Question

If $3 \,sin^{-1}\left(\frac{2x}{1+x^{2}}\right) - 4 \,cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) + 2 \,tan^{-1} \left(\frac{2x}{1-x^{2}}\right) = \frac{\pi}{3}$ where $|x| < 1$, then $x$ is equal to

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $-\frac{\sqrt{3}}{4}$

Answer 1

Answer: (A)

The given equation is
$3(2\,tan^{-1}x) - 4(2\,tan^{-1}x) + 2(2\,tan^{-1}x) = \pi/3$
$\therefore 2 \,tan^{-1}x = \pi/3$
$\therefore tan^{-1} x = \pi/6$
$\therefore x = \frac{1}{\sqrt{3}}$