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If \(cos\: y = x \cos (a + y)\), with $\cos$ \(a \neq \pm 1\), prove that \( \large\frac{dy}{dx}=\frac{\cos^2(a + y)}{\sin \: a}\)

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  • $\big(\large\frac{u}{v}\big)'=\large\frac{u'v-uv'}{v^2}$ wherever $v\neq 0$
Step 1:
$\cos y=x\cos(a+y)$
$x=\Large\frac{\cos y}{\cos(a+y)}$
$\Large\frac{dx}{dy}=\frac{\cos(a+y).\Large\frac{d}{dy}(\cos y)-\cos y.\Large\frac{d}{dy}(\cos(a+y))}{\cos^2(a+y)}$
$\quad\;=\Large\frac{\cos(a+y).(-\sin y)-\cos y.(-\sin(a+y))}{\cos^2(a+y)}$
$\quad\;=\Large\frac{\sin(a+y)\cos y-\cos(a+y)\sin y}{\cos^2(a+y)}$
$\quad\;=\Large\frac{\sin(a+y-y)}{\cos^2(a+y)}$
$\quad\;=\Large\frac{\sin a}{\cos^2(a+y)}$
Step 2:
$\Large\frac{dx}{dy}=\frac{\sin a}{\cos^2(a+y)}$
$\Large\frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}$
Hence proved.
answered May 14, 2013 by sreemathi.v
 

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