Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

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