# Choose the correct alternative in the following:

Question:

Choose the correct alternative in the following:

If $\sin y=x \cos (a+y)$, then $\frac{d y}{d x}$ is equal to

A. $\frac{\cos ^{2}(a+y)}{\cos a}$

B. $\frac{\cos a}{\cos ^{2}(a+y)}$

C. $\frac{\sin ^{2} y}{\cos a}$

D. none of these

Solution:

$\sin y=x \cos (a+y)$

$x=\frac{\sin y}{\cos (a+y)}$

Differentiating w.r.t y we get,

$\frac{d x}{d y}=\frac{\frac{d \sin y}{d x} \cdot \cos (a+y)-\frac{d \cos (a+y)}{d x} \cdot(\sin y)}{\cos ^{2}(a+y)}($ Using quotient rule $)$

$\frac{d x}{d y}=\frac{\cos y \cdot \cos (a+y)-[-\sin (a+y)] \cdot(\sin y)}{\cos ^{2}(a+y)}$

$\frac{d x}{d y}=\frac{\cos (a+y) \cdot \cos y+\sin (a+y) \cdot(\sin y)}{\cos ^{2}(a+y)}$

$\frac{d x}{d y}=\frac{\cos [(a+y)-y]}{\cos ^{2}(a+y)} \cup \operatorname{sing} \cos (a-b)=\cos a \cdot \cos b+\sin a \cdot \sin b$

$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\cos \mathrm{a}}{\cos ^{2}(\mathrm{a}+\mathrm{y})}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos ^{2}(\mathrm{a}+\mathrm{y})}{\cos \mathrm{a}}=(\mathrm{A})$