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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

2 ^ - 1 (1)/(5) + ^ - 1 5 2 7 + 2 ^ - 1 (1)/(8) =

2tan⁡−115+sec⁡−1527+2tan⁡−118=2\tan^{- 1}\dfrac{1}{5} + \sec^{- 1}\dfrac{5\sqrt{2}}{7} + 2\tan^{- 1}\dfrac{1}{8} =2 ^ - 1 (1)/(5) + ^ - 1 5 2 7 + 2 ^ - 1 (1)/(8) =

Options

  1. Aπ6\dfrac{π}{6}
  2. Bπ4\dfrac{π}{4}
  3. Cπ3\dfrac{π}{3}
  4. Dπ8\dfrac{π}{8}

Correct answer

B. π4\dfrac{π}{4}

Step-by-step solution

Given that,

2tan⁡−115+sec⁡−1527+2tan⁡−1182\tan^{- 1}\dfrac{1}{5} + \sec^{- 1}\dfrac{5\sqrt{2}}{7} + 2\tan^{- 1}\dfrac{1}{8}

=2tan⁡−115+2tan⁡−118+sec⁡−1527= 2\tan^{- 1}\dfrac{1}{5} + 2\tan^{- 1}\dfrac{1}{8} + \sec^{- 1}\dfrac{5\sqrt{2}}{7}

=2(tan⁡−115+tan⁡−118)+sec⁡−1527= 2\left(\tan^{- 1}\dfrac{1}{5} + \tan^{- 1}\dfrac{1}{8}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}

=2tan⁡−1(15+181−15⋅18)+sec⁡−1527= 2\tan^{- 1}\left(\dfrac{\dfrac{1}{5} + \dfrac{1}{8}}{1 - \dfrac{1}{5} \cdot \dfrac{1}{8}}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}

\left∴ \tan^{- 1}A + \tan^{- 1}B = \tan^{- 1}\left(\dfrac{A + B}{1 - AB}\right)\right)</p><p></p><p>= 2\tan^{- 1}\left(\dfrac{8 + 5}{40 - 1}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}</p><p></p><p>= 2\tan^{- 1}\left(\dfrac{13}{39}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}</p><p></p><p>= 2\tan^{- 1}\left(\dfrac{1}{3}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}</p><p></p><p>= \tan^{- 1}\left(\dfrac{2 \times \dfrac{1}{3}}{1 - \left(\dfrac{1}{3}\right)2}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}</p><p>(∴2tan⁡−1A=tan⁡−1(2A1−A2))</p><p></p><p>\left(∴ 2\tan^{- 1}A = \tan^{- 1} \left(\dfrac{2A}{1 - A^{2}}\right)\right)</p><p>= \tan^{- 1}\left(\dfrac{6}{8}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}

=tan⁡−1(34)+sec⁡−1527</p><p>= \tan^{- 1}\left(\dfrac{3}{4}\right) + \sec^{- 1}\dfrac{5\sqrt{2}}{7}</p><p>= \tan^{- 1} \left(\dfrac{3}{4}\right) + \tan^{- 1} \left(\dfrac{1}{7}\right)

~

2 ^ - 1 (1)/(5) + ^ - 1 5 2 7 + 2 ^ - 1 (1)/(8) ==tan⁡−1(34+171−34⋅17)= \tan^{- 1} \left(\dfrac{\dfrac{3}{4} + \dfrac{1}{7}}{1 - \dfrac{3}{4} \cdot \dfrac{1}{7}}\right)

=tan⁡−1(21+428−3)</p><p>= \tan^{- 1} \left(\dfrac{21 + 4}{28 - 3}\right)</p><p>= \tan^{- 1} \left(\dfrac{25}{25}\right)

= \tan^{- 1} \left(1\right)</p><p></p><p>∴ 2\tan^{- 1}15+sec⁡−1527+\dfrac{1}{5} + \sec^{- 1}\dfrac{5\sqrt{2}}{7} +2\tan^{- 1}\dfrac{1}{8} = π4\dfrac{π}{4}

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