Question

If $\sin y=x\cos \left(a+y\right),\mathrm{then}\frac{dy}{dx}$ is equal to
(a) $\frac{{\cos }^2\left(a+y\right)}{\cos a}$

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

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

(d) none of these

Answer 1

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

We have, $\sin y=x\cos \left(a+y\right)$

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