JEE Advanced 2022 · Paper 1 · Mathematics · Inverse Trigonometric Functions
Considering only the principal values of the inverse trigonometric functions, the value of 32cos−122+π2+14sin−122π2+π2+tan−12π\dfrac{3}{2}\cos^{- 1}\sqrt{\dfrac{2}{2 + {\pi}^{2}}} + \dfrac{1}{4}\sin^{- 1}\dfrac{2\sqrt{2}\pi}{2 + {\pi}^{2}} + \tan^{- 1}\dfrac{\sqrt{2}}{\pi}23cos−12+π22+41sin−12+π222π+tan−1π2 is _______.
(If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places)
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32cos−122+π2+14sin−122π2+π2+tan−12π\dfrac{3}{2}\cos^{- 1}\sqrt{\dfrac{2}{2 + {\pi}^{2}}} + \dfrac{1}{4}\sin^{- 1}\dfrac{2\sqrt{2}\pi}{2 + {\pi}^{2}} + \tan^{- 1}\dfrac{\sqrt{2}}{\pi}23cos−12+π22+41sin−12+π222π+tan−1π2=32tan−1π2+14tan−1(22ππ2−2)+tan−12π= \dfrac{3}{2}\tan^{- 1}\dfrac{\pi}{\sqrt{2}} + \dfrac{1}{4}\tan^{- 1}\left(\dfrac{2\sqrt{2}\pi}{{\pi}^{2} - 2}\right) + \tan^{- 1}\dfrac{\sqrt{2}}{\pi}=23tan−12π+41tan−1(π2−222π)+tan−1π2=π2+12tan−1π2−14tan−1(22π2−π2)= \dfrac{\pi}{2} + \dfrac{1}{2}\tan^{- 1}\dfrac{\pi}{\sqrt{2}} - \dfrac{1}{4}\tan^{- 1}\left(\dfrac{2\sqrt{2}\pi}{2 - {\pi}^{2}}\right)=2π+21tan−12π−41tan−1(2−π222π)=π2+12tan−1(π2)−14tan−1(2⋅(π2)1−(π2)2)= \dfrac{\pi}{2} + \dfrac{1}{2}\tan^{- 1}\left(\dfrac{\pi}{\sqrt{2}}\right) - \dfrac{1}{4}\tan^{- 1}\left(\dfrac{2 \cdot \left(\dfrac{\pi}{\sqrt{2}}\right)}{1 - {\left(\dfrac{\pi}{\sqrt{2}}\right)}^{2}}\right)=2π+21tan−1(2π)−41tan−11−(2π)22⋅(2π)=π2+12tan−1(π2)−14(−π+2tan−1(π2))= \dfrac{\pi}{2} + \dfrac{1}{2}\tan^{- 1}\left(\dfrac{\pi}{\sqrt{2}}\right) - \dfrac{1}{4}\left(- \pi + 2\tan^{- 1}\left(\dfrac{\pi}{\sqrt{2}}\right)\right)=2π+21tan−1(2π)−41(−π+2tan−1(2π))=π2+π4=3π4= \dfrac{\pi}{2} + \dfrac{\pi}{4} = \dfrac{3\pi}{4}=2π+4π=43π=2.36= 2 . 36=2.36
32cos−122+π2+14sin−122π2+π2+tan−12π\dfrac{3}{2}\cos^{- 1}\sqrt{\dfrac{2}{2 + {\pi}^{2}}} + \dfrac{1}{4}\sin^{- 1}\dfrac{2\sqrt{2}\pi}{2 + {\pi}^{2}} + \tan^{- 1}\dfrac{\sqrt{2}}{\pi}23cos−12+π22+41sin−12+π222π+tan−1π2=32tan−1π2+14tan−1(22ππ2−2)+tan−12π= \dfrac{3}{2}\tan^{- 1}\dfrac{\pi}{\sqrt{2}} + \dfrac{1}{4}\tan^{- 1}\left(\dfrac{2\sqrt{2}\pi}{{\pi}^{2} - 2}\right) + \tan^{- 1}\dfrac{\sqrt{2}}{\pi}=23tan−12π+41tan−1(π2−222π)+tan−1π2
=π2+12tan−1π2−14tan−1(22π2−π2)= \dfrac{\pi}{2} + \dfrac{1}{2}\tan^{- 1}\dfrac{\pi}{\sqrt{2}} - \dfrac{1}{4}\tan^{- 1}\left(\dfrac{2\sqrt{2}\pi}{2 - {\pi}^{2}}\right)=2π+21tan−12π−41tan−1(2−π222π)
=π2+12tan−1(π2)−14tan−1(2⋅(π2)1−(π2)2)= \dfrac{\pi}{2} + \dfrac{1}{2}\tan^{- 1}\left(\dfrac{\pi}{\sqrt{2}}\right) - \dfrac{1}{4}\tan^{- 1}\left(\dfrac{2 \cdot \left(\dfrac{\pi}{\sqrt{2}}\right)}{1 - {\left(\dfrac{\pi}{\sqrt{2}}\right)}^{2}}\right)=2π+21tan−1(2π)−41tan−11−(2π)22⋅(2π)
=π2+12tan−1(π2)−14(−π+2tan−1(π2))= \dfrac{\pi}{2} + \dfrac{1}{2}\tan^{- 1}\left(\dfrac{\pi}{\sqrt{2}}\right) - \dfrac{1}{4}\left(- \pi + 2\tan^{- 1}\left(\dfrac{\pi}{\sqrt{2}}\right)\right)=2π+21tan−1(2π)−41(−π+2tan−1(2π))
=π2+π4=3π4= \dfrac{\pi}{2} + \dfrac{\pi}{4} = \dfrac{3\pi}{4}=2π+4π=43π
=2.36= 2 . 36=2.36
