# If $x\sin(a+y)+\sin a\cos(a+y)=0,prove\;that\;\large {\frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin a}}$

Toolbox:
• To find $\large\frac{dy}{dx}$ for implict function $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
• A function $f(x,y)$ is said to be implicit if it is jumbled in such a way,that it is not possible to write $y$ exclusively s a function of $x$.
Step 1:
$x\sin(a+y)+\sin a\cos(a+y)=0$-----(1)
Differentiating on both sides (Apply product rule to differentiate $x\sin(a+y)$)
$x.[\sin(a+y)]^1+[\sin (a+y)].(x)^1+\sin a[\cos (a+y)]=0$
$\Rightarrow x.\cos(a+y)\large\frac{dy}{dx}$$+\sin(a+y).1+\sin a(-\sin(a+y).\large\frac{dy}{dx}$$=0$
$\Rightarrow x.\cos(a+y)\large\frac{dy}{dx}$$+\sin(a+y).1-\sin a(\sin(a+y).\large\frac{dy}{dx}$$=0$
$\large\frac{dy}{dx}$$[x\cos(a+y)-\sin a.\sin (a+y)]=-\sin(a+y) Step 2: Therefore \large\frac{dy}{dx}=\frac{-\sin(a+y)}{x\cos(a+y)-\sin a.\sin(a+y)} But from equation (1) we get, x=\large\frac{-\sin a\cos(a+y)}{\sin(a+y)} Now substituting for x we get, \large\frac{dy}{dx}=\large\frac{-\sin(a+y)}{2}+\large\frac{\sin a\cos(a+y)}{-\sin(a+y)}-$$\sin a\sin(a+y)$
$\Rightarrow \large\frac{\sin^2(a+y)}{\sin a(\cos^2(a+y)+\sin^2(a+y))}$
But $\sin^2(a+y)+\cos^2(a+y)=1$
Therefore $\large\frac{dy}{dx}=\large\frac{\sin^2(a+y)}{\sin a}$
Hence proved.