#### Please solve RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 30 maths textbook solution

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

Hint:

Differentiate the function w.r.t x

Given:

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

Solution:

\begin{aligned} &\sin y=x \cos (a+y) \\\\ &\frac{d}{d x}(\sin y)=\frac{d}{d x}[x \cos (a+y)] \end{aligned}

\begin{aligned} &\cos y \frac{d y}{d x}=1 \times \cos (a+y)-x \sin (a+y) \frac{d}{d x}(a+y) \\\\ &\cos y \frac{d y}{d x}=\cos (a+y)-x \sin (a+y) \frac{d y}{d x} \end{aligned}

\begin{aligned} &\cos y \frac{d y}{d x}+x \sin (a+y) \frac{d y}{d x}=\cos (a+y) \\\\ &{[\cos y+x \sin (a+y)] \frac{d y}{d x}=\cos (a+y)} \end{aligned}

$\left[\cos y+\frac{\sin y}{\cos (a+y)} \times \sin (a+y)\right] \frac{d y}{d x}=\cos (a+y)$        $\left[\begin{array}{l} \sin y=x \cos (a+y) \\\\ x=\frac{\sin y}{\cos (a+y)} \end{array}\right]$

$\left[\frac{\cos (a+y) \cos y+\sin y \sin (a+y)}{\cos (a+y)}\right] \frac{d y}{d x}=\cos (a+y)$

\begin{aligned} &\frac{\cos (a+y-y)}{\cos (a+y)} \times \frac{d y}{d x}=\cos (a+y) \\\\ &\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\cos a} \end{aligned}