Properties and identities of inverse trigonometric functions

Two ordinary trig functions, sin⁡\sin and cos⁡\cos, are cofunctions of each other: sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos\theta. Their inverses inherit a version of that partnership, and a handful of such identities — complementary-angle pairs, reciprocal swaps, and addition formulae for tan⁡−1\tan^{-1} — let you collapse a clutter of inverse-trig terms into one clean angle without ever going back to first principles. The catch is that every one of these identities is a domain-restricted shortcut. The ordinary tangent addition formula, for instance, has no trouble with any angle, but tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) only holds as written when xy<1xy < 1; push past that boundary and the raw formula lands outside the principal range, and a correction of ±π\pm\pi is required. Learning these identities means learning their fine print as much as their formulas.

The core insight

Complementary-angle, reciprocal, and addition identities let you combine or convert inverse trigonometric expressions into a single term, provided you track the domain restriction attached to each one and correct by ±π\pm\pi when a combined angle falls outside the principal range.

The core idea

Complementary-angle identities. Because each inverse function's principal range is set up so cofunction pairs sum to a right angle:

sin⁡−1x+cos⁡−1x=π2, x∈[−1,1];tan⁡−1x+cot⁡−1x=π2, x∈R;sec⁡−1x+csc⁡−1x=π2, ∣x∣≥1\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \ x \in [-1,1]; \qquad \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \ x \in \mathbb{R}; \qquad \sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}, \ |x| \ge 1

where: each pair on the left consists of a function and its cofunction evaluated at the same argument xx; the identity fails if the two arguments differ.

Reciprocal identities. These relate a function to the inverse of its reciprocal ratio:

sin⁡−1 ⁣(1x)=csc⁡−1x,cos⁡−1 ⁣(1x)=sec⁡−1x,tan⁡−1 ⁣(1x)=cot⁡−1x  (x>0)\sin^{-1}\!\left(\frac1x\right) = \csc^{-1}x, \qquad \cos^{-1}\!\left(\frac1x\right) = \sec^{-1}x, \qquad \tan^{-1}\!\left(\frac1x\right) = \cot^{-1}x \ \ (x>0)

where: the last identity needs x>0x > 0; for x<0x < 0 it becomes tan⁡−1(1x)=−π+cot⁡−1x\tan^{-1}\left(\tfrac1x\right) = -\pi + \cot^{-1}x, because the principal ranges of tan⁡−1\tan^{-1} (an open interval around 00) and cot⁡−1\cot^{-1} (an interval that never touches 00 from below in the same way) sit differently on the negative side.

Sum and difference formulae for tan⁡−1\tan^{-1}, obtained from the ordinary tangent addition formula and then restricted to where the principal values behave consistently:

tan⁡−1x+tan⁡−1y=tan⁡−1 ⁣(x+y1−xy),xy<1;tan⁡−1x+tan⁡−1y=π+tan⁡−1 ⁣(x+y1−xy),x,y>0, xy>1\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right), \quad xy < 1; \qquad \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\!\left(\frac{x+y}{1-xy}\right), \quad x,y>0,\ xy>1

where: x,yx, y are real numbers and the sign of 1−xy1 - xy decides which version applies — this is the single most exam-tested subtlety in the chapter. A companion difference formula holds under its own restriction: tan⁡−1x−tan⁡−1y=tan⁡−1 ⁣(x−y1+xy)\tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\left(\frac{x-y}{1+xy}\right), valid when x,yx, y are suitably restricted so the combined angle stays principal.

Double-angle-style formulae, useful for simplifying 2tan⁡−1x2\tan^{-1}x:

2tan⁡−1x=sin⁡−1 ⁣(2x1+x2), ∣x∣≤1;2tan⁡−1x=tan⁡−1 ⁣(2x1−x2), −1<x<12\tan^{-1}x = \sin^{-1}\!\left(\frac{2x}{1+x^2}\right), \ |x| \le 1; \qquad 2\tan^{-1}x = \tan^{-1}\!\left(\frac{2x}{1-x^2}\right), \ -1 < x < 1

The reliable general strategy: convert everything to tan⁡−1\tan^{-1} of some quantity (often via x=tan⁡θx = \tan\theta), apply the known ordinary-trig addition formula, re-express as an inverse function, and always check the resulting angle sits inside the correct principal range before writing the final answer — adjusting by π\pi if it does not.

Figure
A number line from minus one to one showing where the raw tangent addition formula needs no correction and where a pi correction is required based on the sign of one minus x y
tan⁻¹x + tan⁻¹y: when the raw formula needs a correction xy xy = 1 xy less than 1 use formula directly xy greater than 1 (x, y positive) add pi to the raw arctan result the correction keeps the combined angle inside the principal range of tan inverse
Derivation ACombining two tan inverse terms with the addition formula

The goal is to prove, using only the ordinary tangent addition formula from Class 11 and a careful range check, that tan⁡−1(12)+tan⁡−1(13)\tan^{-1}\left(\tfrac12\right) + \tan^{-1}\left(\tfrac13\right) collapses to the single clean value π4\tfrac{\pi}{4}, illustrating exactly how the general identity tan⁡−1x+tan⁡−1y=tan⁡−1 ⁣(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right) is applied and verified in practice.

Step 1. Set x=12x = \tfrac12 and y=13y = \tfrac13, and check the validity condition before touching the formula: xy=16<1xy = \tfrac16 < 1, so the direct (no-correction) version of the addition formula is the one to use.

Step 2. Let α=tan⁡−1(12)\alpha = \tan^{-1}\left(\tfrac12\right) and β=tan⁡−1(13)\beta = \tan^{-1}\left(\tfrac13\right). By definition tan⁡α=12\tan\alpha = \tfrac12 and tan⁡β=13\tan\beta = \tfrac13, and since both arguments are positive, both α,β∈(0,π4)\alpha, \beta \in \left(0, \tfrac{\pi}{4}\right).

Step 3. Apply the ordinary tangent addition formula tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha+\beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}, substituting the values: tan⁡(α+β)=12+131−16=5656=1\tan(\alpha+\beta) = \dfrac{\tfrac12+\tfrac13}{1-\tfrac16} = \dfrac{\tfrac56}{\tfrac56} = 1.

Step 4. Convert back to inverse form: α+β=tan⁡−1(1)\alpha + \beta = \tan^{-1}(1), provided α+β\alpha+\beta lies in the principal range (−π2,π2)\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right). Since α,β∈(0,π4)\alpha, \beta \in \left(0, \tfrac{\pi}{4}\right), their sum lies in (0,π2)\left(0, \tfrac{\pi}{2}\right), which is comfortably inside that range, so no correction term is needed and the substitution is valid.

Step 5. Read off the known value tan⁡−1(1)=π4\tan^{-1}(1) = \tfrac{\pi}{4}, giving tan⁡−1(12)+tan⁡−1(13)=π4\tan^{-1}\left(\tfrac12\right) + \tan^{-1}\left(\tfrac13\right) = \tfrac{\pi}{4}, exactly as claimed. ■\blacksquare

Worked example

Evaluate sin⁡−1(35)+sin⁡−1(513)\sin^{-1}\left(\dfrac{3}{5}\right) + \sin^{-1}\left(\dfrac{5}{13}\right) as a single inverse sine.

Step 1. Let α=sin⁡−1(35)\alpha = \sin^{-1}\left(\tfrac35\right) and β=sin⁡−1(513)\beta = \sin^{-1}\left(\tfrac{5}{13}\right), so sin⁡α=35\sin\alpha = \tfrac35 and sin⁡β=513\sin\beta = \tfrac{5}{13}, with both α,β∈(0,π2)\alpha, \beta \in \left(0, \tfrac{\pi}{2}\right) since both ratios are positive and less than 1.

Step 2. Find the cosines using cos⁡=1−sin⁡2\cos = \sqrt{1-\sin^2}, positive because both angles are in the first quadrant: cos⁡α=1−925=45\cos\alpha = \sqrt{1 - \tfrac{9}{25}} = \tfrac{4}{5} and cos⁡β=1−25169=1213\cos\beta = \sqrt{1 - \tfrac{25}{169}} = \tfrac{12}{13}.

Step 3. Apply the sine addition formula: sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β=35⋅1213+45⋅513=3665+2065=5665\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \tfrac35\cdot\tfrac{12}{13} + \tfrac45\cdot\tfrac{5}{13} = \tfrac{36}{65} + \tfrac{20}{65} = \tfrac{56}{65}.

Step 4. Check the range: since sin⁡α=0.6\sin\alpha = 0.6 gives α\alpha under 40∘40^\circ and sin⁡β≈0.385\sin\beta \approx 0.385 gives β\beta under 23∘23^\circ, the sum α+β\alpha+\beta is comfortably under 90∘90^\circ, hence inside (0,π2)\left(0, \tfrac{\pi}{2}\right) — safe to write directly as sin⁡−1\sin^{-1} of the combined ratio.

Final answer: sin⁡−1(35)+sin⁡−1(513)=sin⁡−1(5665)\sin^{-1}\left(\tfrac35\right) + \sin^{-1}\left(\tfrac{5}{13}\right) = \sin^{-1}\left(\dfrac{56}{65}\right).

Competitive stretch

A common JEE-style twist reverses the question: given sin⁡−1x+sin⁡−1y=π2\sin^{-1}x + \sin^{-1}y = \tfrac{\pi}{2}, find cos⁡−1x+cos⁡−1y\cos^{-1}x + \cos^{-1}y. Using sin⁡−1x=π2−cos⁡−1x\sin^{-1}x = \tfrac{\pi}{2} - \cos^{-1}x term by term turns the given equation into (π2−cos⁡−1x)+(π2−cos⁡−1y)=π2\left(\tfrac{\pi}{2}-\cos^{-1}x\right) + \left(\tfrac{\pi}{2}-\cos^{-1}y\right) = \tfrac{\pi}{2}, so cos⁡−1x+cos⁡−1y=π2\cos^{-1}x + \cos^{-1}y = \tfrac{\pi}{2} as well — a quick substitution trick, not a coincidence, since the complementary identity applies to each variable independently.

Common traps

  • Applying tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\tfrac{x+y}{1-xy}\right) blindly when xy>1xy > 1: the raw formula gives an angle outside the principal range, and a ±π\pm\pi correction is required.
  • Treating sin⁡−1x+cos⁡−1y\sin^{-1}x + \cos^{-1}y as if it simplifies the same way sin⁡−1x+cos⁡−1x\sin^{-1}x + \cos^{-1}x does: the tidy identity needs the same argument in both terms; different arguments need a full derivation.
  • Skipping the range check after combining two angles, such as forgetting to confirm a sum of two sin⁡−1\sin^{-1} angles still lies in [−π2,π2]\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] before writing it as a single sin⁡−1\sin^{-1}.
  • Using tan⁡−1(1x)=cot⁡−1x\tan^{-1}\left(\tfrac1x\right) = \cot^{-1}x for negative xx: this reciprocal identity only holds as stated for x>0x > 0; for x<0x < 0 a −π-\pi correction is required.

Ask the tutor: Want to work through a few more tan inverse sums where you have to decide for yourself whether the pi correction applies?

Why does the tangent addition identity need a correction term at all, when sine and cosine complementary identities do not?

The complementary identities (sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}, and similarly for the other two pairs) are true for every value in the stated domain with no exceptions, because the principal ranges of the two functions were deliberately chosen to make the sum exact everywhere. The tan⁡−1\tan^{-1} addition formula is different: it is built from the ordinary tangent addition formula, which is periodic and can jump discontinuously (tangent is undefined at 90∘90^\circ). When xy>1xy > 1, the natural combined angle from the ordinary formula lands outside tan⁡−1\tan^{-1}'s narrow principal range of (−π2,π2)\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right), so you must add or subtract π\pi to bring it back inside. Always check the sign of 1−xy1-xy before trusting the raw formula.

Do I need to memorise the reciprocal identity correction for negative x, or can I always just draw a right triangle instead?

Either approach works, and many students prefer the triangle method because it avoids memorising the correction outright. But for negative xx there is no positive-side right triangle to draw directly — you have to reason about which quadrant the angle falls in, which is exactly the same work the memorised correction tan⁡−1(1x)=−π+cot⁡−1x\tan^{-1}\left(\tfrac1x\right) = -\pi + \cot^{-1}x (for x<0x<0) encodes. In practice, the safest habit is to substitute a concrete negative value like x=−1x=-1 into both sides and see the mismatch for yourself once, so the correction becomes intuitive rather than a rule you might misremember under exam pressure.

Is there a shortcut for recognising when a tan inverse sum problem needs the pi correction before I even start calculating?

Yes, a quick pre-check saves time: look at the signs and rough sizes of xx and yy. If both are positive and their product xyxy already exceeds 1 by inspection (for example x=2,y=3x=2, y=3 gives xy=6xy=6), you know immediately the correction term will be needed, without computing the combined fraction first. If xy<1xy < 1, or if xx and yy have opposite signs (which keeps xyxy negative, always less than 1), you can apply the direct formula with confidence. This quick sign-and-size check is worth doing before any algebra, since it tells you which version of the formula to reach for.

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