Question

If $siny=xsin(a+y)$, then $dy/dx$ is

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

Answer 1

Answer: (C)

$siny=xsin\left(a+y\right)$ or $\frac{siny}{sin\left(a+y\right)}=x\dots \left(i\right)$
Differentiating $\left(i\right)$ with respect to $x$, we get
$\frac{sin\left(a+y\right)cosy\frac{dy}{dx}-cos\left(a+y\right)siny\frac{dy}{dx}}{si{n}^2\left(a+y\right)}=1$
$\Rightarrow \frac{dy}{dx}\left[sin\left(a+y-y\right)\right]=si{n}^2\left(a+y\right)$
(Using $sin(A-B)=sinAcosB-cosAsinB$)
$\therefore \frac{dy}{dx}=\frac{si{n}^2\left(a+y\right)}{sina}$