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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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$0$
We have,
$y=\cos ^{-1}\left(\frac{a \cos x-b \sin x}{\sqrt{a^2+b^2}}\right)$
$y=\cos ^{-1}(\cos (x-\theta))\left[\because\right.$ where, $\left.\theta=\tan ^{-1} b / a\right]$
$\begin{aligned} & y=x-\theta \\ & y=x-\tan ^{-1} b / a \Rightarrow d y / d x=1\end{aligned}$
Again differentiating $\frac{d^2 y}{d x^2}=0$
$y=\cos ^{-1}\left(\frac{a \cos x-b \sin x}{\sqrt{a^2+b^2}}\right)$
$y=\cos ^{-1}(\cos (x-\theta))\left[\because\right.$ where, $\left.\theta=\tan ^{-1} b / a\right]$
$\begin{aligned} & y=x-\theta \\ & y=x-\tan ^{-1} b / a \Rightarrow d y / d x=1\end{aligned}$
Again differentiating $\frac{d^2 y}{d x^2}=0$
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