Question

Solve the equation:${\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)+{\sin }^{-1}\left(\frac{2x}{x^2+1}\right)+{\tan }^{-1}\left(\frac{2x}{x^2-1}\right)=\frac{2\pi }{3}$

• A. $\frac{1}{\sqrt{3}}$
• B. $\cot {20}^{\circ }$
• C. $\frac{-1}{\sqrt{3}}$
• D. $\tan {20}^{\circ }$

Answer 1

The correct option is B $\cot {20}^{\circ }$

Let $x=\cot y$.

$\therefore {\cos }^{-1}\left(\frac{{\cot }^{2}y-1}{{\cot }^{2}y+1}\right)+{\sin }^{-1}\left(\frac{2coty}{{\cot }^{2}y+1}\right)+{\tan }^{-1}\left(\frac{2\cot y}{{\cot }^{2}y-1}\right)$

$={\cos }^{-1}\left(\cos 2y\right)+{\sin }^{-1}\left(\sin 2y\right)+{\tan }^{-1}\left(\tan 2y\right)$

$\Rightarrow 3\left(2y\right)=6y=\frac{2\pi }{3}$

$y=\frac{\pi }{9}={20}^0$.

Now, $x=\cot \left(y\right)$

$x=\cot \left({20}^0\right)$.