Question

If ${\sin }^{2}x+{\cos }^{2}y=1$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{\sin 2x}{\sin 2y}$
• B. $\frac{{\sin }^{2}y}{\sin 2x}$
• C. $\frac{{\sin }^{2}x}{{\sin }^{2}y}$
• D. $-\frac{{\sin }^{2}y}{{\sin }^{2}x}$

Answer 1

The correct option is A $\frac{\sin 2x}{\sin 2y}$

Given, ${\sin }^{2}x+{\cos }^{2}y=1$
On differentiating both sides with respect to $x$, we get
$\frac{d}{dx}({\sin }^{2}x+{\cos }^{2}y)=\frac{d}{dx}\left(1\right)$
$\Rightarrow 2\sin x\cos x+2\cos y(-\sin y\frac{dy}{dx})=0$
$[usingchainrule,\frac{d}{dx}f\left\{g\left(x\right)\right\}=f^{{}^{\prime }}\left(x\right)\frac{d}{dx}g\left(x\right)]$
$\Rightarrow -2\sin y\cos y\frac{dy}{dx}=-2\sin x\cos x$
$\Rightarrow \frac{dy}{dx}=\frac{-\sin 2x}{-\sin 2y}=\frac{\sin 2x}{\sin 2y}$ $(\because \sin 2x=2\sin x\cos x)$