If x (a+y)+ (a+y)=0, prove that dy dx = ^2(a+y)

Question

If $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$

✓

Answer

We have, $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0$
Differentiate with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]+\frac{\text{d}}{\text{dx}}\big[\sin\text{a}\cos(\text{a}+\text{y})\big]=0$
$\Rightarrow\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\sin\text{a}\frac{\text{d}}{\text{dx}}\cos(\text{a}+\text{y})=0$
$\Rightarrow\Big[\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})(1)\Big] \\ +\sin\text{a}\Big[-\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})\Big]=0$
$\Rightarrow\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}(\text{a}+\text{y})+\sin(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})-\sin\text{a}\sin(\text{a}+\text{y})\big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[-\sin\text{a}\frac{\cos^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}-\sin\text{a}\sin(\text{a}+\text{y})\Big] \\ =-\sin(\text{a}+\text{y})$
$\Big[\because\ \text{x}=-\sin\text{a}\frac{\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big]$
$ \Rightarrow-\frac{\text{dy}}{\text{dx}}\Big[\frac{\sin\text{a}\cos^2(\text{a}+\text{y})+\sin\text{a}\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big] \\ =-\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\Big[\frac{\sin(\text{a}+\text{y})}{\sin\text{a}\{\cos^2(\text{a}+\text{y})+\sin^2(\text{a}+\text{y})\}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$