Question

If $u=\sin^{-1}(\frac{2x}{1+x^2})$ and $v=\tan^{-1}(\frac{2x}{1-x^2})$ then $\frac{du}{dv}$ is

• A. $\frac{1-x^2}{1+x^2}$
• B. $\frac{1}{2}$
• C. 1
• D. 2

Answer 1

Answer: (C)

1

Answer 2

The correct answer is (C) : 1.