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

Solution

We have,

sin y = x sin ( a + y )

$\Rightarrow x = \frac{\sin y}{\sin (a + y)}$

Differentiating the above function we have,

$1 = \frac{\sin (a + y) \times \cos y \frac{dy}{dx} - \sin y \times \cos (a + y) \frac{dy}{dx}}{\sin^{2} (a + y)} \Rightarrow \sin^{2} (a + y) = [\sin (a + y) \times \cos y - \sin y \times \cos (a + y)] \frac{dy}{dx} \Rightarrow \frac{\sin^{2} (a + y)}{[\sin (a + y) \times \cos y - \sin y \times \cos (a + y)]} = \frac{dy}{dx} \Rightarrow \frac{\sin^{2} (a + y)}{\sin (a + y - y)} = \frac{dy}{dx} \Rightarrow \frac{\sin^{2} (a + y)}{\sin a} = \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{\sin^{2} (a + y)}{\sin a}$

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

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