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Formula Sheet

Inverse Trigonometric Functions Formulas

All key formulas grouped by subtopic. Each one has a quick reminder and common mistakes to watch for.

10 formulas · 7 subtopics

Principal Value Ranges

#1
sin⁡−1x∈[−π2,π2],cos⁡−1x∈[0,π],tan⁡−1x∈(−π2,π2)\sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \quad \cos^{-1}x \in [0, \pi], \quad \tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

💡 Domain of sin inverse and cos inverse is [-1, 1]. Domain of tan inverse is all real numbers. The principal value is the unique value in the specified range. cosec inverse, sec inverse, and cot inverse follow from these.

⚠ Writing sin inverse(sin(5pi/6)) = 5pi/6. The answer must lie in [-pi/2, pi/2], so the correct answer is pi/6.

Complementary Pair Identity

#2
sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2,cosec⁡−1x+sec⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad \cosec^{-1}x + \sec^{-1}x = \frac{\pi}{2}

💡 Valid for all x in the respective domains. These are the most frequently used identities in JEE. They allow you to convert between inverse trig functions quickly.

Sum of Two tan inverse Values

#3
tan⁡−1x+tan⁡−1y={tan⁡−1x+y1−xy,xy<1π+tan⁡−1x+y1−xy,xy>1, x>0−π+tan⁡−1x+y1−xy,xy>1, x<0\tan^{-1}x + \tan^{-1}y = \begin{cases} \tan^{-1}\frac{x+y}{1-xy}, & xy < 1 \\ \pi + \tan^{-1}\frac{x+y}{1-xy}, & xy > 1,\, x > 0 \\ -\pi + \tan^{-1}\frac{x+y}{1-xy}, & xy > 1,\, x < 0 \end{cases}

💡 The condition xy < 1 vs xy > 1 determines whether the sum stays in (-pi/2, pi/2) or shifts by pi. This is the most tested formula in inverse trig.

⚠ Using tan inverse(x) + tan inverse(y) = tan inverse((x+y)/(1-xy)) without checking whether xy < 1. When xy > 1, you must add or subtract pi.

Double Angle: 2 tan inverse x

#4
2tan⁡−1x={sin⁡−12x1+x2,∣x∣≤1cos⁡−11−x21+x2,x≥0tan⁡−12x1−x2,∣x∣<12\tan^{-1}x = \begin{cases} \sin^{-1}\frac{2x}{1+x^2}, & |x| \le 1 \\ \cos^{-1}\frac{1-x^2}{1+x^2}, & x \ge 0 \\ \tan^{-1}\frac{2x}{1-x^2}, & |x| < 1 \end{cases}

💡 These identities connect tan inverse to sin inverse and cos inverse via double-angle substitution. The domain restrictions are critical and are the main source of errors.

sin inverse of 2x sqrt(1 - x^2)

#5
sin⁡−1(2x1−x2)={2sin⁡−1x,∣x∣≤12π−2sin⁡−1x,12<x≤1\sin^{-1}(2x\sqrt{1-x^2}) = \begin{cases} 2\sin^{-1}x, & |x| \le \frac{1}{\sqrt{2}} \\ \pi - 2\sin^{-1}x, & \frac{1}{\sqrt{2}} < x \le 1 \end{cases}

💡 This comes from substituting x = sin(theta). The split at 1/sqrt(2) corresponds to theta = pi/4. JEE often tests this with x = sqrt(3)/2 or x = 1/sqrt(2).

⚠ Applying the formula 2*sin inverse(x) without checking whether x is in the valid range. For x > 1/sqrt(2), the answer is pi - 2*sin inverse(x).

Composition: sin(cos inverse x)

#6
sin⁡(cos⁡−1x)=1−x2,cos⁡(sin⁡−1x)=1−x2,tan⁡(sin⁡−1x)=x1−x2\sin(\cos^{-1}x) = \sqrt{1-x^2}, \quad \cos(\sin^{-1}x) = \sqrt{1-x^2}, \quad \tan(\sin^{-1}x) = \frac{x}{\sqrt{1-x^2}}

💡 Draw a right triangle with the known side to find the other trig ratio. If cos inverse(x) = theta, then cos(theta) = x, so sin(theta) = sqrt(1 - x^2).

Domain Restrictions

#7
sin⁡−1x:x∈[−1,1],cos⁡−1x:x∈[−1,1],tan⁡−1x:x∈R,cosec⁡−1x:∣x∣≥1,sec⁡−1x:∣x∣≥1\sin^{-1}x: x \in [-1,1], \quad \cos^{-1}x: x \in [-1,1], \quad \tan^{-1}x: x \in \mathbb{R}, \quad \cosec^{-1}x: |x| \ge 1, \quad \sec^{-1}x: |x| \ge 1

💡 cosec inverse and sec inverse have domains |x| >= 1 (excludes the interval (-1, 1)). cot inverse has domain all real numbers. These domain checks are often the first step in solving problems.

Derivative of Inverse Trig Functions

#8
ddxsin⁡−1x=11−x2,ddxtan⁡−1x=11+x2,ddxcos⁡−1x=−11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}, \quad \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}}

💡 The derivative of cos inverse is the negative of sin inverse's derivative. Similarly, cot inverse's derivative is the negative of tan inverse's. These are essential for integration as well.

tan inverse Difference Formula

#9
tan⁡−1x−tan⁡−1y=tan⁡−1x−y1+xy,xy>−1\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}, \quad xy > -1

💡 This is the difference version of the sum formula. The condition xy > -1 ensures the result is in (-pi/2, pi/2). Used heavily in telescoping series problems.

Telescoping tan inverse Series

#10
∑r=1ntan⁡−111+r+r2=∑r=1n[tan⁡−1(r+1)−tan⁡−1r]=tan⁡−1(n+1)−π4\sum_{r=1}^{n} \tan^{-1}\frac{1}{1+r+r^2} = \sum_{r=1}^{n} \left[\tan^{-1}(r+1) - \tan^{-1}r\right] = \tan^{-1}(n+1) - \frac{\pi}{4}

💡 The key trick: 1/(1 + r + r^2) = ((r+1) - r)/(1 + r(r+1)). This makes each term a difference of two tan inverse values, leading to telescoping cancellation.

⚠ Not recognizing the telescoping pattern. Always try to express the general term as tan inverse(something) - tan inverse(something else) using the difference formula.
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