Question

If $3si{n}^{-1}\left(\frac{2x}{1+x^2}\right)-4co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$, where $|x|<1$, then $x$ is equal to

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $-\frac{\sqrt{3}}{4}$
• E. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

Since $|x|<1$
Substituting $x=tan\theta$
The above expression reduces to
$3si{n}^{-1}\left(sin2\theta \right)-4co{s}^{-1}\left(cos2\theta \right)+2ta{n}^{-1}\left(tan2\theta \right)$
$=6\theta -8\theta +4\theta$
$=2\theta$
$=\frac{\pi }{3}$
Therefore
$\theta =\frac{\pi }{6}$
$ta{n}^{-1}x=\frac{\pi }{6}$
$x=tan\left(\frac{\pi }{6}\right)$
$x=\frac{1}{\sqrt{3}}$