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JEE Main 2019 · 8 April · Shift I

JEE Main 2019 Mathematics question on Inverse Trigonometric Functions

If α=cos⁡−1(35)\alpha = \cos^{-1}\left(\frac{3}{5}\right), β=tan⁡−1(13)\beta = \tan^{-1}\left(\frac{1}{3}\right), where 0<α,β<π20 < \alpha, \beta < \frac{\pi}{2}, then α−β\alpha - \beta is equal to
  1. A.
    tan⁡−1(9510)\tan^{-1}\left(\frac{9}{5\sqrt{10}}\right)
  2. B.
    cos⁡−1(9510)\cos^{-1}\left(\frac{9}{5\sqrt{10}}\right)
  3. C.
    tan⁡−1(914)\tan^{-1}\left(\frac{9}{14}\right)
  4. D.
    sin⁡−1(9510)\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)
Show answer and solution

Correct option: D

Solution

From α = cos⁻¹(3/5): sin α = 4/5, cos α = 3/5, tan α = 4/3.
From β = tan⁻¹(1/3): tan β = 1/3.

tan(α - β) = (tan α - tan β)/(1 + tan α · tan β) = (4/3 - 1/3)/(1 + 4/9) = (3/3)/(13/9) = 1·(9/13) = 9/13.

Now find sin(α - β). With tan(α - β) = 9/13, the hypotenuse is √(81 + 169) = √250 = 5√10, so sin(α - β) = 9/(5√10).

Since α, β ∈ (0, π/2) and α > β (as tan α = 4/3 > 1/3 = tan β), α - β is a positive angle in (0, π/2), so α - β = sin⁻¹(9/(5√10)).

[JEE Main 2019 · 8 April · Shift I]

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