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If $x\sin (a+y)+\sin a \cos(a+y)=0$,Prove that $\large\frac{dy}{dx}=\large\frac{\sin^2(a+y)}{\sin a}$

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xsin(a+y)+sin acos(a+y)=0
(i.e) xsin(a+y)=-sin acos(a+y).
$\Rightarrow x=\Large \frac{-sin acos(a+y)}{sin(a+y)}$
$x=-sin a cot (a+y)$
Differentiating with respect to y on both sides
We know that $\frac{d}{dy}(cot(a+y))=-cosec^2(a+y)$
$\frac{dx}{dy}=-sin a(-cosec^2(a+y))$
$\quad\quad=sin a (cosec^2(a+y)$
But $cosec\theta=\frac{1}{sin\theta}$
$\Large \frac{dx}{dy}=\frac{sin a}{sin^2(a+y)}$
$\Large \frac{dy}{dx}=\frac{sin^2(a+y)}{sin a}$
answered Mar 25, 2013 by sreemathi.v

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