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If \( k = \tan\left( \frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left( \frac{2}{3} \right) \right) + \tan\left( \frac{1}{2}\sin^{-1}\left( \frac{2}{3} \right) \right) \), then the number of solutions of the equation\( \sin^{-1}(kx - 1) = \sin^{-1} x - \cos^{-1} x \) is ____________.

The number of solutions of \( \tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6} \), where \( -\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}} \), is equal to:

Let the maximum value of \( (\sin^{-1}x)^2+(\cos^{-1}x)^2 \) for \( x\in\left[-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\right] \) be \( \frac{m}{n}\pi^2 \), where \( \gcd(m,n)=1 \). Then \( m+n \) is equal to ____________.

Let \( 0<\alpha<1, \; \beta=\frac{1}{3\alpha} \) and \( \tan^{-1}(1-\alpha)+\tan^{-1}(1-\beta)=\frac{\pi}{4} \). Then \( 6(\alpha+\beta) \) is equal to:

If \( \frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1}\!\left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) = \alpha \), then \( \tan \alpha \) is equal to ______.

Let \( \alpha=3\sin^{-1}\!\left(\dfrac{6}{11}\right) \) and \( \beta=3\cos^{-1}\!\left(\dfrac{4}{9}\right) \), where inverse trigonometric functions take only the principal values. Given below are two statements:
Statement-I: \( \cos(\alpha+\beta)>0 \).
Statement-II: \( \cos(\alpha)<0 \).
In the light of the above statements, choose the correct answer from the options given below:

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JEE MAIN 2026 22nd Jan Morning Shift

The number of solutions of \( \tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6} \), where \( -\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}} \), is equal to: