#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 22

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

Hint:

Use chain rule

Given:

$x \sin (a+y)+\sin a \cos (a+y)=0$

Solution:

\begin{aligned} &x \sin (a+y)+\sin a \cos (a+y)=0 \\ &x \sin (a+y)=-\sin a \cos (a+y) \end{aligned}

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

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

Differentiating  equation w.r.t x

$\frac{d}{d x}\left(\frac{-x}{\sin a}\right)=\frac{d}{d x} \cot (a+y)$

$\frac{-1}{\sin a} \cdot \frac{d x}{d x}=\frac{d \cot (a+y)}{d(a+y)} \times \frac{d(a+y)}{d x}$                [Use chain rule]

$\frac{-1}{\sin a}=-\cos e c^{2}(a+y) \times\left(\frac{d a}{d x}+\frac{d y}{d x}\right)$            $\left[\because \frac{d(\cot \theta)}{d \theta}=-\cos e c^{2} \theta\right]$

$-\frac{1}{\sin a}=-\cos e c^{2}(a+y)\left[0+\frac{d y}{d x}\right]$

$-\frac{1}{\sin a}=-\operatorname{cosec}^{2}(a+y) \cdot \frac{d y}{d x}$

$\frac{d y}{d x}=\frac{1}{\sin a \cdot \cos e c^{2}(a+y)}$

$\frac{d y}{d x}=\frac{1}{\sin a \cdot\left(\frac{1}{\sin ^{2}(a+y)}\right)}$                            $\left[\because \cos e c \theta=\frac{1}{\sin \theta}\right]$

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

Thus, proved