Question

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

• A. 2
• B. cos a
• C. sin a
• D. None of these

Answer 1

The correct option is C sin a

$x=\frac{siny}{sin(a+y)}.....\left(i\right)$
$\Rightarrow \frac{dx}{dy}=\frac{sin\left(a\right)}{si{n}^2(a+y)}$
$\therefore \frac{dy}{dx}=\frac{si{n}^2(a+y)}{sin\left(a\right)}=\frac{A}{1+x^2-2xcosa}$
$Putx=0,y=0,$
$thenA=sina$