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NTA Abhyas JEE Main2020MathematicsInverse Trigonometric FunctionsPractice

If α = sin - 1 ⁡ 3 2 + sin - 1 ⁡ 1 3 and β = cos - 1 ⁡ 3 2 + cos - 1 ⁡ 1 3 , then

Options

  1. Aα > β
  2. Bα = β
  3. Cα < β
  4. Dα + β = 2 π

Correct answer

C. α < β

Step-by-step solution

α + β = sin - 1 ⁡ 3 2 + cos - 1 ⁡ 3 2 + sin - 1 ⁡ 1 3 + cos - 1 ⁡ 1 3 = π 2 + π 2 = π Also, α = π 3 + sin - 1 ⁡ 1 3 < π 3 + sin - 1 ⁡ 1 2 As sin ⁡ θ is increasing in 0 , π 2 ∴ α < π 3 + π 6 = π 2 ⇒ β > π 2 > α ⇒ α < β

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