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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2019 · 8 April · Morning Shift · Mathematics · Inverse Trigonometric Functions

If α = ^ - 1 ((3)/(5)) , β = ^ - 1 ((1)/(3)) , where 0 < α, β <...

If α=cos⁡−1(35)\alpha = \cos^{- 1}\left(\dfrac{3}{5}\right) , β=tan⁡−1(13)\beta = \tan^{- 1}\left(\dfrac{1}{3}\right) , where 0<α,β<π2,0 < \alpha, \beta < \dfrac{\pi}{2}, then α−β\alpha - \beta is equal to

Options

  1. Atan⁡−1(914)\tan^{- 1}\left(\dfrac{9}{14}\right)
  2. Bcos⁡−1(9510)\cos^{- 1}\left(\dfrac{9}{5\sqrt{10}}\right)
  3. Csin⁡−1(9510)\sin^{- 1} \left(\dfrac{9}{5\sqrt{10}}\right)
  4. Dtan⁡−1(9510)\tan^{- 1}\left(\dfrac{9}{5\sqrt{10}}\right)

Correct answer

C. sin⁡−1(9510)\sin^{- 1} \left(\dfrac{9}{5\sqrt{10}}\right)

Step-by-step solution

α=cos⁡−135=tan⁡−143\alpha = \cos^{- 1}\dfrac{3}{5} = \tan^{- 1}\dfrac{4}{3}
∴a−β=tan⁡−143−tan⁡−113∴ a - \beta = \tan^{- 1}\dfrac{4}{3} - \tan^{- 1}\dfrac{1}{3}
=tan⁡−143−131+43×13= \tan^{- 1}\dfrac{\dfrac{4}{3} - \dfrac{1}{3}}{1 + \dfrac{4}{3} \times \dfrac{1}{3}}
=tan⁡−11139= \tan^{- 1}\dfrac{1}{\dfrac{13}{9}}
=tan−1913= ta{n}^{- 1}\dfrac{9}{13}

=sin⁡−1(9510)= \sin^{- 1} \left(\dfrac{9}{5\sqrt{10}}\right)

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