Question

If $ab<1$ and ${\cos }^{-1}\left(\frac{1-a^2}{1+a^2}\right)+{\cos }^{-1}\left(\frac{1-b^2}{1+b^2}\right)=2{\tan }^{-1}x$, then $x$ is equal to

• A. $\frac{a}{1+ab}$
• B. $\frac{a}{1-ab}$
• C. $\frac{a-b}{1+ab}$
• D. $\frac{a+b}{1+ab}$
• E. $\frac{a+b}{1-ab}$

Answer 1

The correct option is E $\frac{a+b}{1-ab}$

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

Similarly, ${\cos }^{-1}\left(\frac{1-b^2}{1+b^2}\right)={\tan }^{-1}\left(\frac{2b}{1-b^2}\right)=2{\tan }^{-1}b$

$\therefore {\cos }^{-1}\left(\frac{1-a^2}{1+a^2}\right)+{\cos }^{-1}\left(\frac{1-b^2}{1+b^2}\right)=2{\tan }^{-1}a+2{\tan }^{-1}b=2{\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$

$\therefore x=\frac{a+b}{1-ab}$