If cos y = x cos (a+y) with cos a neq pm 1 | vedclass

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(B) Given the equation: $ \cos y = x \cos (a+y) $We can express $ x $ as: $ x = \frac{\cos y}{\cos (a+y)} $Differentiating both sides with respect to

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$ \frac{\cos ^{2}(a+y)}{\sin a} $

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(B) Given the equation: $ \cos y = x \cos (a+y) $We can express $ x $ as: $ x = \frac{\cos y}{\cos (a+y)} $Differentiating both sides with respect to $ x $ using the quotient rule: $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2} $$ 1 = \frac{\cos (a+y) \cdot (-\sin y \frac{dy}{dx}) - \cos y \cdot (-\sin (a+y) \frac{dy}{dx})}{\cos^2 (a+y)} $$ 1 = \frac{dy}{dx} \left[ \frac{\sin (a+y) \cos y - \cos (a+y) \sin y}{\cos^2 (a+y)} \right] $Using the trigonometric identity $ \sin (A-B) = \sin A \cos B - \cos A \sin B $: $ 1 = \frac{dy}{dx} \left[ \frac{\sin (a+y-y)}{\cos^2 (a+y)} \right] $$ 1 = \frac{dy}{dx} \left[ \frac{\sin a}{\cos^2 (a+y)} \right] $Therefore,$ \frac{dy}{dx} = \frac{\cos^2 (a+y)}{\sin a} $

If $ \cos y = x \cos (a+y) $ with $ \cos a \neq \pm 1 $,then $ \frac{d y}{d x} $ is equal to

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