Question

If $x \sin (a+ y)+\sin a \cos (a +y)=0$, then $\frac{d y}{d x}$ is equal to

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

Answer 1

Answer: (A)

Here, $x \sin (a +y)+\sin a \cos (a+ y)=0$...(i)
On differentiating w.r.t. $x$, we get
$\frac{d}{d x}[x \sin (a+ y)]+\frac{d}{d x}[\sin a \cos (a+ y)]=0$
$\Rightarrow \left[x \frac{d}{d x}\{\sin (a +y)\}+\sin (a+ y) \frac{d}{d x}(x)\right]$
$+\sin a \frac{d}{d x} \cos (a +y)=0$
(using product rule and chain rule)
$\Rightarrow \left[x \cos (a +y) \frac{d}{d x}(a+ y)+\sin (a +y)\right]$
$+\sin a\left[-\sin (a +y) \frac{d}{d x}(a +y)\right]=0$
$\Rightarrow [x \cos (a+ y)(\left.\left(a+\frac{d y}{d x}\right)+\sin (a+ y)\right]$
$-\sin a \sin (a+ y)\left(0+\frac{d y}{d x}\right)=0$
$\Rightarrow x \cos (a +y) \frac{d y}{d x}+\sin (a+ y )$
$-\sin a \sin (a+ y) \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}[x \cos (a +y)-\sin a \sin (a +y)]$
$=-\sin (a+ y)$
[from Eq. (i), putting $\left. x=-\sin a \frac{\cos (a+ y)}{\sin (a +y)}\right]$
$\Rightarrow \frac{d y}{d x}\left[-\sin a \frac{\cos ^{2}(a +y)}{\sin (a+ y)}-\sin a \sin (a +y)\right]$
$=-\sin (a +y)$
$\Rightarrow -\frac{d y}{d x}\left[\frac{\sin a \cos ^{2}(a +y)+\sin a \sin ^{2}(a+ y)}{\sin (a+ y)}\right]$
$=-\sin (a +y)$
$\Rightarrow \frac{d y}{d x}=\sin (a +y)$
$\left[\frac{\sin (a+ y)}{\sin a\left\{\cos ^{2}(a+ y)+\sin ^{2}(a +y)\right\}}\right]$
$\Rightarrow \frac{d y}{d x}=\frac{\sin ^{2}(a+ y)}{\sin a}$
$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$