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


We are given with an equation $\cos y=x \cos (a+y)$, we have to prove that $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$ by using the given

equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,

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

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

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

We can further solve it by using the given equation,

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

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

By using $\sin A \cos B-\cos A \sin B=\sin (A-B)$

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

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