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Let $f(x) = \cos \left( 2 \tan^{-1} \sin \left( \cot^{-1} \sqrt{\frac{1-x}{x}} \right) \right), 0 < x < 1$. Then:

A possible value of \( \tan\!\left(\frac{1}{4}\sin^{-1}\!\frac{\sqrt{63}}{8}\right) \) is:

The real values function \( f(x) = \dfrac{\text{cosec}^{-1} x}{\sqrt{x - [x]}} \), where \( [x] \) denotes the greatest integer less than or equal to \( x \), is defined for all \( x \) belonging to:

\( \cos^{-1}(\cos(-5)) + \sin^{-1}(\sin(6)) - \tan^{-1}(\tan(12)) \) is equal to:
(The inverse trigonometric functions take the principal values)

If \( \tan\left( \dfrac{\pi}{9} \right), x, \tan\left( \dfrac{7\pi}{18} \right) \) are in arithmetic progression and \( \tan\left( \dfrac{\pi}{9} \right), y, \tan\left( \dfrac{5\pi}{18} \right) \) are also in arithmetic progression, then \( |x - 2y| \) is equal to:

If \( y(x) = \cot^{-1}\!\left(\sqrt{\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}}\right),\ x \in \left( \frac{\pi}{2}, \pi \right) \) then \( \frac{dy}{dx} \) at \( x = \frac{5\pi}{6} \) is:

If \( \displaystyle \sum_{r=1}^{50} \tan^{-1}\!\left(\frac{1}{2r^{2}}\right) = p \), then the value of \( \tan p \) is:

The domain of the function \( f(x) = \sin^{-1}\!\left(\dfrac{3x^{2}+x-1}{(x-1)^{2}}\right) + \cos^{-1}\!\left(\dfrac{x-1}{x+1}\right) \) is:

Let \( M \) and \( m \) respectively be the maximum and minimum values of the function \( f(x)=\tan^{-1}(\sin x+\cos x) \) in \( \left[0,\frac{\pi}{2}\right] \). Then the value of \( \tan(M-m) \) is equal to:

The sum of possible values of \( x \) for \( \tan^{-1}(x+1) + \cot^{-1}\!\left(\dfrac{1}{x-1}\right) = \tan^{-1}\!\left(\dfrac{8}{31}\right) \) is:

Given that the inverse trigonometric functions take principal values only. Then the number of real values of \( x \) which satisfy \( \sin^{-1}\!\left(\dfrac{3x}{5}\right)+\sin^{-1}\!\left(\dfrac{4x}{5}\right)=\sin^{-1}x \) is equal to:

The number of real roots of the equation \( \tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^{2}+x+1} = \dfrac{\pi}{4} \) is :

The value of \( \tan\!\left(2\tan^{-1}\!\left(\dfrac{3}{5}\right)+\sin^{-1}\!\left(\dfrac{5}{13}\right)\right) \) is equal to:

If \( (\sin^{-1} x)^2 - (\cos^{-1} x)^2 = a; \, 0 < x < 1, \, a \neq 0 \), then the value of \( 2x^2 - 1 \) is:

If the domain of the function \( f(x) = \dfrac{ \cos^{-1}\sqrt{x^2 - x + 1} }{ \sqrt{ \sin^{-1}\!\left( \dfrac{2x - 1}{2} \right) } } \) is the interval \( (\alpha, \beta) \), then \( \alpha + \beta \) is equal to:

If \( \cot^{-1}(\alpha) = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 18 + \cot^{-1} 32 + \ldots \) upto \( 100 \) terms, then \( \alpha \) is:

Question 1 of 16Next →

JEE Main 2021 26th August Morning Shift

Let $f(x) = \cos \left( 2 \tan^{-1} \sin \left( \cot^{-1} \sqrt{\frac{1-x}{x}} \right) \right), 0 < x < 1$. Then: