Question

If $siny = x \,sin(a + y)$, then $dy/dx$ is

• A. $sin(a + y)$
• B. $sin^2(a + y)$
• C. $\frac{sin^{2}\left(a+y\right)}{sin\,a}$
• D. $\frac{sin\left(a+y\right)}{sin\,a}$

Answer 1

Answer: (C)

$sin\,y=x\,sin\left(a+y\right)$ or $\frac{sin\,y}{sin\left(a+y\right)}=x\quad\ldots\left(i\right)$
Differentiating $\left(i\right)$ with respect to $x$, we get
$\frac{sin\left(a+y\right)cos\,y \frac{dy}{dx}-cos\left(a+y\right)sin\,y \frac{dy}{dx}}{sin^{2}\left(a+y\right)}=1$
$\Rightarrow \frac{dy}{dx}\left[sin\left(a+y-y\right)\right]=sin^{2}\left(a+y\right)$
(Using $sin(A - B) = sinA \,cosB - cosAsinB$)
$\therefore \frac{dy}{dx}=\frac{sin^{2}\left(a+y\right)}{sin\,a}$