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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Inverse Trigonometric Functions

Considering the principal values of inverse trigonometric...

Considering the principal values of inverse trigonometric functions, the value of the expression tan⁡\tan\left(2 \sin ^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos ^{-1}\left(\frac{3}{\sqrt{10}}\right)\right) is equal to :

Options

  1. A1663\frac{16}{63}
  2. B−3356-\frac{33}{56}
  3. C−1663-\frac{16}{63}
  4. D3356\frac{33}{56}

Correct answer

D. 3356\frac{33}{56}

Step-by-step solution

Let α=sin⁡−1(213)\alpha = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right).
Then sin⁡α=213\sin\alpha = \frac{2}{\sqrt{13}}, cos⁡α=313\cos\alpha = \frac{3}{\sqrt{13}}, tan⁡α=23\tan\alpha = \frac{2}{3}. So tan⁡(2α)=2⋅231−49=125\tan(2\alpha) = \frac{2 \cdot \frac{2}{3}}{1 - \frac{4}{9}} = \frac{12}{5}.
Let γ=cos⁡−1(310)\gamma = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right).
Then cos⁡γ=310\cos\gamma = \frac{3}{\sqrt{10}}, sin⁡γ=110\sin\gamma = \frac{1}{\sqrt{10}}, tan⁡γ=13\tan\gamma = \frac{1}{3}. So tan⁡(2γ)=2⋅131−19=34\tan(2\gamma) = \frac{2 \cdot \frac{1}{3}}{1 - \frac{1}{9}} = \frac{3}{4}.
Therefore tan⁡(2α−2γ)=125−341+125⋅34=33205620=3356\tan(2\alpha - 2\gamma) = \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}} = \frac{\frac{33}{20}}{\frac{56}{20}} = \frac{33}{56}
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