List of top Mathematics Questions on Inverse Trigonometric Functions

$\tan^{-1} \left( \frac{1}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \cdot 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n \cdot (n+1)} \right) =$

If $f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)$, then $f'\left(\frac{1}{2}\right) =$

If $\sin^{-1} x + \sin^{-1} y = \pi/2$, then $x^2$ is equal to

The value of $\tan^{-1}\left(\frac{\cos x-\sqrt{3}\sin x}{\sqrt{3}\cos x+\sin x}\right)$ , where $0<x<\frac{\pi}{2}$ is

The value of $\tan^{2}(\sec^{-1}(3))$ is

The value of $\cos^{-1}\left(\cos\frac{2\pi}{3}\right)+\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$ is equal to

If \( y = \sin^{-1}\!\left(\dfrac{5x + 12\sqrt{1-x^2}}{13}\right) \), then \( \dfrac{dy}{dx} \) is equal to:

The value of $\tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}\left(-\frac{1}{2}\right)$ is

For the curve \(y = 3x^3 - 3x^2 + 1\) at \(x = 1\), find the equation of the tangent.

A plane is formed by the axes whose centroid is \(\left(2, -\frac{2}{3}, \frac{1}{2}\right)\). Find the distance of the plane from the origin.

If the domain of the function $f(x)=\sin^{-1}\!\left(\dfrac{1}{x^2-2x-2}\right)$ is $(-\infty,\alpha)\cup[\beta,\gamma]\cup[\delta,\infty)$, then $\alpha+\beta+\gamma+\delta$ is equal to

If domain of $f(x)=\sin^{-1}\!\left(\dfrac{1}{x^2-2x-2}\right)$ is $(-\infty,\alpha]\cup[\beta,\gamma]\cup[\delta,\infty)$, then $(\alpha+\beta+\gamma+\delta)$ is

If the domain of the function \(f(x) = \cos^{-1}\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^2-3x+1)\) is the interval \([\alpha, \beta]\), then \(\alpha + 2\beta\) is equal to:}

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 _________.

If the domain of the function \[ \cos^{-1}\!\left(\frac{2x-5}{11x-7}\right) + \sin^{-1}\!\left(2x^2 - 3x + 1\right) \] is \[ [0,a] \cup \left[\frac{12}{13},\, b\right], \] then the value of \( \dfrac{1}{ab} \) is:

Let \[ k = \tan\!\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\!\frac{2}{3}\right) + \tan^{-1}\!\left(\frac{1}{2}\sin^{-1}\!\frac{2}{3}\right). \] Then the number of solutions of the equation \[ \sin^{-1}(kx - 1) = \sin^{-1}x - \cos^{-1}x \] is:

If domain of $f(x)=\sin^{-1}\!\left(\dfrac{1}{x^2-2x-2}\right)$ is $(-\infty,\alpha]\cup[\beta,\gamma]\cup[\delta,\infty)$, then $(\alpha+\beta+\gamma+\delta)$ is

The number of values of \(x\) satisfying \[ \tan^{-1}(4x)+\tan^{-1}(6x)=\frac{\pi}{6}, \quad x\in\left[-\frac{1}{2\sqrt{6}},\,\frac{1}{2\sqrt{6}}\right] \] is

The range of \( x \) for which the equation \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}(x) \) holds true

\( -\frac{2\pi}{5} \) is the principal value of

for $|x| < 1$, sin(tan-1x) equal to

for $|x| < 1$, sin(tan-1x) equal to

\[ \sum_{k=1}^{n} \left( \alpha^k + \frac{1}{\alpha^k} \right)^2 = 20, \quad \alpha \text{ is one of the roots of } x^2 + x + 1 = 0, \text{ then } n = ? \]