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JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)

JEE Main 2023 Mathematics question on Inverse Trigonometric Functions

tan⁡−1(1+33+3)+sec⁡−1(8+436+33)\tan^{-1}\left(\dfrac{1+\sqrt{3}}{3+\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\dfrac{8+4\sqrt{3}}{6+3\sqrt{3}}}\right) is equal to :
  1. A.
    π3\dfrac{\pi}{3}
  2. B.
    π6\dfrac{\pi}{6}
  3. C.
    π4\dfrac{\pi}{4}
  4. D.
    π2\dfrac{\pi}{2}
Show answer and solution

Correct option: A

Solution

Simplify the first term: (1+√3)/(3+√3) = (1+√3)/(√3(√3+1)) = 1/√3, so tan⁻¹(1/√3) = π/6.

For the second term, simplify the argument of sec⁻¹: (8+4√3)/(6+3√3) = 4(2+√3)/(3(2+√3)) = 4/3, so sec⁻¹(√(4/3)) = sec⁻¹(2/√3). Since sec θ = 2/√3 means cos θ = √3/2, so θ = π/6.

Adding: π/6 + π/6 = π/3.

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

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