Question

If $x\sin (a+y)+\sin a\cos (a+y)=0$, then $\frac{dy}{dx}$ 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
$\frac{d}{dx}[x\sin (a+y)]+\frac{d}{dx}[\sin a\cos (a+y)]=0$
$\Rightarrow \left[x\frac{d}{dx}\{\sin (a+y)\}+\sin (a+y)\frac{d}{dx}(x)\right]$
$+\sin a\frac{d}{dx}\cos (a+y)=0$
(using product rule and chain rule)
$\Rightarrow \left[x\cos (a+y)\frac{d}{dx}(a+y)+\sin (a+y)\right]$
$+\sin a\left[-\sin (a+y)\frac{d}{dx}(a+y)\right]=0$
$\Rightarrow [x\cos (a+y)(\left(a+\frac{dy}{dx}\right)+\sin (a+y)]$
$-\sin a\sin (a+y)\left(0+\frac{dy}{dx}\right)=0$
$\Rightarrow x\cos (a+y)\frac{dy}{dx}+\sin (a+y)$
$-\sin a\sin (a+y)\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}[x\cos (a+y)-\sin a\sin (a+y)]$
$=-\sin (a+y)$
[from Eq. (i), putting $x=-\sin a\frac{\cos (a+y)}{\sin (a+y)}]$
$\Rightarrow \frac{dy}{dx}\left[-\sin a\frac{{\cos }^2(a+y)}{\sin (a+y)}-\sin a\sin (a+y)\right]$
$=-\sin (a+y)$
$\Rightarrow -\frac{dy}{dx}\left[\frac{\sin a{\cos }^2(a+y)+\sin a{\sin }^2(a+y)}{\sin (a+y)}\right]$
$=-\sin (a+y)$
$\Rightarrow \frac{dy}{dx}=\sin (a+y)$
$\left[\frac{\sin (a+y)}{\sin a\left\{{\cos }^2(a+y)+{\sin }^2(a+y)\right\}}\right]$
$\Rightarrow \frac{dy}{dx}=\frac{{\sin }^2(a+y)}{\sin a}$
$\left(\because {\sin }^2\theta +{\cos }^2\theta =1\right)$