JEE Advanced Inverse Trigonometric Functions — practice questions
56 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice JEE Advanced Inverse Trigonometric Functions in the app →The principal value of $ in^{-1}(x)$ lies in the interval:The value of $ in^{-1}\left(\dfrac{1}{2}\right)$ is:$ in^{-1}( in x) = x$ holds for $x$ in which interval?$\cos^{-1}(-x) = $If $ in^{-1}(x) + in^{-1}(y) = \dfrac{\pi}{2}$, then $x^2 + y^2$ equals:If $\tan^{-1}\left(\dfrac{1}{2}\right) + \tan^{-1}\left(\dfrac{1}{3}\right) = ?$Range of arcsin(x) (principal value):Range of arccos(x):Range of arctan(x):Value of arcsin(1):Value of arctan(1):arcsin(x) + arccos(x) =arctan(x) + arctan(1/x) (for x > 0) =Domain of f(x) = arccos(x):d/dx arcsin(x) =d/dx arctan(x) =Value of arcsin(-1/2):Value of arccos(-1):sin(arccos(x)) =tan(arcsin(x)) =2 arcsin(x) for |x| ≤ 1/√2 equals:Value of cos(arcsin(3/5)):cos⁻¹(cos(5π/4)) =tan⁻¹(tan(3π/4)) =Simplify: arcsin(sin(x)) for x ∈ [π/2, π]:Solve 2 arctan(x) = π/3:Range of f(x) = arcsin(x) + arccos(x):d/dx [arctan(x²)] =Simplify arctan(sin x / (1+cos x)) for x ∈ (-π, π):arccos(x) + arccos(-x) =arctan(x) - arctan(y) = (for xy > -1):cos(2 arctan(x)) =d/dx arcsin(sqrt(x)) =Number of solutions to arctan(x) = π/3 in real x:Maximum value of sin(2 arctan(x)) over real x:Integral ∫ 1/sqrt(1 - x²) dx =Integral ∫ 1/(1 + x²) dx =Range of f(x) = arcsin(x) - arccos(x):arctan(x) is an _____ function (odd/even):Maximum value of x² + (arcsin(x))² for x in [-1, 1]:The domain of the function $f(x) = in^{-1}(x)$ isThe range of $\cos^{-1}(x)$ isFor every $x \in [-1, 1]$, the value of $ in^{-1}(x) + \cos^{-1}(x)$ isThe principal value of $\tan^{-1}(1)$ is$\tan^{-1}\left(\tfrac{1}{2}\right) + \tan^{-1}\left(\tfrac{1}{3}\right)$ equals$\cot^{-1}\left(-\tfrac{1}{ qrt{3}}\right) = $$ in^{-1}\left( in\tfrac{2\pi}{3}\right)$ equals$\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equals$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $$\displaystyle um_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equalsFor $x = -2$, the value of $\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$ equalsAll real $x$ satisfying $ in^{-1}(x) + in^{-1}(1-x) = \cos^{-1}(x)$ areThe principal value of $ in^{-1}(1/2)$ is:For $x \in [-1, 1]$, the identity $ in^{-1}x + \cos^{-1}x$ equals:The value of $\cos^{-1}(-1)$ is:The value of $ in(\tan^{-1}(3/4))$ is: