JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsEasyQuestionIf a=sin−1(sin(5))a=\sin ^{-1}(\sin (5))a=sin−1(sin(5)) and b=cos−1(cos(5))b=\cos ^{-1}(\cos (5))b=cos−1(cos(5)), then a2+b2a^2+b^2a2+b2 is equal toOptionsA25B4π2+254 \pi^2+254π2+25C8π2−40π+508 \pi^2-40 \pi+508π2−40π+50D4π2−20π+504 \pi^2-20 \pi+504π2−20π+50Check AnswerHide SolutionSolutiona=sin−1(sin5)=5−2π and b=cos−1(cos5)=2π−5∴a2+b2=(5−2π)2+(2π−5)2=8π2−40π+50\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}a=sin−1(sin5)=5−2π and b=cos−1(cos5)=2π−5∴a2+b2=(5−2π)2+(2π−5)2=8π2−40π+50