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\( y = \cos\left(\frac{\pi}{3} + \cos^{-1}\frac{x}{2}\right) \), then \( (x - y)^2 + 3y^2 \) is equal to ____________.

If \( \alpha>\beta>\gamma>0 \), then the expression \( \displaystyle \cot^{-1} \left\{ \beta + \frac{(1 + \beta^2)}{(\alpha - \beta)} \right\} + \cot^{-1} \left\{ \gamma + \frac{(1 + \gamma^2)}{(\beta - \gamma)} \right\} + \cot^{-1} \left\{ \alpha + \frac{(1 + \alpha^2)}{(\gamma - \alpha)} \right\} \) is equal to:

Considering the principal values of the inverse trigonometric functions, \( \sin^{-1}\!\left(\dfrac{\sqrt{3}}{2}\,x + \dfrac{1}{2}\sqrt{1-x^2}\right),\; -\dfrac{1}{2} < x < \dfrac{1}{\sqrt{2}} \) is equal to:

If for some \( \alpha, \beta : \alpha \le \beta \), \( \alpha + \beta = 8 \) and \( \sec^{2}\!\left(\tan^{-1}\! \alpha \right) + cosec^{2}\!\left(\cot^{-1}\! \beta \right) = 36 \), then \( \alpha^{2} + \beta \) is ______.

The value of \( \cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}(2)}-1}{\tan(2)}\right) - \cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}\!\left(\dfrac{1}{2}\right)}+1}{\tan\!\left(\dfrac{1}{2}\right)}\right) \) is equal to:

\( \cos\left(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{33}{65}\right) \) is equal to:

If \( \frac{\pi}{2} \leq x \leq \frac{3\pi}{4} \), then \( \cos^{-1}\!\left( \frac{12}{13}\cos x + \frac{5}{13}\sin x \right) \) is equal to:

The sum of the infinite series \( \cot^{-1}\!\left(\dfrac{7}{4}\right)+\cot^{-1}\!\left(\dfrac{19}{4}\right)+\cot^{-1}\!\left(\dfrac{39}{4}\right)+\cot^{-1}\!\left(\dfrac{67}{4}\right)+\ldots \) is:

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of \( 16\left( (\sec^{-1} x)^{2} + (cosec^{-1} x)^{2} \right) \) is:

Let \( S = \{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1) \} \). Then \( \sum\limits_{x \in S} (2x - 1)^2 \) is equal to ________.

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JEE MAIN 2025 4th April Morning Shift

Considering the principal values of the inverse trigonometric functions, \( \sin^{-1}\!\left(\dfrac{\sqrt{3}}{2}\,x + \dfrac{1}{2}\sqrt{1-x^2}\right),\; -\dfrac{1}{2} < x < \dfrac{1}{\sqrt{2}} \) is equal to: