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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

The value of 2 ^ -1 (2)/(3) - 2 ^ -1 6 + 1 2 3 is equal to

The value of 2cos⁡−123−2cos⁡−16+1232\cos^{-1} \sqrt{\dfrac{2}{3}} - 2\cos^{-1} \dfrac{\sqrt{6} + 1}{2\sqrt{3}} is equal to

Options

  1. Aπ3\dfrac{\pi}{3}
  2. Bπ4\dfrac{\pi}{4}
  3. Cπ2\dfrac{\pi}{2}
  4. Dπ6\dfrac{\pi}{6}

Correct answer

A. π3\dfrac{\pi}{3}

Step-by-step solution

2\cos^{- 1} \left(\sqrt{\dfrac{2}{3}}\right) -2cos⁡−1(6+123)<br/>2\cos^{- 1}\left(\dfrac{\sqrt{6} + 1}{2\sqrt{3}}\right)<br />= 2tan^{- 1}\left(\dfrac{1}{2}\right) -2tan−1(3−21+6)=2tan^{- 1}\left(\dfrac{\sqrt{3} - \sqrt{2}}{1 + \sqrt{6}}\right) =2tan^{- 1} \left(\dfrac{1}{2}\right) - 2 \left[tan−1tan^{- 1}\sqrt{3} -tan−1tan^{- 1}\sqrt{2}\right]
=2(tan⁡−112+tan⁡−12)−= 2\left(\tan^{- 1}\dfrac{1}{\sqrt{2}} + \tan^{- 1}\sqrt{2}\right) -2\tan^{- 1} 3<br/>\sqrt{3}<br />= \pi - 2π3=π3\dfrac{2\pi}{3} = \dfrac{\pi}{3}
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