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JEE Main 202120 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
20 July
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2021 · 20 July · Evening Shift · Mathematics · Inverse Trigonometric Functions

The value of tan (2 ^ - 1 ((3)/(5)) + ^ - 1 ((5)/(13))) is equal to:

The value of tantan\left(2\tan^{- 1}\left(\dfrac{3}{5}\right) + \sin^{- 1}\left(\dfrac{5}{13}\right)\right) is equal to:

Options

  1. A−18169\dfrac{- 181}{69}
  2. B22021\dfrac{220}{21}
  3. C−29176\dfrac{- 291}{76}
  4. D15163\dfrac{151}{63}

Correct answer

B. 22021\dfrac{220}{21}

Step-by-step solution

We have,

tan⁡(2tan⁡−1(35)+sin⁡−1(513))\tan \left(2\tan^{- 1}\left(\dfrac{3}{5}\right) + \sin^{- 1}\left(\dfrac{5}{13}\right)\right)

=tan⁡(tan⁡−1(35)+tan⁡−1(35)+sin⁡−1(513))= \tan \left(\tan^{- 1}\left(\dfrac{3}{5}\right) + \tan^{- 1}\left(\dfrac{3}{5}\right) + \sin^{- 1}\left(\dfrac{5}{13}\right)\right)

=tan⁡(tan⁡−1(35+351−925)+sin⁡−1(513))= \tan \left(\tan^{- 1}\left(\dfrac{\dfrac{3}{5} + \dfrac{3}{5}}{1 - \dfrac{9}{25}}\right) + \sin^{- 1}\left(\dfrac{5}{13}\right)\right)

=tan⁡(tan⁡−1(158)+tan⁡−1(512))= \tan \left(\tan^{- 1}\left(\dfrac{15}{8}\right) + \tan^{- 1}\left(\dfrac{5}{12}\right)\right)

=tan⁡−1[158+5121−158⋅512]=tan⁡−122021= \tan^{- 1}\left[\dfrac{\dfrac{15}{8} + \dfrac{5}{12}}{1 - \dfrac{15}{8} \cdot \dfrac{5}{12}}\right] = \tan^{- 1}\dfrac{220}{21}

=tan⁡(tan⁡−122021)=22021= \tan \left(\tan^{- 1}\dfrac{220}{21}\right) = \dfrac{220}{21}

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