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NDA Maths · Inverse Trigonometry

Identities, Properties & Sum-Difference Formulas

Inverse trig functions each return an angle in a fixed principal range, obey a handful of odd/even and complementary identities, and combine through the tan⁻¹a ± tan⁻¹b sum formula.

All Inverse Trigonometry notesNDA Maths strategy

Why this matters

The chapter's largest pocket (17 PYQs). Almost every question is an identity in disguise: the complementary rule (sin⁻¹x + cos⁻¹x = π/2) and the tan⁻¹ sum formula crack the majority. Get the principal range right and the rest is substitution.

Concept 1 of 4

Principal Values & Basic Properties

Intuition

Sine, cosine, and tangent each hit every value infinitely often, so to invert them we pick ONE branch — the principal range. Every inverse-trig answer must land in its range, and the odd/even rules tell you how a negative input shifts the result.

Definition

Principal-value ranges:

  • sin⁡−1x∈[−π2,π2]\sin^{-1}x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], cos⁡−1x∈[0,π]\cos^{-1}x \in [0, \pi] (for −1≤x≤1-1 \le x \le 1).
  • tan⁡−1x∈(−π2,π2)\tan^{-1}x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right), cot⁡−1x∈(0,π)\cot^{-1}x \in (0, \pi) (for all real xx).

Sign rules: sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x) = -\sin^{-1}x and tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x (odd); but cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x and cot⁡−1(−x)=π−cot⁡−1x\cot^{-1}(-x) = \pi - \cot^{-1}x.

Principal ranges

sin⁡−1x∈[−π2,π2],cos⁡−1x∈[0,π],tan⁡−1x∈(−π2,π2)\sin^{-1}x \in [-\tfrac{\pi}{2},\tfrac{\pi}{2}], \quad \cos^{-1}x \in [0,\pi], \quad \tan^{-1}x \in (-\tfrac{\pi}{2},\tfrac{\pi}{2})
−π−π/20π/2πsin⁻¹cos⁻¹tan⁻¹open ends on tan⁻¹ — value never reaches ±π/2

Worked example

Evaluate cos⁡−1 ⁣(−12)\cos^{-1}\!\left(-\tfrac{1}{2}\right).
  1. Use cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x: cos⁡−1 ⁣(−12)=π−cos⁡−112\cos^{-1}\!\left(-\tfrac12\right) = \pi - \cos^{-1}\tfrac12.
  2. cos⁡−112=π3\cos^{-1}\tfrac12 = \tfrac{\pi}{3}, so the value is π−π3\pi - \tfrac{\pi}{3}.
Answer:2π3\dfrac{2\pi}{3}.
Practice this concept2 quick reps

Practice — Level 1 (2 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    sin⁡−1(−1)=?\sin^{-1}(-1) = ?
  2. 2.
    Range of tan⁡−1x\tan^{-1}x?

From the bank · past-year question

Example 1Inverse TrigonometryMODERATE
Consider the following in respect of inverse circular functions: I. sin⁡−1(−x)=−sin⁡−1x\sin^{-1}(-x)=-\sin^{-1}x II. cos⁡−1(−x)=cos⁡−1x\cos^{-1}(-x)=\cos^{-1}x III. tan⁡−1(−x)=π−tan⁡−1x\tan^{-1}(-x)=\pi-\tan^{-1}x IV. cot⁡−1(−x)=π−cot⁡−1x\cot^{-1}(-x)=\pi-\cot^{-1}x. How many of the above are correct?

[Q34 · Apr · 2026]

cos⁻¹ and cot⁻¹ are NOT odd

cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x, not −cos⁡−1x-\cos^{-1}x — because the range [0,π][0,\pi] has no negative angles. The same holds for cot⁡−1\cot^{-1}. Treating them as odd is the classic error.
Drill 2 more on principal values & basic properties

Concept 2 of 4

Complementary Identities

Intuition

An inverse function and its co-function add to a right angle for any valid input. This single fact converts mixed sin⁻¹/cos⁻¹ (or tan⁻¹/cot⁻¹) expressions into one variable and solves a whole class of 'find the value' questions in one step.

Definition

For all valid xx:

sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2,sec⁡−1x+csc⁡−1x=π2.\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}, \quad \tan^{-1}x + \cot^{-1}x = \tfrac{\pi}{2}, \quad \sec^{-1}x + \csc^{-1}x = \tfrac{\pi}{2}.
Also tan⁡−1x+tan⁡−11x=π2\tan^{-1}x + \tan^{-1}\tfrac{1}{x} = \tfrac{\pi}{2} for x>0x > 0. Use these to replace one inverse function by π2\tfrac{\pi}{2} minus the other, collapsing an equation to a single unknown.

Complementary pairs

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

Worked example

If 2sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \pi, find sin⁡−1x\sin^{-1}x.
  1. Write cos⁡−1x=π2−sin⁡−1x\cos^{-1}x = \tfrac{\pi}{2} - \sin^{-1}x.
  2. Then 2sin⁡−1x+π2−sin⁡−1x=π⇒sin⁡−1x=π22\sin^{-1}x + \tfrac{\pi}{2} - \sin^{-1}x = \pi \Rightarrow \sin^{-1}x = \tfrac{\pi}{2}.
Answer:sin⁡−1x=π2\sin^{-1}x = \dfrac{\pi}{2} (so x=1x = 1).

From the bank · past-year question

Example 2Inverse TrigonometryMODERATE
If 4sin⁡−1x+cos⁡−1x=π4\sin^{-1}x+\cos^{-1}x=\pi, then what is sin⁡−1x+4cos⁡−1x\sin^{-1}x+4\cos^{-1}x equal to?

[Q21 · Sep · 2024]

Drill 5 more on complementary identities

Concept 3 of 4

Sum & Difference Formulas

Intuition

Two arctangents combine into one by the tangent-of-a-sum formula. The same idea handles sin⁻¹ + tan⁻¹ etc. once you convert everything to a tangent. The only catch is a validity check on the combined formula.

Definition

The workhorse identities:

tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab(ab<1),\tan^{-1}a + \tan^{-1}b = \tan^{-1}\dfrac{a+b}{1-ab} \quad (ab < 1),
tan⁡−1a−tan⁡−1b=tan⁡−1a−b1+ab.\tan^{-1}a - \tan^{-1}b = \tan^{-1}\dfrac{a-b}{1+ab}.
When ab>1ab > 1 (with a,b>0a,b>0) add π\pi; when ab>1ab > 1 with both negative, subtract π\pi. To combine a sin⁡−1\sin^{-1} or cos⁡−1\cos^{-1} with a tan⁡−1\tan^{-1}, first rewrite each as a tan⁡−1\tan^{-1} (draw the right triangle).

Arctangent sum

tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab(ab<1)\tan^{-1}a + \tan^{-1}b = \tan^{-1}\dfrac{a+b}{1-ab}\quad (ab<1)

Worked example

Evaluate tan⁡−112+tan⁡−113\tan^{-1}\tfrac{1}{2} + \tan^{-1}\tfrac{1}{3}.
  1. ab=16<1ab = \tfrac16 < 1, so =tan⁡−112+131−16=tan⁡−15/65/6= \tan^{-1}\dfrac{\tfrac12 + \tfrac13}{1 - \tfrac16} = \tan^{-1}\dfrac{5/6}{5/6}.
  2. =tan⁡−1(1)= \tan^{-1}(1).
Answer:π4\dfrac{\pi}{4}.

From the bank · past-year question

Example 3Inverse TrigonometryMODERATE
What is tan⁡−1 ⁣(14)+tan⁡−1 ⁣(35)\tan^{-1}\!\left(\frac{1}{4}\right) + \tan^{-1}\!\left(\frac{3}{5}\right) equal to?

[Q44 · Apr · 2018]

Check ab < 1 before using the sum formula

tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab} only when ab<1ab<1. If ab>1ab>1 (positive a,ba,b) the true value exceeds π2\tfrac{\pi}{2} and you must add π\pi. Skipping this gives an answer in the wrong quadrant.
Drill 6 more on sum & difference formulas

Concept 4 of 4

The 2 tan⁻¹ Substitutions

Intuition

Expressions like 2x/(1+x²) and (1−x²)/(1+x²) are exactly the double-angle formulas in disguise — so an inverse trig of them collapses to 2 tan⁻¹x. Recognising the pattern turns a fearsome equation into simple arctangent algebra.

Definition

For suitable xx:

  • sin⁡−12x1+x2=2tan⁡−1x\sin^{-1}\dfrac{2x}{1+x^2} = 2\tan^{-1}x
  • cos⁡−11−x21+x2=2tan⁡−1x\cos^{-1}\dfrac{1-x^2}{1+x^2} = 2\tan^{-1}x
  • tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x

These come from sin⁡2θ,cos⁡2θ,tan⁡2θ\sin 2\theta, \cos 2\theta, \tan 2\theta with θ=tan⁡−1x\theta = \tan^{-1}x. They reduce a tangled equation to a linear one in arctangents.

Double-angle substitution

tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x

Worked example

Simplify cos⁡−11−x21+x2\cos^{-1}\dfrac{1-x^2}{1+x^2} for x≥0x \ge 0.
  1. Set x=tan⁡θx = \tan\theta with θ=tan⁡−1x\theta = \tan^{-1}x. Then 1−x21+x2=cos⁡2θ\dfrac{1-x^2}{1+x^2} = \cos 2\theta.
  2. So the expression is cos⁡−1(cos⁡2θ)=2θ\cos^{-1}(\cos 2\theta) = 2\theta (valid since 2θ∈[0,π]2\theta \in [0,\pi] for x≥0x\ge0).
Answer:2tan⁡−1x2\tan^{-1}x.

From the bank · past-year question

Example 4Inverse TrigonometryHARD
If sin⁡−12p1+p2−cos⁡−11−q21+q2=tan⁡−12x1−x2\sin^{-1}\dfrac{2p}{1+p^2} - \cos^{-1}\dfrac{1-q^2}{1+q^2} = \tan^{-1}\dfrac{2x}{1-x^2}, then what is x equal to?

[Q40 · Apr · 2019]

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

  • Principal Values & Basic Properties

    Principal ranges

    sin⁡−1x∈[−π2,π2],cos⁡−1x∈[0,π],tan⁡−1x∈(−π2,π2)\sin^{-1}x \in [-\tfrac{\pi}{2},\tfrac{\pi}{2}], \quad \cos^{-1}x \in [0,\pi], \quad \tan^{-1}x \in (-\tfrac{\pi}{2},\tfrac{\pi}{2})
  • Complementary Identities

    Complementary pairs

    sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}, \quad \tan^{-1}x + \cot^{-1}x = \tfrac{\pi}{2}
  • Sum & Difference Formulas

    Arctangent sum

    tan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab(ab<1)\tan^{-1}a + \tan^{-1}b = \tan^{-1}\dfrac{a+b}{1-ab}\quad (ab<1)
  • The 2 tan⁻¹ Substitutions

    Double-angle substitution

    tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x

Watch out for (2)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Inverse TrigonometryEASY
If cos⁡−1x=sin⁡−1x\cos^{-1}x=\sin^{-1}x, then which one of the following is correct?

[Q39 · Apr · 2024]

Example 2Inverse TrigonometryEASY
If kk is a root of x2−4x+1=0x^2 - 4x + 1 = 0, then what is tan⁡−1k+tan⁡−11k\tan^{-1}k + \tan^{-1}\frac{1}{k} equal to?

[Q49 · Apr · 2025]

Example 3Inverse TrigonometryMODERATE
What is tan⁡−1 ⁣ab−tan⁡−1 ⁣a−ba+b\tan^{-1}\!\dfrac{a}{b}-\tan^{-1}\!\dfrac{a-b}{a+b} equal to?

[Q44 · Apr · 2024]

Example 4Inverse TrigonometryEASY
The principal value of sin⁡−1x\sin^{-1} x lies in the interval

[Q40 · Sep · 2017]

Example 5Inverse TrigonometryEASY
Let sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{3\pi}{2} for 0≤x,y,z≤10\le x,y,z\le1. What is the value of x1000+y1001+z1002x^{1000}+y^{1001}+z^{1002}?

[Q28 · Sep · 2021]

Drill every past-year question on this subtopic

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