If y = cos^{-1} left{ frac{a cos x - b sin x} | vedclass

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(D) Given,$y = \cos^{-1} \left( \frac{a \cos x - b \sin x}{\sqrt{a^2 + b^2}} ight)$.Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos 	heta$ and $\frac{b}{\sq

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(D) Given,$y = \cos^{-1} \left( \frac{a \cos x - b \sin x}{\sqrt{a^2 + b^2}} ight)$.Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos 	heta$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin 	heta$,where $	heta = 	an^{-1} \left( \frac{b}{a} ight)$.Then,$y = \cos^{-1} (\cos 	heta \cos x - \sin 	heta \sin x)$.Using the trigonometric identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get:$y = \cos^{-1} (\cos(x + 	heta))$.$y = x + 	heta$.Since $	heta = 	an^{-1} \left( \frac{b}{a} ight)$ is a constant,differentiating with respect to $x$ gives:$\frac{dy}{dx} = \frac{d}{dx} (x + 	heta) = 1 + 0 = 1$.Differentiating again with respect to $x$:$\frac{d^2 y}{dx^2} = \frac{d}{dx} (1) = 0$.

If $y = \cos^{-1} \left\{ \frac{a \cos x - b \sin x}{\sqrt{a^2 + b^2}} \right\}$,then $\frac{d^2 y}{d x^2}$ is equal to

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