Question

If $x\sin y=\sin (y+a)$ and $\frac{dy}{dx}=\frac{A}{1+x^2-2x\cos a}$ then the value of $A$ is

• A. $2$
• B. $\cos a$
• C. $-\sin a$
• D. $-2$

Answer 1

The correct option is C $-\sin a$

$xsiny=sin(y+a);\frac{dy}{dx}=\frac{A}{1+x^2-2xcosa}$

L.H.S

$xsiny=sin(y+a)$

$\Rightarrow xsiny=sinacosy+cosasiny$

$siny(a-cosa)=cosysina$

$\frac{siny}{cosy}=\frac{sina}{x-cosa}$

$tany=\frac{sina}{x-cosa};y=ta{n}^{-1}\left(\frac{sina}{x-cosa}\right)$

$\therefore \frac{dy}{dx}=\frac{1}{1+(\frac{sina}{x-cosa}{)}^2}\times \left[\frac{(x-cosa)\left(0\right)\left(sina\right)(1-0)}{(x-cosa{)}^2}\right]$

$\Rightarrow \frac{(x-cosa{)}^2}{(x^2+co{s}^{2}a+si{n}^{2}a-2xcosa)}$

$\Rightarrow \frac{-sin}{1+x^2-2xcosa}$

RHS $\Rightarrow \frac{A}{1+x^2-2xcosa}$

$\therefore A=-sina,$ option C is correct