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. $\frac{p-q}{pq+1}$

Answer 1

Answer: (D)

$\frac{p-q}{pq+1}$

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

$\Rightarrow 2.{\tan }^{-1}p-2{\tan }^{-1}q={\tan }^{-1}\frac{2x}{1-x^2}=2{\tan }^{-1}x$, Identities

$\Rightarrow 2({\tan }^{-1}p-{\tan }^{-1}q)=2{\tan }^{-1}x$

$\Rightarrow {\tan }^{-1}\left(\frac{p-q}{1+pq}\right)={\tan }^{-1}x$

$\therefore x=\frac{p-q}{1+pq}$