Question

The algebraic expression for $\tan \left({\sin }^{-1}\cos {\tan }^{-1}\frac{x}{2}\right)$ is

• A. $\frac{2}{x}$
• B. $\frac{x}{2}$
• C. $\frac{1}{x}$
• D. $\frac{2}{|x|}$

Answer 1

The correct option is A $\frac{2}{x}$

$\tan \left({\sin }^{-1}\cos {\tan }^{-1}\frac{x}{2}\right)$

$\because {\tan }^{-1}\frac{x}{2}={\cos }^{-1}\frac{2}{\sqrt{4+x^2}}$

$=\tan \left({\sin }^{-1}\cos \left(co{s}^{-1}\frac{2}{\sqrt{4+x^2}}\right)\right)$

$=\tan \left({\sin }^{-1}\left(\frac{2}{\sqrt{4+x^2}}\right)\right)$

$\because {\sin }^{-1}\frac{2}{\sqrt{4+x^2}}={\tan }^{-1}\frac{x}{2}$

$=\tan \left({\tan }^{-1}\left(\frac{2}{x}\right)\right)$

$=\frac{2}{x}.$

Hence, the answer is $\frac{2}{x}.$