Question

The derivative of $si{n}^{-1}\frac{2x}{1+x^2}$ with respect to $co{s}^{-1}\frac{1-x^2}{1+x^2}$ is

• A. $-1$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

Let $P=si{n}^{-1}\frac{2x}{1+x^2}=2ta{n}^{-1}x$
And $q=co{s}^{-1}\frac{1-x^2}{1+x^2}=2ta{n}^{-1}x$
$\therefore \frac{dp}{dx}=\frac{2}{1+x^2}=and\frac{dq}{dx}=\frac{2}{1+x^2}$
$\Rightarrow \frac{dp}{dq}=\frac{\frac{dp}{dx}}{\frac{dq}{dx}}=\frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}}=1$