Question

If $xsin\left(a+y\right)=siny$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

Answer: (A)

Given $xsin\left(a+y\right)=siny$
$\Rightarrow x=\frac{siny}{sin\left(a+y\right)}$
Differentiate both sides $w$.$r$.$t$. $x$, we get
$1=\frac{sin\left(a+y\right)cosy\frac{dy}{dx}-cos\left(a+y\right)siny\frac{dy}{dx}}{si{n}^2\left(a+y\right)}$
$\Rightarrow si{n}^2\left(a+y\right)=sin\left(a+y-y\right)\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{si{n}^2\left(a+y\right)}{sina}$