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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2019 · 12 April · Morning Shift · Mathematics · Inverse Trigonometric Functions

The value of ^ - 1 ((12)/(13)) - ^ - 1 ((3)/(5)) is equal to:

The value of sin⁡−1\sin^{- 1}\left(\dfrac{12}{13}\right) - \sin^{- 1} (35)\left(\dfrac{3}{5}\right) is equal to:

Options

  1. Aπ−cos⁡−1(3365)\pi - \cos^{- 1} \left(\dfrac{33}{65}\right)
  2. Bπ2−cos⁡−1(965)\dfrac{\pi}{2} - \cos^{- 1} \left(\dfrac{9}{65}\right)
  3. Cπ−sin⁡−1(6365)\pi - \sin^{- 1} \left(\dfrac{63}{65}\right)
  4. Dπ2−sin⁡−1(5665)\dfrac{\pi}{2} - \sin^{- 1} \left(\dfrac{56}{65}\right)

Correct answer

D. π2−sin⁡−1(5665)\dfrac{\pi}{2} - \sin^{- 1} \left(\dfrac{56}{65}\right)

Step-by-step solution

sin⁡−1(1213)−sin⁡−1(35)\sin^{- 1} \left(\dfrac{12}{13}\right) - \sin^{- 1} \left(\dfrac{3}{5}\right)
=tan⁡−1(125)−tan⁡−1(34)= \tan^{- 1} \left(\dfrac{12}{5}\right) - \tan^{- 1} \left(\dfrac{3}{4}\right)\left[\sin^{- 1} x = \tan^{- 1} \left(\dfrac{x}{\sqrt{1 - {x}^{2}}}\right)\right]<br/><br />= \tan^{- 1} \left(\dfrac{\dfrac{12}{5} - \dfrac{3}{4}}{1 + \dfrac{12 \times 3}{5 \times 4}}\right)<br/><br />= \tan^{- 1} (3356)\left(\dfrac{33}{56}\right)
=sin⁡−1(3365)= \sin^{- 1} \left(\dfrac{33}{65}\right)
=cos⁡−1(5665)=π2−sin⁡−1(5665)= \cos^{- 1} \left(\dfrac{56}{65}\right) = \dfrac{\pi}{2} - \sin^{- 1} \left(\dfrac{56}{65}\right)
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