Question

\[\text{ If }\cos y = x\cos\left( a + y \right),\text{  where } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?

Answer 1

\[\cos y = x\cos\left( a + y \right)\]
\[ \Rightarrow - \sin y\frac{dy}{dx} = \cos\left( a + y \right) - x\sin\left( a + y \right)\frac{dy}{dx}\]
\[ \Rightarrow - \sin y\frac{dy}{dx} + x\sin\left( a + y \right)\frac{dy}{dx} = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\cos y}{\cos\left( a + y \right)}\sin\left( a + y \right) - \sin y \right] = \cos\left( a + y \right) \left[ \because x = \frac{\cos y}{\cos\left( a + y \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\cos y\sin\left( a + y \right) - \sin y\cos\left( a + y \right)}{\cos\left( a + y \right)} \right] = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\sin\left( a + y - y \right)}{\cos\left( a + y \right)} \right] = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{\sin a}{\cos\left( a + y \right)} \right] = \cos\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\]