Question

If $\sin y=x\sin \left(a+y\right),\mathrm{then}\frac{dy}{dx}\mathrm{is}$

(a) $\frac{\sin a}{\sin a{\sin }^2\left(a+y\right)}$

(b) $\frac{{\sin }^2\left(a+y\right)}{\sin a}$

(c) $\sin a{\sin }^2\left(a+y\right)$

(d) $\frac{{\sin }^2\left(a-y\right)}{\sin a}$

Answer 1

(b) $\frac{{\sin }^2\left(a+y\right)}{\sin a}$

$\mathrm{We}\mathrm{have},\sin y=x\sin \left(a+y\right)$

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(\sin y\right)=\frac{d}{dx}\left[x\sin \left(a+y\right)\right] \\ \Rightarrow \cos y\frac{dy}{dx}=\sin \left(a+y\right)\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left\{\sin \left(a+y\right)\right\} \\ \Rightarrow \cos y\frac{dy}{dx}=\sin \left(a+y\right)\times 1+x\cos \left(a+y\right)\frac{dy}{dx} \\ \Rightarrow \cos y\frac{dy}{dx}=\sin \left(a+y\right)+x\cos \left(a+y\right)\frac{dy}{dx} \\ \Rightarrow \cos y\frac{dy}{dx}-x\cos \left(a+y\right)\frac{dy}{dx}=\sin \left(a+y\right) \\ \Rightarrow \left\{\cos y-x\cos \left(a+y\right)\right\}\frac{dy}{dx}=\sin \left(a+y\right) \\ \Rightarrow \left\{\cos y-\frac{\sin y}{\sin \left(a+y\right)}\times \cos \left(a+y\right)\right\}\frac{dy}{dx}=\sin \left(a+y\right)\left[\begin{array}{c}\because \sin y=2\sin x\cos x\\ \therefore x=\frac{\sin y}{\sin \left(a+y\right)}\end{array}\right] \\ \Rightarrow \left\{\frac{\sin \left(a+y\right)\cos y-\sin y\cos \left(a+y\right)}{\sin \left(a+y\right)}\right\}\frac{dy}{dx}=\sin \left(a+y\right) \\ \Rightarrow \frac{\sin \left(a+y-y\right)}{\sin \left(a+y\right)}\times \frac{dy}{dx}=\sin \left(a+y\right) \\ \Rightarrow \frac{dy}{dx}=\frac{{\sin }^2\left(a+y\right)}{\sin a} \end{gathered}$$