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Question

If $$\cos y = x \cos ( a+y)$$, with $$\cos$$ a $$\neq$$  $$\pm $$1, prove that $$\dfrac{dy}{dx}$$ = $$\dfrac{\cos^2(a+y)}{\sin a}.$$


Solution

We have,
$$\cos y=x\cos(a+y)$$

On differentiating w,r,t $$x$$, we get
$$=\sin y\cfrac{dy}{dx}=\cos(a+y)-x\sin (a+y)\cfrac{dy}{dx}$$
$$\implies \cfrac{dy}{dx}=\cfrac{\cos(a+y)}{x\sin(a+y)-\sin y}$$
$$\implies \cfrac{dy}{dx}=\cfrac{\cos^2(a+y)}{\cos y \sin(a+y)-\sin y\cos(a+y)}$$
$$\implies \cfrac{dy}{dx}=\cfrac{\cos^2(a+y)}{\sin a}$$

Hence, proved.

Mathematics

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