If $\sin ^2 x+\cos ^2 y=1$, then $\frac{d y}{d x}=$

If $\sin ^2 x+\cos ^2 y=1$, then $\frac{d y}{d x}=$
  1. $\frac{\sin ^2 x}{\sin ^2 y}$
  2. $\frac{\sin ^2 y}{\sin ^2 x}$
  3. $\frac{\sin 2 x}{\sin 2 y}$
  4. $\frac{-\sin ^2 y}{\sin ^2 x}$

Solution

$\begin{aligned} & \sin ^2 x+\cos ^2 y=1 \\ & \therefore 2 \sin x \cos x-2 \cos y \sin y \frac{d y}{d x}=0 \\ & \therefore \frac{d y}{d x}=\frac{2 \sin x \cos x}{2 \sin y \cos y}=\frac{\sin 2 x}{\sin 2 y}\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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