# Siny=sin (a+y)x then proof that dy/dx=sin a /(1-2xcosa+x^2)how?

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#### Explanation

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#### Explanation:

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1
Mar 8, 2018

We use
color(red)((1)d/(dx)(tan^(-1)X)=1/(1+X^2)
$\textcolor{red}{\left(2\right) \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot {u}^{'} - u \cdot {v}^{'}}{v} ^ 2}$

#### Explanation:

Here,
$\sin y = \sin \left(a + y\right) \cdot x = x \left[\sin a \cos y + \cos a \sin y\right]$
$\implies \sin y = x \sin a \cos y + x \cos a \sin y$
$\implies \sin y - x \cos a \sin y = x \sin a \cos y$
$\implies \sin y \left(1 - x \cos a\right) = x \sin a \cos y$
$\implies \sin \frac{y}{\cos} y = \frac{x \sin a}{1 - x \cos a}$$\implies \tan y = \frac{x \sin a}{1 - x \cos a}$
$\implies y = {\tan}^{- 1} \left(\frac{x \sin a}{1 - x \cos a}\right)$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {\left(\frac{x \sin a}{1 - x \cos a}\right)}^{2}} \frac{d}{\mathrm{dx}} \left(\frac{x \sin a}{1 - x \cos a}\right)$
$= \frac{\cancel{{\left(1 - x \cos a\right)}^{2}}}{{\left(1 - x \cos a\right)}^{2} + {\left(x \sin a\right)}^{2}} \frac{\left(1 - x \cos a\right) \cdot \sin a - x \cdot \sin a \cdot \left(- \cos a\right)}{\cancel{{\left(1 - x \cos a\right)}^{2}}}$
=(sina-xcosasina+xsinacosa)/(1-2xcosa+x^2cos^2a+x^2sin^2a)=(sina)/(1-2xcosa+x^2(cos^2a+sin^2a))=(sina)/(1-2xcosa+x^2

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