If $y=\cos ^{-1}\left(\frac{5 \cos x-12 \sin x}{13}\right), x \in\left(0, \frac{\pi}{2}\right)$, then $\frac{d y}{d x}$ is equal to

1

$y=\cos ^{-1}\left(\frac{5}{13} \cos x-\frac{12}{13} \sin x\right)=\cos ^{-1}(\cos \alpha \cos x-\sin \alpha \sin x)$

$=\cos ^{-1}(\cos (x+\alpha))=x+\alpha$

∴  $\frac{d y}{d x}=1$

Hence (1) is correct answer.