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प्रश्न
If \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?
उत्तर
\[\text{ We have, } x\sin\left( a + y \right) + \sin a\cos\left( a + y \right) = 0 \]
Differentiate with respect to x,
\[\Rightarrow \frac{d}{dx}\left[ x \sin\left( a + y \right) \right] + \frac{d}{dx}\left[ \sin a \cos\left( a + y \right) \right] = 0\]
\[ \Rightarrow \left[ x\frac{d}{dx}\sin \left( a + y \right) + \sin\left( a + y \right)\frac{d}{dx}\left( x \right) \right] + \sin a\frac{d}{dx}\cos\left( a + y \right) = 0 \]
\[ \Rightarrow \left[ x \cos\left( a + y \right)\frac{d}{dx}\left( a + y \right) + \sin\left( a + y \right)\left( 1 \right) \right] + \sin a\left[ - \sin\left( a + y \right)\frac{d}{dx}\left( a + y \right) \right] = 0\]
\[ \Rightarrow x \cos\left( a + y \right)\frac{d y}{d x} + \sin\left( a + y \right) - \sin a\sin\left( a + y \right)\frac{d y}{d x} = 0\]
\[ \Rightarrow \frac{d y}{d x}\left[ x \cos\left( a + y \right) - \sin a \sin\left( a + y \right) \right] = - \sin\left( a + y \right)\]
\[ \Rightarrow \frac{d y}{d x}\left[ - \sin a\frac{\cos^2 \left( a + y \right)}{\sin\left( a + y \right)} - \sin a \sin\left( a + y \right) \right] = - \sin\left( a + y \right) \left[ \because x = - \sin a\frac{\cos\left( a + y \right)}{\sin\left( a + y \right)} \right]\]
\[ \Rightarrow - \frac{d y}{d x}\left[ \frac{\sin a \cos^2 \left( a + y \right) + \sin a \sin^2 \left( a + y \right)}{\sin\left( a + y \right)} \right] = - \sin\left( a + y \right)\]
\[ \Rightarrow \frac{d y}{d x} = \sin\left( a + y \right)\left[ \frac{\sin\left( a + y \right)}{\sin a\left\{ \cos^2 \left( a + y \right) + \sin^2 \left( a + y \right) \right\}} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\sin^2 \left( a + y \right)}{\sin a} \]
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