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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

If α = ^ -1 3 2 + ^ -1 (1)/(3) and β = ^ -1 3 2 + ^ -1 (1)/(3) ,...

If α=sin⁡−132+sin⁡−113\alpha = \sin^{-1} \dfrac{\sqrt{3}}{2} + \sin^{-1} \dfrac{1}{3} and β=cos⁡−132+cos⁡−113\beta = \cos^{-1} \dfrac{\sqrt{3}}{2} + \cos^{-1} \dfrac{1}{3}, then

Options

  1. Aα>β\alpha > \beta
  2. Bα=β\alpha = \beta
  3. Cα<β\alpha < \beta
  4. Dα+β=2π\alpha + \beta = 2\pi

Correct answer

C. α<β\alpha < \beta

Step-by-step solution

α+β=sin⁡−132+cos⁡−132+sin⁡−113+cos⁡−113\alpha + \beta = \sin^{- 1} \dfrac{\sqrt{3}}{2} + \cos^{- 1} \dfrac{\sqrt{3}}{2} + \sin^{- 1} \dfrac{1}{3} + \cos^{- 1} \dfrac{1}{3}

=π2+π2=π= \dfrac{\pi}{2} + \dfrac{\pi}{2} = \pi

Also, α=π3+sin⁡−113<π3+sin⁡−112\alpha = \dfrac{\pi}{3} + \sin^{- 1} \dfrac{1}{3} < \dfrac{\pi}{3} + \sin^{- 1} \dfrac{1}{2}

As sinθsin \theta is increasing in \left[0, \dfrac{\pi}{2}\right]<br/><br/><br/><br/>∴ \alpha < \dfrac{\pi}{3} + \dfrac{\pi}{6} = \dfrac{\pi}{2}<br/><br/><br/><br/>⇒ \beta > \dfrac{\pi}{2} > \alpha<br/><br/><br/><br/>⇒ \alpha < \beta
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