MHT CET · Maths · Differentiation
If \(\sin ^2 x+\cos ^2 y=1\), then \(\frac{d y}{d x}=\)
- A \(\frac{\sin ^2 x}{\sin ^2 y}\)
- B \(\frac{\sin ^2 y}{\sin ^2 x}\)
- C \(\frac{\sin 2 x}{\sin 2 y}\)
- D \(\frac{-\sin ^2 y}{\sin ^2 x}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sin 2 x}{\sin 2 y}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin ^2 x+\cos ^2 y=1 \\ & \therefore 2 \sin x \cos x-2 \cos y \sin y \frac{d y}{d x}=0 \\ & \therefore \frac{d y}{d x}=\frac{2 \sin x \cos x}{2 \sin y \cos y}=\frac{\sin 2 x}{\sin 2 y}\end{aligned}\)
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