Q126 / 200 · Manipal (MET) 2014 · Mathematics
If $x \sin (a+y)+\sin a \cos (a+y)=0$, then $\frac{d y}{d x}$ is equal to
- Correct answerA$\frac{\sin ^2(a+y)}{\sin a}$
- B$\frac{\cos ^2(a+y)}{\cos a}$
- C$\frac{\sin ^2(a+y)}{\cos a}$
- D$\frac{\cos ^2(a+y)}{\sin a}$
Solution
Here, $x \sin (a+y)+\sin a \cos (a+y)=0$...(i)
On differentiating w.r.t. $x$, we get
$\begin{aligned}
& \frac{d}{d x}[x \sin (a+y)]+\frac{d}{d x}[\sin a \cos (a+y)]=0 \\
& \begin{aligned}
\Rightarrow\left[x \frac{d}{d x}\{\sin (a+y)\}\right. & \left.+\sin (a+y) \frac{d}{d x}(x)\right] \\
& +\sin a \frac{d}{d x} \cos (a+y)=0
\end{aligned}
\end{aligned}$
(using product rule and chain rule)
$\begin{aligned}
& \begin{array}{r}
\Rightarrow \\
{\left[x \cos (a+y) \frac{d}{d x}(a+y)+\sin (a+y)\right]} \\
\\
+\sin a\left[-\sin (a+y) \frac{d}{d x}(a+y)\right]=0
\end{array} \\
& \begin{array}{r}
\Rightarrow\left[x \cos (a+y)\left(0+\frac{d y}{d x}\right)+\sin (a+y)\right] \\
-\sin a \sin (a+y)\left(0+\frac{d y}{d x}\right)=0
\end{array} \\
& \begin{array}{r}
\Rightarrow x\cos (a+y) \frac{d y}{d x}+\sin (a+y) \\
-\sin a \sin (a+y) \frac{d y}{d x}=0
\end{array}
\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}[x \cos (a+y)-\sin a \sin (a+y)] \\ &=-\sin (a+y) \\ & {[\text { from Eq. (i), putting } x}\left.=-\sin a \frac{\cos (a+y)}{\sin (a+y)}\right]\end{aligned}$
$\begin{gathered}\Rightarrow \frac{d y}{d x}\left[-\sin a \frac{\cos ^2(a+y)}{\sin (a+y)}-\sin a \sin (a+y)\right] \\ \Rightarrow-\frac{d y}{d x}\left[\frac{\sin a \cos ^2(a+y)+\sin a \sin ^2(a+y)}{\sin (a+y)}\right] \\ =-\sin (a+y)\end{gathered}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}=\sin (a+y) \\ & \quad\left[\frac{\sin (a+y)}{\sin a\left\{\cos ^2(a+y)+\sin ^2(a+y)\right\}}\right]\end{aligned}$
$\Rightarrow \frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a} \quad\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)$
Hence, the correct answer is option: 'A'.