Question

If ${\sin }^{-1}\left(\frac{2p}{1+p^2}\right)-{\cos }^{-1}\left(\frac{1-q^2}{1+q^2}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ then the value of $x$ is equal to

• A. $\frac{p+q}{1+pq}$
• B. $\frac{p-q}{1-pq}$
• C. $\frac{p-q}{pq-1}$
• D. $\frac{p-q}{1+pq}$

Answer 1

Answer: (D)

Putting $p=\tan \theta ,q=\tan \varphi ,x=\tan \Psi$, we get
${\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)-{\cos }^{-1}\left(\frac{1-{\tan }^2\varphi }{1+{\tan }^2\varphi }\right)={\tan }^{-1}\left(\frac{2\tan \Psi }{1-{\tan }^2\Psi }\right)$
$\Rightarrow {\sin }^{-1}\left(\sin 2\theta \right)-{\cos }^{-1}\left(\cos 2\varphi \right)$
$\Rightarrow \theta -\varphi =\Psi \Rightarrow {\tan }^{-1}p-{\tan }^{-1}q={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left(\frac{p-q}{1+pq}\right)={\tan }^{-1}x\Rightarrow x=\frac{p-q}{1+pq}$