If a=sin ^{-1}(sin (5)) and b=cos ^{-1}(cos ( | vedclass

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(B) We know that $\sin^{-1}(\sin x) = x - 2n\pi$ where $x \in [2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}]$.Since $5$ radians lies in the interval $

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$8 \pi^2-40 \pi+50$

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(B) We know that $\sin^{-1}(\sin x) = x - 2n\pi$ where $x \in [2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}]$.Since $5$ radians lies in the interval $[\frac{3\pi}{2}, \frac{5\pi}{2}]$,we have $a = \sin^{-1}(\sin 5) = 5 - 2\pi$.We know that $\cos^{-1}(\cos x) = 2n\pi - x$ where $x \in [(2n-1)\pi, 2n\pi]$.Since $5$ radians lies in the interval $[\pi, 2\pi]$,we have $b = \cos^{-1}(\cos 5) = 2\pi - 5$.Now,calculate $a^2 + b^2$:$a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2$$= (5 - 2\pi)^2 + (5 - 2\pi)^2$$= 2(5 - 2\pi)^2$$= 2(25 - 20\pi + 4\pi^2)$$= 50 - 40\pi + 8\pi^2$$= 8\pi^2 - 40\pi + 50$.

If $a=\sin ^{-1}(\sin (5))$ and $b=\cos ^{-1}(\cos (5))$,then $a^2+b^2$ is equal to

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