Inverse Trigonometry Questions - JEE

If $\sum_{n=0}^{\infty} 2\cot^{-1}\left(\frac{n^2 + n + 4}{2}\right) = k\pi$, then find the value of $k$.

The value of $\displaystyle\sum_{m=1}^{\infty}\left(\tan^{-1}\left(\dfrac{3m^2-3m+1}{m^6-3m^5+3m^4-m^3+1}\right)\right)$ equals:

Complete solution set of [\cot^{-1}x] + 2[\tan^{-1}x] = 0, where [\cdot] denotes the greatest integer function, is equal to

The value of $ \displaystyle\sum_{\omega=1}^{\infty} \sin^{-1}\left[\dfrac{2\omega+1}{\omega(\omega+1)(\sqrt{\omega^2+2\omega}+\sqrt{\omega^2-1})}\right] $ is equal to:

$2\cos x - 3\sin x = a$ has real solutions for x if:

If $2\tan^{-1}\frac{1}{5} - \sin^{-1}\frac{1}{5} = -\cos^{-1}\frac{63}{\lambda}$, then $\lambda =$

Given $0 \leq x \leq \frac{1}{2}$, then the value of $\sin^{-1}\left(\frac{x + \sqrt{1-x^2}}{2}\right) - \sin^{-1}x$ is

Let $\cos^{-1}(4x^3 - 3x) = a + b\cos^{-1}x$.If $x \in \left(\frac{1}{2}, 1\right]$, then $\lim_{y \to a} b\cos y$ is:

Let a_1 = 1, a_2, a_3, a_4, \ldots be consecutive natural numbers. Then \tan^{-1}\left(\frac{1}{1+a_1 a_2}\right) + \tan^{-1}\left(\frac{1}{1+a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+a_{2021}a_{2022}}\right) \text{ is equal to}

The upper $\left(\dfrac{3}{4}\right)$th portion of a vertical pole subtends an angle $\tan^{-1}\left(\dfrac{3}{5}\right)$ at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is:

If the sum and product of four positive consecutive terms of a G.P. are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

Let (a, b) \subset (0, 2\pi) be the largest interval for which \sin^{-1}(\sin\theta) - \cos^{-1}(\sin\theta) > 0, \theta \in (0, 2\pi) holds. If \alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0 and \alpha - \beta = b - a, then \alpha is equal to:

The sum of the roots of the equation \[\cos^{-1}(\cos x) = [x]\] where $[x]$ denotes the greatest integer function, is

The value of $\cos^{-1} x + \cos^{-1}\left(\frac{x}{2} + \frac{1}{2}\sqrt{3-3x^2}\right)$ is equal to: $\left(\frac{1}{2} \leq x \leq 1\right)$

If $ \cos^{-1}\!\left(\dfrac{2}{3x}\right) + \cos^{-1}\!\left(\dfrac{3}{4x}\right) = \dfrac{\pi}{2} \left(x > \dfrac{3}{4}\right) $, then $ x $ is equal to

Range of f(x) = \sin^{-1}\log_{[x]} + \log(\sin^{-1}[x]), where [] denotes GIF is

Find the maximum value of $3\cos\theta + 5\sin\left(\theta - \dfrac{\pi}{6}\right)$.

Ex. 65: Let $f(x) = ab \sin x + b\sqrt{1 - a^2} \cos x + c$, where $|a| 0$ then

The value of $\tan\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right)$ is

$\tan^{-1}(\tan \theta) = \theta$, for all $\theta$ belonging to

Solve: tan⁻¹((x+1)/(x-1)) + tan⁻¹((x-1)/x) = tan⁻¹(-7)

Considering only the principal values of inverse functions, the set $ A = \left\{x \geq 0;\, \tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}\right\} $

$\cos[\tan^{-1}\{\sin(\cot^{-1}x)\}]=$

934. Find the number of integers not in the domain of $f(x) = \cos^{-1}\!\left(\dfrac{2-x}{2x}\right)$.

Let $S_1$ is the complete solution set of the inequality $\cos^{-1}(x) > \cos^{-1}\left(x^2\right)$ and $S_2$ is the complete solution set of the inequality $\left(\cos^{-1} x^2\right) > 0$, then $S_1 \cap S_2$ is

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{-1}(x)+\tan^{-1}(2x)=\dfrac{\pi}{4}$ is:

If $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{2}$, find $x$.

If $\tan^{-1} y : \tan^{-1} x = 4:1$, express $y$ as an algebraic function of $x$. Hence or otherwise prove that $22\frac{1}{2}$ is a root of the equation $x^4 + 1 = 6x^2$.

The range of the function $ f(x) = \sin^{-1}\!\left(\log_2 \dfrac{x^2}{2}\right) $ is:

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

Find the value of $\cos^{-1}(\cos 13)$.

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2\sin^{-1}x+3\cos^{-1}x=\dfrac{2\pi}{5}$, is

The value of $\tan^{-1}\left(\frac{x\cos\theta}{1-x\sin\theta}\right) - \cot^{-1}\left(\frac{\cos\theta}{x-\sin\theta}\right)$ is

Find the value of $\tan^{-1}\left(\dfrac{1}{2}\tan 2A\right) + \tan^{-1}(\cot A) + \tan^{-1}(\cot^3 A)$ for \(0

Given expression = $1 + 2^2 + 1 + 3^2 + \text{cosec}\left(\tan^{-1}\dfrac{4}{3} + \tan^{-1}\dfrac{4}{3}\right)$. Find the value of the expression.

The value of \cos^{-1}\left(\cot\left(\sin^{-1}\sqrt{\frac{1-x^2}{4}}\right)\right) + \sec^{-1}\left(\sqrt{1+x^2}\right)

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

The value of $\sin\left(\cos^{-1}\dfrac{1}{2} + \sin^{-1}\dfrac{\sqrt{3}}{2}\right)$ is ______.

In a triangle ABC the expression $a\cos B\cos C + b\cos C\cos A + c\cos A\cos B$ equals to:

The numerical value of $\cos[\tan^{-1}(-3) + \cot^{-1}(-3)]$ = ______.

If $\displaystyle\sum_{n=1}^{\infty} \cot^{-1}\!\left(2 + \frac{n(n+1)}{2}\right) = \tan^{-1} a$, then $a$ is equal to

Suppose in $\triangle ABC$ with sides a, b, c the following equation holds true \[\frac{\cos A}{a} + k_1 = \frac{\cos B}{b} + k_2 = \frac{\cos C}{c} + k_3 = \frac{a^2 + b^2 + c^2}{8}.\] If $abc = 4$, then the value of $k_1 k_2 k_3$ is:

Since f(x) is onto, the range of f(x) equals co-domain. The range of f(x) = cos−1(4x2 + 3x) is $\left[\frac{\pi}{2}, \pi - \cos^{-1}\frac{9}{16}\right]$. What is the answer? (Integer answer: 25)

Find the value of $\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ$.

In a triangle $ABC$, $a = 4$, $b = 3$, $\angle A = 60°$, then $c$ is the root of the equation

If $\cos^{-1}\!\sqrt{p}+\cos^{-1}\!\sqrt{1-p}+\cos^{-1}\!\sqrt{1-q}=\dfrac{3\pi}{4}$, then $q$ is

The domain of the function $ f(x) = \sin^{-1}\left[\log_3\left(\dfrac{x}{3}\right)\right] $ is:

If $\cos\theta + \sec\theta = 2$ then $\cos^n\theta + \sec^n\theta$ is equal to

If $\sin(\alpha + \beta) = 1$ and $\sin(\alpha - \beta) = \dfrac{1}{2}$, then $\tan(\alpha + 2\beta) \cdot \tan(2\alpha + \beta)$ is equal to:

From a point on the ground, the angle of elevation of the top of a tower is $ \tan^{-1}\left(\dfrac{3}{5}\right) $. The tower is 40 m away from the point. A flag is hoisted at the top of the tower and the angle of elevation of the bottom of the flag from the same point is $ \alpha $ where $ \tan\alpha = \dfrac{3}{5} $. If $ \tan(\alpha + \beta) = \dfrac{x}{40} $ (where $ \beta $ is the angle subtended by the flag at the point on the ground), find the height $ x $ of the flag (in metres).

Inverse Trigonometry Questions (1043)