If y = frac{1}{sqrt{a^2 - b^2}} cos^{-1} left | vedclass

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(A) Let $	heta = a + b \cos(x - \alpha)$. The given expression is $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left( \frac{a \cos(x - \alpha) + b}{	he

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$\frac{1}{a + b \cos(x - \alpha)}$

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(A) Let $	heta = a + b \cos(x - \alpha)$. The given expression is $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left( \frac{a \cos(x - \alpha) + b}{	heta} ight)$.Taking $\cos$ on both sides: $\cos(y \sqrt{a^2 - b^2}) = \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)}$.Differentiating both sides with respect to $x$:$-\sqrt{a^2 - b^2} \sin(y \sqrt{a^2 - b^2}) \frac{dy}{dx} = \frac{d}{dx} \left( \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} ight)$.Using the quotient rule,the derivative of the right side is $\frac{-a \sin(x - \alpha)(a + b \cos(x - \alpha)) - (a \cos(x - \alpha) + b)(-b \sin(x - \alpha))}{(a + b \cos(x - \alpha))^2} = \frac{-(a^2 - b^2) \sin(x - \alpha)}{	heta^2}$.Thus,$-\sqrt{a^2 - b^2} \sin(y \sqrt{a^2 - b^2}) \frac{dy}{dx} = \frac{-(a^2 - b^2) \sin(x - \alpha)}{	heta^2}$.Simplifying,$\sin(y \sqrt{a^2 - b^2}) \frac{dy}{dx} = \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta^2}$.Since $\sin(y \sqrt{a^2 - b^2}) = \sqrt{1 - \cos^2(y \sqrt{a^2 - b^2})} = \sqrt{1 - \left( \frac{a \cos(x - \alpha) + b}{	heta} ight)^2} = \frac{\sqrt{	heta^2 - (a \cos(x - \alpha) + b)^2}}{	heta} = \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta}$.Substituting this back: $\left( \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta} ight) \frac{dy}{dx} = \frac{\sqrt{a^2 - b^2} \sin(x - \alpha)}{	heta^2}$.Therefore,$\frac{dy}{dx} = \frac{1}{	heta} = \frac{1}{a + b \cos(x - \alpha)}$.

If $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left[ \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} \right]$,then $\frac{dy}{dx} = $

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