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

The correct option is A $\frac{{\sin }^2(a+y)}{\sin a}$

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