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JEE Main 20203 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
3 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2020 · 3 September · Morning Shift · Mathematics · Inverse Trigonometric Functions

2π - ( ^ - 1 (4)/(5) + ^ - 1 (5)/(13) + ^ - 1 (16)/(65)) is equal...

2\pi - (sin⁡−145+sin⁡−1513+sin⁡−11665)\left(\sin^{- 1}\dfrac{4}{5} + \sin^{- 1}\dfrac{5}{13} + \sin^{- 1}\dfrac{16}{65}\right) is equal to :

Options

  1. Aπ2\dfrac{\pi}{2}
  2. B5π4\dfrac{5\pi}{4}
  3. C3π2\dfrac{3\pi}{2}
  4. D7π4\dfrac{7\pi}{4}

Correct answer

C. 3π2\dfrac{3\pi}{2}

Step-by-step solution

sin⁡−145+sin⁡−1513+sin⁡−11665\sin^{- 1}\dfrac{4}{5} + \sin^{- 1}\dfrac{5}{13} + \sin^{- 1}\dfrac{16}{65}

=tan⁡−143+tan⁡−1512+tan⁡−11663= \tan^{- 1}\dfrac{4}{3} + \tan^{- 1}\dfrac{5}{12} + \tan^{- 1}\dfrac{16}{63}

=tan⁡−1(43+5121−43⋅512)+tan⁡−11663= \tan^{- 1}\left(\dfrac{\dfrac{4}{3} + \dfrac{5}{12}}{1 - \dfrac{4}{3} \cdot \dfrac{5}{12}}\right) + \tan^{- 1}\dfrac{16}{63}

=tan⁡−16316+tan⁡−11663=π2= \tan^{- 1}\dfrac{63}{16} + \tan^{- 1}\dfrac{16}{63} = \dfrac{\pi}{2}

∴2π−π2=3π2∴ 2\pi - \dfrac{\pi}{2} = \dfrac{3\pi}{2}

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