If \sin y= x\, \sin \left ( a+y \right ),  prove that \frac{dy}{dx}= \frac{\sin ^{2}\left ( a+y \right )}{\sin a}
 

 

 

 

 
 
 
 
 

Answers (1)

\left ( \sin y \right )= x\left ( \sin \left ( a+y \right ) \right )\left [ given \right ]
To prove: \frac{dy}{dx}= \frac{\sin ^{2}\left ( a+y \right )}{\sin a}
\sin y= x\sin \left ( a+y \right )
          \Rightarrow x= \frac{\sin y}{\sin \left ( a+y \right )} 
Differentiating both sides w.r.t x
\Rightarrow 1= \frac{\sin \left ( a+y \right )\cos y-\cos \left ( a+y \right )\sin y}{\sin ^{2}\left ( a+y \right )}\frac{dy}{dx}
\sin \left ( m-n \right )= \sin m\: \cos n-\sin n\: \cos m
\Rightarrow \frac{dy}{dx}= \frac{\sin ^{2}\left ( a+y \right )}{\sin \left ( a+y\: -y \right )} \Rightarrow \frac{dy}{dx}= \frac{\sin ^{2}\left ( a+y \right )}{\sin a}
 

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