1. Use principal value ranges

For inverse trigonometric functions:

• $\sin^{-1}x \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$
• $\cos^{-1}x \in [0,\pi]$

We need to evaluate: $a=\sin^{-1}(\sin 5), \qquad b=\cos^{-1}(\cos 5)$ where $5$ is in radians.

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2. Find $a=\sin^{-1}(\sin 5)$

We must bring angle $5$ to an equivalent angle lying in the principal range of $\sin^{-1}$, i.e. $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Since $\frac{3\pi}{2}<5<2\pi,$ we use the identity: $\sin 5 = \sin(\pi-5') \text{ form is not convenient here, so reduce directly.}$

Note that $5-2\pi \in \left[-\frac{\pi}{2},0\right]$ because $2\pi \approx 6.283$, so $5-2\pi \approx -1.283,$ which lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Hence, $a=\sin^{-1}(\sin 5)=5-2\pi.$

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3. Find $b=\cos^{-1}(\cos 5)$

We need the principal value of cosine inverse, which must lie in $[0,\pi]$.

Since $\pi<5<2\pi,$ we use: $\cos(5)=\cos(2\pi-5).$

Now, $2\pi-5 \in [0,\pi]$ (since $2\pi-5\approx 1.283$), so it is already in the principal range of $\cos^{-1}$.

Thus, $b=\cos^{-1}(\cos 5)=2\pi-5.$

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4. Compute $a^2+b^2$

We have $a=5-2\pi, \qquad b=2\pi-5.$

Since these are negatives of each other, $a^2=b^2=(2\pi-5)^2.$

Therefore, $a^2+b^2=2(2\pi-5)^2.$

Expand: $(2\pi-5)^2=4\pi^2-20\pi+25$ so $a^2+b^2=2\left(4\pi^2-20\pi+25\right)=8\pi^2-40\pi+50.$

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5. Match with options

$8\pi^2-40\pi+50$ corresponds to Option C.

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6. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So the answer agrees.