Jee Main 2024 31 01 2024 Shift2 - Question1
Question 1
$a=\sin ^{-1}(\sin 5), b=\cos ^{-1}(\cos 5)$ then $a^{2}+b^{2}$ is equal to
(1) $8 \pi^{2}-40 \pi+50$
(2) $4 \pi^{2}+25$
(3) $8 \pi^{2}-50$
(4) $8 \pi^{2}+40 \pi+50$
Show Answer
Answer: (1)
Solution:
$a=\sin ^{-1}(\sin 5)=5-2 \pi$
and $b=\cos ^{-1}(\cos 5)=2 \pi-5$
$\therefore a^{2}+b^{2}=(5-2 \pi)^{2}+(2 \pi-5)^{2}$
$=8 \pi^{2}-40 \pi+50$
BETA
