Question

The value of $2{\tan }^{-1}(-2)$ is equal to

• A. ${\sin }^{-1}(-\frac{4}{5})$
• B. $-{\sin }^{-1}\left(\frac{4}{5}\right)$
• C. ${\sin }^{-1}\left(\frac{4}{5}\right)-\pi$
• D. $-{\cos }^{-1}\left(\frac{4}{5}\right)-\frac{\pi }{2}$

Answer 1

Answer: (C), (D)

${\sin }^{-1}\left(\frac{4}{5}\right)-\pi$
$-{\cos }^{-1}\left(\frac{4}{5}\right)-\frac{\pi }{2}$

Assume $2{\tan }^{-1}(-2)=\theta$,

$\theta$ will lie in somewhere in range $(-\frac{\pi }{2},0)$
$\therefore \tan \frac{\theta }{2}=-2$

$\Rightarrow \tan \theta =-\frac{4}{3}$
[We used $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$ for the above equation]

we know that:

$\because \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$
From above,

$\theta ={\tan }^{-1}\left(\frac{4}{3}\right)-\pi ={\sin }^{-1}\left(\frac{4}{5}\right)-\pi$ --------(i)
Also we know that,

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}x=\frac{\pi }{2}-{\cos }^{-1}x$

Substitute $x=\frac{4}{5}$ in the above equation, we get

$\Rightarrow {\sin }^{-1}\frac{4}{5}=-\frac{\pi }{2}-{\cos }^{-1}\frac{4}{5}$