JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(a=\sin ^{-1}(\sin (5))\) and \(b=\cos ^{-1}(\cos (5))\), then \(a^2+b^2\) is equal to
- A \(4 \pi^2+25\)
- B \(8 \pi^2-40 \pi+50\)
- C \(4 \pi^2-20 \pi+50\)
- D \(25\)
Answer & Solution
Correct Answer
(B) \(8 \pi^2-40 \pi+50\)
Step-by-step Solution
Detailed explanation
\(a=\sin ^{-1}(\sin 5)=5-2 \pi \) \(\text { and } b=\cos ^{-1}(\cos 5)=2 \pi-5 \) \(\therefore a^2+b^2=(5-2 \pi)^2+(2 \pi-5)^2 \) \(=8 \pi^2-40 \pi+50\)
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