Question

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

Answer 1

$\mathrm{We}\mathrm{have},x\sin \left(a+y\right)+\sin a\cos \left(a+y\right)=0$
Differentiate with respect to x,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left[x\sin \left(a+y\right)\right]+\frac{d}{dx}\left[\sin a\cos \left(a+y\right)\right]=0 \\ \Rightarrow \left[x\frac{d}{dx}\sin \left(a+y\right)+\sin \left(a+y\right)\frac{d}{dx}\left(x\right)\right]+\sin a\frac{d}{dx}\cos \left(a+y\right)=0 \\ \Rightarrow \left[x\cos \left(a+y\right)\frac{d}{dx}\left(a+y\right)+\sin \left(a+y\right)\left(1\right)\right]+\sin a\left[-\sin \left(a+y\right)\frac{d}{dx}\left(a+y\right)\right]=0 \\ \Rightarrow x\cos \left(a+y\right)\frac{\mathit{d}y}{\mathit{d}x}+\sin \left(a+y\right)-\sin a\sin \left(a+y\right)\frac{\mathit{d}y}{\mathit{d}x}=0 \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}\left[x\cos \left(a+y\right)-\sin a\sin \left(a+y\right)\right]=-\sin \left(a+y\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}\left[-\sin a\frac{{\cos }^2\left(a+y\right)}{\sin \left(a+y\right)}-\sin a\sin \left(a+y\right)\right]=-\sin \left(a+y\right)\left[\because x=-\sin a\frac{\cos \left(a+y\right)}{\sin \left(a+y\right)}\right] \\ \Rightarrow -\frac{\mathit{d}y}{\mathit{d}x}\left[\frac{\sin a{\cos }^2\left(a+y\right)+\sin a{\sin }^2\left(a+y\right)}{\sin \left(a+y\right)}\right]=-\sin \left(a+y\right) \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\sin \left(a+y\right)\left[\frac{\sin \left(a+y\right)}{\sin a\left\{{\cos }^2\left(a+y\right)+{\sin }^2\left(a+y\right)\right\}}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{{\sin }^2\left(a+y\right)}{\sin a} \end{gathered}$$