Question

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

• A. $\frac{sin\sqrt{a}}{sin\left(a+y\right)}$
• B. $\frac{si{n}^2\left(a+y\right)}{sina}$
• C. $sin\left(a+y\right)$
• D. None of these

Answer 1

Answer: (B)

Given, $siny=xsin\left(a+y\right)$
$\Rightarrow x=\frac{siny}{sin\left(a+y\right)}$
On differentiating w.r. to y, we get
$\frac{dx}{dy}=\frac{d}{dy}\left[\frac{siny}{sin\left(a+y\right)}\right]$
$=\frac{sin\left(a+y\right)cosy-sinycos\left(a+y\right)}{si{n}^2\left(a+y\right)}$
$=\frac{sina}{si{n}^2\left(a+y\right)}\Rightarrow \frac{dy}{dx}=\frac{si{n}^2\left(a+y\right)}{sina}$