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 $p = \tan \theta, q = \tan \phi, x = \tan \Psi$, we get
$\sin^{-1} \left(\frac{2 \tan \theta }{1 +\tan^{2} \theta }\right)- \cos^{-1} \left(\frac{1-\tan^{2} \phi }{1+\tan^{2} \phi }\right) = \tan^{-1}\left(\frac{2 \tan\Psi}{1-\tan^{2} \Psi }\right) $
$\Rightarrow \sin^{-1}\left(\sin 2\theta \right) -\cos^{-1} \left(\cos2\phi\right) $
$ \Rightarrow \theta-\phi = \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}$