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Question
If \[\cos y = x \cos \left( a + y \right), \text{ with } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?
Solution
\[\text{ We have }, \cos y = x \cos\left( a + y \right)\]
\[\text{ Differentiating with respect to x, we get }, \]
\[\frac{d}{dx}\left( \cos y \right) = \frac{d}{dx}\left\{ x\cos\left( a + y \right) \right\}\]
\[ \Rightarrow - \sin y\frac{dy}{dx} = \cos\left( a + y \right)\frac{d}{dx}\left( x \right) + x\frac{d}{dx} \cos\left( a + y \right) \]
\[ \Rightarrow - \sin y\frac{dy}{dx} = \cos\left( a + y \right) + x\left[ - \sin\left( a + y \right) \right]\frac{dy}{dx}\]
\[ \Rightarrow \left[ x\sin\left( a + y \right) - \sin y \right]\frac{dy}{dx} = \cos\left( a + y \right) \]
\[ \Rightarrow \left[ \frac{\cos y}{\cos\left( a + y \right)}\sin\left( a + y \right) - \sin y \right]\frac{dy}{dx} = \cos\left( a + y \right) \left[ \because \cos y = x \cos\left( a + y \right) \Rightarrow x = \frac{\cos y}{\cos\left( a + y \right)} \right]\]
\[ \Rightarrow \left[ \cos y\sin\left( a + y \right) - \sin y\cos\left( a + y \right) \right]\frac{dy}{dx} = \cos^2 \left( a + y \right)\]
\[ \Rightarrow \sin\left( a + y - y \right)\frac{dy}{dx} = \cos^2 \left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\]
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