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Question
If \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] , prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?
Solution
\[\text{ We have}, x \sin\left( a + y \right) + \sin a \cos\left( a + y \right) = 0\]
Differentiating with respect to x using chain rule,
\[\frac{d}{dx}\left[ x \sin\left( a + y \right) + \sin a \cos\left( a + y \right) \right] = 0\]
\[ \Rightarrow x\frac{d}{dx}\sin\left( a + y \right) + \sin\left( a + y \right)\frac{d}{dx}\left( x \right) + \sin a\frac{d}{dx}\cos\left( a + y \right) + \cos\left( a + y \right)\frac{d}{dx}\sin a = 0\]
\[ \Rightarrow x \cos\left( a + y \right)\left( 0 + \frac{dy}{dx} \right) + \sin\left( a + y \right) + \sin a\left\{ - \sin\left( a + y \right)\frac{dy}{dx} \right\} + 0 = 0\]
\[ \Rightarrow \left[ x \cos\left( a + y \right) - \sin a \sin\left( a + y \right) \right]\frac{dy}{dx} + \sin\left( a + y \right) = 0\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{\sin\left( a + y \right)}{x \cos\left( a + y \right) - \sin a \sin\left( a + y \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sin\left( a + y \right)}{\left\{ - \frac{\sin a \cos\left( a + y \right)}{\sin\left( a + y \right)} \right\}\cos\left( a + y \right) - \sin a \sin\left( a + y \right)} .................\left[ \because x = - \frac{\sin a \cos\left( a + y \right)}{\sin\left( a + y \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a \cos^2 \left( a + y \right) + \sin a \sin^2 \left( a + y \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a\left[ \cos^2 \left( a + y \right) + \sin^2 \left( a + y \right) \right]}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a} ..............\left[ \because \cos^2 \left( a + y \right) + \sin^2 \left( a + y \right) = 1 \right]\]
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