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

$\frac{{\sin }^2(a+y)}{\sin a}$

Explanation for the correct answer.

Step 1: Simplify the given equation.

$\begin{array}{rcl}x\sin (a+y)+\sin a\cos (a+y)& =& 0\\ \Rightarrow x\sin (a+y)& =& -\sin a\cos (a+y)\\ \Rightarrow x& =& \frac{-\sin a\cos (a+y)}{\sin (a+y)}\\ \Rightarrow x& =& -\sin a\cot \left(a+y\right)\end{array}$

Step 2: Find $\frac{dy}{dx}$.

Differentiate $\begin{array}{rcl}x& =& -\sin a\cot \left(a+y\right)\end{array}$ with respect to $y$.

$\begin{array}{rcl}\frac{dx}{dy}& =& -\sin a·-\cos e{c}^2\left(a+y\right)\\ & =& \frac{\sin a}{{\sin }^2\left(a+y\right)}\end{array}$

Therefore, $\frac{dy}{dx}=\frac{{\sin }^2(a+y)}{\sin a}$

Hence, option A is correct.