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Question

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


Solution

$$\sin y = x\sin(a+y)$$
$$\Rightarrow x = \dfrac{\sin y}{\sin(a+y)}$$
Differentiating with sides w.r.t.$$y$$, we get
$$\dfrac{dx}{dy} = \dfrac{\cos y\sin(a+y) - \sin y\cos(a+y)}{\sin^2(a+y)}$$ [Quotient Rule]

$$\Rightarrow \dfrac{dx}{dy} = \dfrac{\cos y(\sin a\cos y +\ cos a\sin y) - \sin y(\cos a\cos y - \sin a\sin y)}{\sin^2(a+y)}$$

$$\Rightarrow \dfrac{dx}{dy} = \dfrac{\cos^2 y\sin a +\ cos a\sin y\cos y - \sin y\cos a\cos y + \sin a\sin^2 y}{\sin^2(a+y)}$$

$$\Rightarrow \dfrac{dx}{dy} = \dfrac{\cos^2 y\sin a + \sin a\sin^2 y}{\sin^2(a+y)}$$

$$\Rightarrow \dfrac{dx}{dy} = \dfrac{\sin a(\cos^2 y + \sin^2 y)}{\sin^2(a+y)}$$

$$\Rightarrow \dfrac{dx}{dy} = \dfrac{\sin a}{\sin^2(a+y)}$$

$$\Rightarrow \dfrac{dy}{dx} = \dfrac{\sin^2(a+y)}{\sin a}$$
Hence Proved.

Mathematics

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