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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.

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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.

Test yourself — a quick 14-question Inverse Trigonometry recall quiz, instant score.

Concept 1 of 4: Principal Values & Basic Properties

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).
Practice this concept5 quick reps

The same idea in a real exam question:

NDA · Apr 2026 · Q34Moderate

Example 1 · Inverse Trigonometry · Identities, Properties, and Sum-Difference Formulas

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?

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.

tan⁻¹ is odd, but its range is OPEN

tan⁡−1x∈(−π2,π2)\tan^{-1}x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) — the endpoints are never attained, because tan⁡(±π2)\tan(\pm\tfrac{\pi}{2}) is undefined. Writing tan⁡−1x=π2\tan^{-1}x = \tfrac{\pi}{2} for any finite xx is wrong.

Inverse-trig answers must LAND in the principal range

After any manipulation, check the result lies in the function's principal range: sin⁡−1∈[−π2,π2]\sin^{-1}\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}], cos⁡−1∈[0,π]\cos^{-1}\in[0,\pi]. An angle like 5π6\tfrac{5\pi}{6} is a valid cos⁡−1\cos^{-1} output but can NEVER be a sin⁡−1\sin^{-1} output.
Drill 2 more on principal values & basic properties

Concept 2 of 4: Complementary Identities

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 but =−π2= -\tfrac{\pi}{2} for x<0x < 0 — because cot⁡−1x=tan⁡−11x\cot^{-1}x=\tan^{-1}\tfrac1x holds only for x>0x>0; for x<0x<0, cot⁡−1x=π+tan⁡−11x\cot^{-1}x=\pi+\tan^{-1}\tfrac1x. So when the two arguments are the roots of a quadratic, Vieta decides the branch: roots with product 1 and NEGATIVE sum (e.g. of x2+5x+1=0x^2+5x+1=0) are both negative, and tan⁡−1α+tan⁡−1β=−π2\tan^{-1}\alpha+\tan^{-1}\beta=-\tfrac\pi2, not +π2+\tfrac\pi2. 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.
Practice this concept3 quick reps

The same idea in a real exam question:

NDA · Sep 2024 · Q21Moderate

Example 2 · Inverse Trigonometry · Identities, Properties, and Sum-Difference Formulas

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?

sin⁻¹x + cos⁻¹x = π/2 always — don't evaluate term by term

sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2} for EVERY valid xx, not just nice values. Students waste time evaluating each inverse separately; the sum is fixed. The same fixed-sum holds for tan⁡−1x+cot⁡−1x\tan^{-1}x+\cot^{-1}x and sec⁡−1x+csc⁡−1x\sec^{-1}x+\csc^{-1}x.
Drill 5 more on complementary identities

Concept 3 of 4: Sum & Difference Formulas

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}.
Practice this concept3 quick reps

The same idea in a real exam question:

NDA · Apr 2018 · Q44Moderate

Example 3 · Inverse Trigonometry · Identities, Properties, and Sum-Difference Formulas

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

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.

Difference formula uses 1 + ab in the denominator

tan⁡−1a−tan⁡−1b=tan⁡−1a−b1+ab\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab} — the denominator is 1+ab1+ab, NOT 1−ab1-ab. Mixing up the sign of the abab term between the sum and difference forms is a frequent slip.
Drill 6 more on sum & difference formulas

Concept 4 of 4: The 2 tan⁻¹ Substitutions

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.
Practice this concept2 quick reps

The same idea in a real exam question:

NDA · Apr 2019 · Q40Hard

Example 4 · Inverse Trigonometry · Identities, Properties, and Sum-Difference Formulas

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?

The 2 tan⁻¹ substitutions need a validity range

sin⁡−12x1+x2=2tan⁡−1x\sin^{-1}\dfrac{2x}{1+x^2} = 2\tan^{-1}x holds only for ∣x∣≤1|x|\le 1; tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x only for ∣x∣<1|x|<1. Outside the range you must add or subtract π\pi — applying the identity blindly gives an out-of-range angle.

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 (7)

Test yourself on Inverse Trigonometry

15 past NDA questions from this chapter, timed at 19 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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