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

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

If k= ((π)/(4)+(1)/(2) ^ -1 ((2)/(3)))+ ((1)/(2) ^ -1 ((2)/(3))) ,...

If k=tan⁡(π4+12cos⁡−1(23))+tan⁡(12sin⁡−1(23))k=\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{2}{3}\right)\right)+\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{2}{3}\right)\right), then the number of solutions of the equation sin⁡−1(kx−1)=sin⁡−1x−cos⁡−1x\sin ^{-1}(k x-1)=\sin ^{-1} x-\cos ^{-1} x is ____\_\_\_\_

Correct answer

1

Step-by-step solution

Let θ=12sin⁡−123\theta = \dfrac{1}{2} \sin^{-1} \dfrac{2}{3}, then 12cos⁡−113=(π4−θ)\dfrac{1}{2} \cos^{-1} \dfrac{1}{3} = \left( \dfrac{\pi}{4} - \theta \right)
k=tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ=2sin⁡2θk = \tan\theta + \cot\theta = \dfrac{1}{\sin\theta \cos\theta} = \dfrac{2}{\sin 2\theta}
k=223=3k = \dfrac{2}{\dfrac{2}{3}} = 3
sin⁡−1(3x−1)=sin⁡−1x−cos⁡−1x\sin^{-1}(3x - 1) = \sin^{-1}x - \cos^{-1}x
sin⁡−1(3x−1)=π2−2cos⁡−1x\sin^{-1}(3x - 1) = \dfrac{\pi}{2} - 2\cos^{-1}x
3x−1=sin⁡(π2−2cos⁡−1x)3x - 1 = \sin\left( \dfrac{\pi}{2} - 2\cos^{-1}x \right)
3x−1=2x2−1⇒x=0,323x - 1 = 2x^2 - 1 \Rightarrow x = 0, \dfrac{3}{2} (rejected)
No. of solution =1= 1
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