JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 5)
If $$a=\sin ^{-1}(\sin (5))$$ and $$b=\cos ^{-1}(\cos (5))$$, then $$a^2+b^2$$ is equal to
25
$$4 \pi^2+25$$
$$8 \pi^2-40 \pi+50$$
$$4 \pi^2-20 \pi+50$$
Explanation
$$\begin{aligned}
& a=\sin ^{-1}(\sin 5)=5-2 \pi \\
& \text { and } b=\cos ^{-1}(\cos 5)=2 \pi-5 \\
& \therefore a^2+b^2=(5-2 \pi)^2+(2 \pi-5)^2 \\
& =8 \pi^2-40 \pi+50
\end{aligned}$$
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