Question

If $ \cos ^{-1} x=\alpha,(0<\,x<\,1) \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=\frac{2 \pi}{3}$, then $\tan ^{-1}(2 x) $ equals:

• A. $\frac {\pi}{2}$
• B. $\frac {\pi}{3}$
• C. $\frac {\pi}{4}$
• D. $\frac {\pi}{6}$

Answer 1

Answer: (B)

Given that $\cos^{-1} \, x =\alpha , 0 < \,x <\, 1\,\,\,\,\,\,\dots(i)$
$ \Rightarrow x =\cos\alpha$ $\therefore \sin^{-1} \left(2x \sqrt{1-x^{2}}\right) +\sec^{-1} \left(\frac{1}{2x^{2} -1}\right) = \frac{2\pi}{3} $
$ \Rightarrow \sin^{-1} \left(2 \cos \alpha \sqrt{1- \cos^{2}\alpha}\right) +\sec^{-1} \left(\frac{1}{2 \cos^{2} \alpha-1}\right) = \frac{2\pi}{3}$
$ \Rightarrow \sin^{-1} \left(\sin2\alpha\right) +\sec^{-1} \left(\sec2\alpha\right) = \frac{2\pi}{3} $
$ \Rightarrow 2\alpha + 2 \alpha = \frac{2\pi}{3}$
$ \Rightarrow \alpha = \frac{\pi}{6} $ From (i)
Now , $ x = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $
$\Rightarrow 2x = \sqrt{3} $
$\therefore \tan^{-1} \left(2x\right) = \tan^{-1} \left(\sqrt{3}\right)$
$ = \frac{\pi}{3} $