Question

Find the values of each of the following:
(i) ${\tan }^{-1}\left\{2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right\}$
(ii) $\cos \left({\sec }^{-1}x+{\mathrm{cosec}}^{-1}x\right),\left|x\right|\ge 1$

Answer 1

(i) Let ${\sin }^{-1}\frac{1}{2}=y$
Then,
$\sin y=\frac{1}{2}$

$$\begin{gathered} \therefore {\tan }^{-1}\left\{2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right\}={\tan }^{-1}\left\{2\cos 2y\right\} \\ ={\tan }^{-1}\left(2\left(1-2{\sin }^{2}y\right)\right)\left[\because \cos 2x=1-2{\sin }^{2}x\right] \\ ={\tan }^{-1}\left\{2\left(1-2\times \frac{1}{4}\right)\right\}\left[\because \sin y=\frac{1}{2}\right] \\ ={\tan }^{-1}\left\{2\times \frac{1}{2}\right\} \\ ={\tan }^{-1}1 \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

$\therefore {\tan }^{-1}\left\{2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right\}=\frac{\mathrm{\pi }}{4}$
(ii)
We have
$$\begin{gathered} \cos \left({\sec }^{-1}x+{\mathrm{cosec}}^{-1}x\right) \\ =\cos \frac{\mathrm{\pi }}{2}\left[\because {\sec }^{-1}x+{\mathrm{cosec}}^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =0 \end{gathered}$$

$\therefore \cos \left({\sec }^{-1}x+{\mathrm{cosec}}^{-1}x\right)=0,\left|x\right|\ge 1$