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

Question

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

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Answer

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

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