If $y=\sin^{-1}\sqrt{\frac{x}{x+1}}+sec^{-1}\sqrt{\frac{x+1}{x}}$, then $\frac{dy}{dx}$ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Given:
$y = \sin^{-1} \left( \sqrt{ \frac{x}{x+1} } \right) + \sec^{-1} \left( \sqrt{ \frac{x+1}{x} } \right)$
Observe that:
$\sqrt{ \frac{x}{x+1} } = \cos \theta \quad \Rightarrow \quad \sqrt{ \frac{x+1}{x} } = \sec \theta$
So, define $\theta = \cos^{-1} \left( \sqrt{ \frac{x}{x+1} } \right)$
Then:
$y = \sin^{-1}(\cos \theta) + \sec^{-1}(\sec \theta) = (\frac{\pi}{2} - \theta) + \theta = \frac{\pi}{2}$
Hence, $y$ is constant.
$\Rightarrow \frac{dy}{dx} = 0$