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
Consider, $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0$
$\Rightarrow\ \text{x}\sin(\text{a}+\text{y})=-\sin\text{a}\cdot\cos(\text{a}+\text{y})$
$\Rightarrow\ \text{x}=\frac{-\sin\text{a}\cdot\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}$
$\Rightarrow\ \text{x}=-\sin\text{a}\cdot\cot(\text{a}+\text{y})$
$\therefore\ \frac{\text{dx}}{\text{dy}}=-\sin\text{a}\cdot\big[-\text{cosec}^2(\text{a}+\text{y})\big]\cdot\frac{\text{d}}{\text{dy}}(\text{a}+\text{y})$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\sin\text{a}\cdot\frac{1}{\sin^2(\text{a}+\text{y})}\cdot1$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
Hence proved.
$\Rightarrow\ \text{x}\sin(\text{a}+\text{y})=-\sin\text{a}\cdot\cos(\text{a}+\text{y})$
$\Rightarrow\ \text{x}=\frac{-\sin\text{a}\cdot\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}$
$\Rightarrow\ \text{x}=-\sin\text{a}\cdot\cot(\text{a}+\text{y})$
$\therefore\ \frac{\text{dx}}{\text{dy}}=-\sin\text{a}\cdot\big[-\text{cosec}^2(\text{a}+\text{y})\big]\cdot\frac{\text{d}}{\text{dy}}(\text{a}+\text{y})$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\sin\text{a}\cdot\frac{1}{\sin^2(\text{a}+\text{y})}\cdot1$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
Hence proved.
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