Question

If x sin (a + y) + sin a cos (a + y) = 0, then the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ is

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

Answer 1

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

Given that,
$x\sin \left(a+y\right)+\sin a\cos \left(a+y\right)=0\Rightarrow x=\frac{-\sin a\cos \left(a+y\right)}{\sin \left(a+y\right)}\Rightarrow x=-\sin a\cot \left(a+y\right)\mathrm{Differentiating}\mathrm{both}\mathrm{sides}w.r.t.y,\mathrm{we}\mathrm{get}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\sin a{\mathrm{cosec}}^2\left(a+y\right)\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\sin a}{{\sin }^2(a+y)}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\sin }^2(a+y)}{\sin a}$