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Question
If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .
Options
\[\frac{\sin a}{\sin a \sin^2 \left( a + y \right)}\]
\[\frac{\sin^2 \left( a + y \right)}{\sin a}\]
\[\sin a \sin^2 \left( a + y \right)\]
\[\frac{\sin^2 \left( a - y \right)}{\sin a}\]
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Solution
\[\frac{\sin^2 \left( a + y \right)}{\sin a}\]
\[\text { We have,} \sin y = x \sin\left( a + y \right)\]
\[\Rightarrow \frac{d}{dx}\left( \sin y \right) = \frac{d}{dx}\left[ x \sin\left( a + y \right) \right]\]
\[ \Rightarrow \cos y\frac{dy}{dx} = \sin\left( a + y \right)\frac{d}{dx}\left( x \right) + x\frac{d}{dx}\left\{ \sin\left( a + y \right) \right\}\]
\[ \Rightarrow \cos y\frac{dy}{dx} = \sin\left( a + y \right) \times 1 + x \cos\left( a + y \right)\frac{dy}{dx}\]
\[ \Rightarrow \cos y\frac{dy}{dx} = \sin\left( a + y \right) + x \cos\left( a + y \right)\frac{dy}{dx}\]
\[ \Rightarrow \cos y\frac{dy}{dx} - x \cos\left( a + y \right)\frac{dy}{dx} = \sin\left( a + y \right)\]
\[ \Rightarrow \left\{ \cos y - x \cos \left( a + y \right) \right\}\frac{dy}{dx} = \sin\left( a + y \right)\]
\[ \Rightarrow \left\{ \cos y - \frac{\sin y}{\sin\left( a + y \right)} \times \cos\left( a + y \right) \right\}\frac{dy}{dx} = \sin\left( a + y \right) .............\binom{\because \sin y = 2 \sin x \cos x}{ \therefore x = \frac{\sin y}{\sin\left( a + y \right)}}\]
\[ \Rightarrow \left\{ \frac{\sin\left( a + y \right) \cos y - \sin y \cos\left( a + y \right)}{\sin\left( a + y \right)} \right\}\frac{dy}{dx} = \sin\left( a + y \right)\]
\[ \Rightarrow \frac{\sin\left( a + y - y \right)}{\sin\left( a + y \right)} \times \frac{dy}{dx} = \sin\left( a + y \right) \]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\]
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