The differential coefficient of $cosec^{-1} \frac{1}{2 x^2-1}$ with respect to $\sqrt{1-x^2}$ at x = 1/2 is

-4

Let  $x = \cos \theta$;  then  $y = \cos ^{-1}\left(\frac{1}{2 x^2-1}\right)=\cos ^{-1}(\sec 2 \theta)=\frac{\pi}{2}-2 \theta$

$Z=\sqrt{1-x^2}=\sqrt{1-\cos ^2 \theta}=\sin \theta$

∴  $\frac{d y}{d z}=\frac{\frac{d y}{d \theta}}{\frac{d z}{d \theta}}=\frac{-2}{\cos \theta}=\frac{-2}{x}$

Hence (1) is correct answer.