Question

If $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sec }^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ , then $\frac{dy}{dx}$ is equal to

• A. $\frac{7}{1+x^2}$
• B. $\frac{4}{1+x^2}$
• C. $\frac{1}{x}$
• D. None of these

Answer 1

Answer: (A)

We have, $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sec }^{-1}\left(\frac{1+x^2}{1-x^2}\right)$
Let $x=\tan \theta$ Then, $y={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)+{\sec }^{-1}\left(\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }\right)$
$=2\theta +2\theta =4{\tan }^{-1}x$
$\therefore$ $\frac{dy}{dx}=\frac{4}{1+x^2}$