If $\cos y=x \cos (a+y)$, with $\cos a \neq \pm 1$, prove that $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$


It is given that, $\cos y=x \cos (a+y)$

$\therefore \frac{d}{d x}[\cos y]=\frac{d}{d x}[x \cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y) \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}[\cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y)+x \cdot[-\sin (a+y)] \frac{d y}{d x}$

$\Rightarrow[x \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y)$    ...(1)

Since $\cos y=x \cos (a+y), x=\frac{\cos y}{\cos (a+y)}$

Then, equation (1) reduces to

$\left[\frac{\cos y}{\cos (a+y)} \cdot \sin (a+y)-\sin y\right] \frac{d y}{d x}=\cos (a+y)$

$\Rightarrow[\cos y \cdot \sin (a+y)-\sin y \cdot \cos (a+y)] \cdot \frac{d y}{d x}=\cos ^{2}(a+y)$

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

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

Hence, proved.

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