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NDA Mathematics · Formula sheet

Inverse Trigonometry formulas

9 formulas and 13 common traps for NDA Mathematics Inverse Trigonometry, grouped by subtopic.

Full notes with worked examples

Identities, Properties & Sum-Difference Formulas

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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

Common traps

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.

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.

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.

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.

Evaluating Composite Inverse Expressions

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Inner-to-Outer Evaluation & sin⁻¹(sin x)

Principal-range reduction

sin⁡−1(sin⁡x)=x  only if x∈[−π2,π2]\sin^{-1}(\sin x) = x \ \text{ only if } x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]

Double- & Half-Angle Compositions

Double-angle tangent

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

Converting Everything to a Tangent

Triangle → tangent

sin⁡−135=tan⁡−134,cot⁡−132=tan⁡−123\sin^{-1}\tfrac{3}{5} = \tan^{-1}\tfrac{3}{4}, \quad \cot^{-1}\tfrac{3}{2} = \tan^{-1}\tfrac{2}{3}

Common traps

sin⁻¹(sin x) ≠ x outside the principal range

sin⁡−1(sin⁡2π3)\sin^{-1}(\sin\tfrac{2\pi}{3}) is NOT 2π3\tfrac{2\pi}{3} (that is outside [−π2,π2][-\tfrac{\pi}{2},\tfrac{\pi}{2}]). Reduce the inner angle into the principal range first.

Each cancellation uses a DIFFERENT reduction rule

For xx just past the range: sin⁡−1(sin⁡x)=π−x\sin^{-1}(\sin x)=\pi-x, but cos⁡−1(cos⁡x)=2π−x\cos^{-1}(\cos x)=2\pi-x (for x∈[π,2π]x\in[\pi,2\pi]) and tan⁡−1(tan⁡x)=x−π\tan^{-1}(\tan x)=x-\pi. Don't reuse the π−x\pi-x rule for all three — match the reduction to the function's own principal range.

Double-angle tangent has 1 − tan²θ, not 1 + tan²θ

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}. The denominator is 1−tan⁡2θ1-\tan^2\theta; using 1+tan⁡2θ1+\tan^2\theta (which is sec⁡2θ\sec^2\theta) is a common confusion that produces the wrong value.

Convert sin⁻¹/cos⁻¹ to tan⁻¹ via the TRIANGLE, not the value

For sin⁡−135\sin^{-1}\tfrac35, the tangent is 34\tfrac34 (opposite 3, adjacent 52−32=4\sqrt{5^2-3^2}=4) — NOT 35\tfrac35. Build the right triangle and read off the missing side before taking the tangent.

Solving Equations & Geometric Applications

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Solving Inverse-Trig Equations

Collapse with the complementary identity

asin⁡−1x+bcos⁡−1x=c →cos⁡−1x=π2−sin⁡−1x one unknowna\sin^{-1}x + b\cos^{-1}x = c \ \xrightarrow{\cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x}\ \text{one unknown}

Geometric Applications

Subtended angle

tan⁡−1h2d−tan⁡−1h1d=tan⁡−1(h2−h1) dd2+h1h2\tan^{-1}\dfrac{h_2}{d} - \tan^{-1}\dfrac{h_1}{d} = \tan^{-1}\dfrac{(h_2-h_1)\,d}{d^2 + h_1 h_2}

Common traps

Reject roots that break the sum-formula validity

Forming f+g1−fg=1\frac{f+g}{1-fg}=1 can introduce a root where fg>1fg>1 — there the real sum is π4+π\tfrac{\pi}{4}+\pi, not π4\tfrac{\pi}{4}. Always test each algebraic root against the original equation.

Subtended-angle denominator is d² + h₁h₂

For the angle between two heights at distance dd: tan⁡−1h2d−tan⁡−1h1d=tan⁡−1(h2−h1)dd2+h1h2\tan^{-1}\dfrac{h_2}{d}-\tan^{-1}\dfrac{h_1}{d}=\tan^{-1}\dfrac{(h_2-h_1)d}{d^2+h_1h_2}. The denominator is d2+h1h2d^2+h_1h_2 (the 1+ab1+ab of the difference formula), not d2−h1h2d^2-h_1h_2. Getting that sign wrong flips the answer.

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