cot ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right), x>1$ in the simplest form.
Solution

Let $x=\sec \theta$, then $\sqrt{x^2-1}=\sqrt{\sec ^2 \theta-1}=\tan \theta$
Therefore, $\cot ^{-1} \frac{1}{\sqrt{x^2-1}}=\cot ^{-1}(\cot \theta)=\theta=\sec ^{-1} x$, which is the simplest form.

Example 6

Find the value of $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$
Solution

We know that $\sin ^{-1}(\sin x)=x$. Therefore, $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\frac{3 \pi}{5}$
But $\quad \frac{3 \pi}{5} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, which is the principal branch of $\sin ^{-1} x$
However $\quad \sin \left(\frac{3 \pi}{5}\right)=\sin \left(\pi-\frac{3 \pi}{5}\right)=\sin \frac{2 \pi}{5}$ and $\frac{2 \pi}{5} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Therefore $\quad \sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)=\frac{2 \pi}{5}$