Inverse Trigonometric Function formulas
Master Inverse Trigonometric Function through 36 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.
Inverse Trigonometric Function, every formula
36 formulas, typeset and free. Print it, or keep it open beside your practice.
Range of sin⁻¹
Q1MCQRange sin⁻¹$\sin^{-1}(1)$ equals:- A$\dfrac{\pi}{2}$
- B$\pi$
- C$0$
- D$-\dfrac{\pi}{2}$
- A
Range of cos⁻¹
Q1MCQRange cos⁻¹$\cos^{-1}(-1)$ equals:- A$\pi$
- B$0$
- C$\dfrac{\pi}{2}$
- D$-\pi$
- A
Range of tan⁻¹
Q1MCQRange tan⁻¹$\tan^{-1}(1)$ equals:- A$\dfrac{\pi}{4}$
- B$\dfrac{\pi}{2}$
- C$\dfrac{3\pi}{4}$
- D$-\dfrac{\pi}{4}$
- A
Range of cot⁻¹
Q1MCQRange cot⁻¹The range of $\cot^{-1}x$ is:- A$(0,\pi)$
- B$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
- C$[0,\pi]$
- D$\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$
- A
Range of sec⁻¹
Q1MCQRange sec⁻¹The value $\sec^{-1}x=\dfrac{\pi}{2}$ is:- Anever attained
- Battained at $x=0$
- Cattained at $x=1$
- Dattained at $x=-1$
- A
Range of cosec⁻¹
Q1MCQRange cosec⁻¹$\csc^{-1}(1)$ equals:- A$\dfrac{\pi}{2}$
- B$0$
- C$\pi$
- D$-\dfrac{\pi}{2}$
- A
Reciprocal: sin⁻¹ ↔ cosec⁻¹
Q1MCQReciprocal sin$\sin^{-1}x$ equals:- A$\csc^{-1}\dfrac{1}{x}$
- B$\sec^{-1}\dfrac{1}{x}$
- C$\cos^{-1}\dfrac{1}{x}$
- D$\csc^{-1}x$
- A
Reciprocal: cos⁻¹ ↔ sec⁻¹
Q1MCQReciprocal cos$\cos^{-1}x$ equals:- A$\sec^{-1}\dfrac{1}{x}$
- B$\csc^{-1}\dfrac{1}{x}$
- C$\sec^{-1}x$
- D$\cos^{-1}\dfrac{1}{x}$
- A
Reciprocal: tan⁻¹ ↔ cot⁻¹
Q1MCQReciprocal tanFor $x>0$, $\tan^{-1}x$ equals:- A$\cot^{-1}\dfrac{1}{x}$
- B$\cot^{-1}x$
- C$\dfrac{\pi}{2}-\cot^{-1}x$
- D$\tan^{-1}\dfrac{1}{x}$
- A
Negative argument: sin⁻¹
Q1MCQsin⁻¹(−x)$\sin^{-1}(-x)$ equals:- A$-\sin^{-1}x$
- B$\pi-\sin^{-1}x$
- C$\sin^{-1}x$
- D$\dfrac{\pi}{2}-\sin^{-1}x$
- A
Negative argument: cos⁻¹
Q1MCQcos⁻¹(−x)$\cos^{-1}(-x)$ equals:- A$\pi-\cos^{-1}x$
- B$-\cos^{-1}x$
- C$\cos^{-1}x$
- D$\dfrac{\pi}{2}-\cos^{-1}x$
- A
Negative argument: tan⁻¹
Q1MCQtan⁻¹(−x)$\tan^{-1}(-x)$ equals:- A$-\tan^{-1}x$
- B$\pi-\tan^{-1}x$
- C$\tan^{-1}x$
- D$\dfrac{\pi}{2}-\tan^{-1}x$
- A
Negative argument: cot⁻¹
Q1MCQcot⁻¹(−x)$\cot^{-1}(-x)$ equals:- A$\pi-\cot^{-1}x$
- B$-\cot^{-1}x$
- C$\cot^{-1}x$
- D$\dfrac{\pi}{2}-\cot^{-1}x$
- A
Negative argument: sec⁻¹
Q1MCQsec⁻¹(−x)$\sec^{-1}(-x)$ equals:- A$\pi-\sec^{-1}x$
- B$-\sec^{-1}x$
- C$\sec^{-1}x$
- D$\dfrac{\pi}{2}-\sec^{-1}x$
- A
sin⁻¹x + cos⁻¹x
Q1MCQsin⁻¹+cos⁻¹$\sin^{-1}x+\cos^{-1}x$ equals:- A$\dfrac{\pi}{2}$
- B$\pi$
- C$0$
- D$\dfrac{\pi}{4}$
- A
tan⁻¹x + cot⁻¹x
Q1MCQtan⁻¹+cot⁻¹$\tan^{-1}x+\cot^{-1}x$ equals:- A$\dfrac{\pi}{2}$
- B$\pi$
- C$0$
- D$\dfrac{\pi}{4}$
- A
sec⁻¹x + cosec⁻¹x
Q1MCQsec⁻¹+cosec⁻¹$\sec^{-1}x+\csc^{-1}x$ equals:- A$\dfrac{\pi}{2}$
- B$\pi$
- C$0$
- D$\dfrac{\pi}{4}$
- A
sin(sin⁻¹x)
Q1Numericalsin(sin⁻¹x)The value of $\sin\!\left(\sin^{-1}\dfrac12\right)$ is (as a decimal):sin⁻¹(sin θ)
Q1MCQsin⁻¹(sin θ)$\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)$ equals:- A$\dfrac{\pi}{3}$
- B$\dfrac{2\pi}{3}$
- C$-\dfrac{\pi}{3}$
- D$\pi$
- A
tan⁻¹x + tan⁻¹y
Q1MCQtan⁻¹ sum$\tan^{-1}\dfrac12+\tan^{-1}\dfrac13$ equals:- A$\dfrac{\pi}{4}$
- B$\dfrac{\pi}{2}$
- C$\dfrac{\pi}{3}$
- D$\dfrac{3\pi}{4}$
- A
tan⁻¹x − tan⁻¹y
Q1MCQtan⁻¹ difference$\tan^{-1}x-\tan^{-1}y$ equals:- A$\tan^{-1}\dfrac{x-y}{1+xy}$
- B$\tan^{-1}\dfrac{x+y}{1-xy}$
- C$\tan^{-1}(x-y)$
- D$\tan^{-1}\dfrac{x-y}{1-xy}$
- A
sin⁻¹x + sin⁻¹y
for x²+y²≤1
Q1MCQsin⁻¹ sum$\sin^{-1}x+\sin^{-1}y$ (for $x^{2}+y^{2}\le1$) equals:- A$\sin^{-1}\!\left(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\right)$
- B$\sin^{-1}(x+y)$
- C$\sin^{-1}(xy)$
- D$\cos^{-1}\!\left(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)$
- A
cos⁻¹x + cos⁻¹y
for x+y≥0
Q1MCQcos⁻¹ sum$\cos^{-1}x+\cos^{-1}y$ (for $x+y\ge0$) equals:- A$\cos^{-1}\!\left(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)$
- B$\cos^{-1}(xy)$
- C$\cos^{-1}(x+y)$
- D$\sin^{-1}\!\left(xy+\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)$
- A
2 tan⁻¹x (sin form)
Q1MCQ2tan⁻¹ (sin)For $|x|\le1$, $2\tan^{-1}x$ equals:- A$\sin^{-1}\dfrac{2x}{1+x^{2}}$
- B$\sin^{-1}\dfrac{2x}{1-x^{2}}$
- C$\cos^{-1}\dfrac{2x}{1+x^{2}}$
- D$\tan^{-1}\dfrac{2x}{1+x^{2}}$
- A
2 tan⁻¹x (cos form)
Q1MCQ2tan⁻¹ (cos)For $x\ge0$, $2\tan^{-1}x$ equals:- A$\cos^{-1}\dfrac{1-x^{2}}{1+x^{2}}$
- B$\cos^{-1}\dfrac{1+x^{2}}{1-x^{2}}$
- C$\sin^{-1}\dfrac{1-x^{2}}{1+x^{2}}$
- D$\tan^{-1}\dfrac{1-x^{2}}{1+x^{2}}$
- A
2 tan⁻¹x (tan form)
Q1MCQ2tan⁻¹ (tan)For $|x|<1$, $2\tan^{-1}x$ equals:- A$\tan^{-1}\dfrac{2x}{1-x^{2}}$
- B$\tan^{-1}\dfrac{2x}{1+x^{2}}$
- C$\sin^{-1}\dfrac{2x}{1-x^{2}}$
- D$\cos^{-1}\dfrac{2x}{1-x^{2}}$
- A
2 sin⁻¹x
Q1MCQ2sin⁻¹For $|x|\le\dfrac{1}{\sqrt2}$, $2\sin^{-1}x$ equals:- A$\sin^{-1}\!\left(2x\sqrt{1-x^{2}}\right)$
- B$\sin^{-1}(2x)$
- C$\cos^{-1}\!\left(2x^{2}-1\right)$
- D$\sin^{-1}\!\left(x\sqrt{1-x^{2}}\right)$
- A
3 sin⁻¹x
Q1MCQ3sin⁻¹For $|x|\le\dfrac12$, $3\sin^{-1}x$ equals:- A$\sin^{-1}(3x-4x^{3})$
- B$\sin^{-1}(4x^{3}-3x)$
- C$3\sin^{-1}(x^{3})$
- D$\sin^{-1}(3x)$
- A
3 cos⁻¹x
Q1MCQ3cos⁻¹For $\dfrac12\le x\le1$, $3\cos^{-1}x$ equals:- A$\cos^{-1}(4x^{3}-3x)$
- B$\cos^{-1}(3x-4x^{3})$
- C$3\cos^{-1}(x^{3})$
- D$\cos^{-1}(3x)$
- A
3 tan⁻¹x
Q1MCQ3tan⁻¹For $|x|<\dfrac{1}{\sqrt3}$, $3\tan^{-1}x$ equals:- A$\tan^{-1}\dfrac{3x-x^{3}}{1-3x^{2}}$
- B$\tan^{-1}\dfrac{3x+x^{3}}{1+3x^{2}}$
- C$3\tan^{-1}(x^{3})$
- D$\tan^{-1}(3x)$
- A
tan⁻¹ chain (=π/2)
Q1MCQtan⁻¹ chain π/2$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\dfrac{\pi}{2}$ holds when:- A$xy+yz+zx=1$
- B$x+y+z=xyz$
- C$xyz=1$
- D$x+y+z=0$
- A
tan⁻¹ chain (=π)
Q1MCQtan⁻¹ chain π$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi$ holds when:- A$x+y+z=xyz$
- B$xy+yz+zx=1$
- C$xyz=0$
- D$x+y+z=0$
- A
θ∈[0,π]: sin⁻¹(cosθ)
Q1MCQsin⁻¹(cos θ)For $\theta\in[0,\pi]$, $\sin^{-1}(\cos\theta)$ equals:- A$\dfrac{\pi}{2}-\theta$
- B$\theta-\dfrac{\pi}{2}$
- C$\pi-\theta$
- D$\theta$
- A
Trig substitution √(a²−x²)
Q1MCQ√(a²−x²) subTo simplify $\sqrt{a^{2}-x^{2}}$, substitute:- A$x=a\sin\theta$
- B$x=a\tan\theta$
- C$x=a\sec\theta$
- D$x=a\cos\theta$ only
- A
Trig substitution √(a²+x²)
Q1MCQ√(a²+x²) subTo simplify $\sqrt{a^{2}+x^{2}}$, substitute:- A$x=a\tan\theta$
- B$x=a\sin\theta$
- C$x=a\sec\theta$
- D$x=a\cos\theta$
- A
Range of (sin⁻¹x)²+(cos⁻¹x)²
Q1MCQMin of squaresThe minimum value of $(\sin^{-1}x)^{2}+(\cos^{-1}x)^{2}$ is:- A$\dfrac{\pi^{2}}{8}$
- B$\dfrac{\pi^{2}}{4}$
- C$\dfrac{5\pi^{2}}{4}$
- D$\dfrac{\pi^{2}}{2}$
- A
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