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MHT-CET Maths · Trigonometric Functions

Inverse Trigonometric Identities — Complementary Pairs, the Addition Formula, Substitution and Telescoping

Sums of inverse trigonometric values collapse through four tools: complementary pairs that add to π/2, the arctan addition formula, a substitution that turns an algebraic argument into a single angle, and a telescoping split of each term into a difference.

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Why this matters

30 PYQs, 63% of them HARD — the hardest page of the chapter. Fourteen are the addition formula (three-term sums, 2 tan⁻¹ forms, and the identity a + b + c = abc when three arctangents sum to π), eight are substitution simplifications, four telescope and four use a complementary pair. Each tool has one signal in the stem, and seeing it first is the whole difficulty.

Concept 1 of 4: Complementary Pairs — sin⁻¹x + cos⁻¹x = π/2

If an angle has sine xx, its complement has cosine xx. So sin⁡−1x\sin^{-1}x and cos⁡−1x\cos^{-1}x are complementary for every xx in [−1,1][-1, 1], and any expression mixing the two for the same argument collapses to a constant.

Definition

  • sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} (∣x∣≤1)(|x| \le 1), tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, sec⁡−1x+csc⁡−1x=π2\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2} (∣x∣≥1)(|x| \ge 1).
  • Swap pairs: cos⁡−1x+sin⁡−1y=π−(sin⁡−1x+cos⁡−1y)\cos^{-1}x + \sin^{-1}y = \pi - (\sin^{-1}x + \cos^{-1}y).
  • Reciprocals: cot⁡−1x=tan⁡−11x\cot^{-1}x = \tan^{-1}\frac1x for x>0x > 0; csc⁡−1x=sin⁡−11x\csc^{-1}x = \sin^{-1}\frac1x.
  • cot⁡−1u−tan⁡−1u=x\cot^{-1}u - \tan^{-1}u = x becomes π2−2tan⁡−1u=x\frac{\pi}{2} - 2\tan^{-1}u = x, so sin⁡x=cos⁡(2tan⁡−1u)=1−u21+u2\sin x = \cos(2\tan^{-1}u) = \dfrac{1 - u^2}{1 + u^2}. With u=cos⁡αu = \cos\alpha or u=cos⁡αu = \sqrt{\cos\alpha} that becomes a half-angle expression.
  • Three cosines summing to π\pi: if cos⁡−1x+cos⁡−1y+cos⁡−1z=π\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi, then x2+y2+z2+2xyz=1x^2 + y^2 + z^2 + 2xyz = 1.

The three complementary pairs

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

Worked example

If sin⁡−1x+cos⁡−1y=2π5\sin^{-1}x + \cos^{-1}y = \frac{2\pi}{5}, find cos⁡−1x+sin⁡−1y\cos^{-1}x + \sin^{-1}y.
Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q146Moderate

Example 1 · Trigonometric Functions · Inverse Trigonometric Identities — Sums, Substitution and Telescoping

If sin⁡−1x+cos⁡−1y=3π10\sin^{-1}x+\cos^{-1}y=\frac{3\pi}{10}, then the value of cos⁡−1x+sin⁡−1y\cos^{-1}x+\sin^{-1}y is
Drill 3 more on complementary pairs — sin⁻¹x + cos⁻¹x = π/2

Concept 2 of 4: The Addition Formula — tan⁻¹x ± tan⁻¹y, 2 tan⁻¹x and Three-Term Sums

tan⁡−1x+tan⁡−1y\tan^{-1}x + \tan^{-1}y is the angle whose tangent is tan⁡(α+β)\tan(\alpha + \beta), which is the compound-angle formula with x,yx, y in place of tan⁡α,tan⁡β\tan\alpha, \tan\beta. Chain it for three terms, and convert any sine or cosine term to a tangent first.

Definition

  • tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\tan^{-1}x + \tan^{-1}y = \tan^{-1}\dfrac{x + y}{1 - xy} while xy<1xy < 1; tan⁡−1x−tan⁡−1y=tan⁡−1x−y1+xy\tan^{-1}x - \tan^{-1}y = \tan^{-1}\dfrac{x - y}{1 + xy}.
  • 2tan⁡−1x=tan⁡−12x1−x2=sin⁡−12x1+x2=cos⁡−11−x21+x22\tan^{-1}x = \tan^{-1}\dfrac{2x}{1 - x^2} = \sin^{-1}\dfrac{2x}{1 + x^2} = \cos^{-1}\dfrac{1 - x^2}{1 + x^2} (for 0≤x≤10 \le x \le 1).
  • Sine sums: sin⁡−1a+sin⁡−1b=sin⁡−1(a1−b2+b1−a2)\sin^{-1}a + \sin^{-1}b = \sin^{-1}\left(a\sqrt{1 - b^2} + b\sqrt{1 - a^2}\right). Or convert: sin⁡−145=tan⁡−143\sin^{-1}\frac45 = \tan^{-1}\frac43, sin⁡−1513=tan⁡−1512\sin^{-1}\frac{5}{13} = \tan^{-1}\frac{5}{12}, and their sum is tan⁡−16316\tan^{-1}\frac{63}{16}, whose complement is sin⁡−11665\sin^{-1}\frac{16}{65}.
  • Three arctangents summing to π\pi: a+b+c=abca + b + c = abc. Summing to π2\frac{\pi}{2}: ab+bc+ca=1ab + bc + ca = 1.
  • Difference of two inverse cosines: cos⁡−1x−cos⁡−1yk=α\cos^{-1}x - \cos^{-1}\frac{y}{k} = \alpha leads (after taking cosines and squaring) to k2x2−2kxycos⁡α+y2=k2sin⁡2αk^2x^2 - 2kxy\cos\alpha + y^2 = k^2\sin^2\alpha.

Addition formula

tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy (xy<1)2tan⁡−1x=tan⁡−12x1−x2\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\ (xy<1) \qquad 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}

Worked example

Evaluate tan⁡−113+tan⁡−114+tan⁡−129\tan^{-1}\frac13 + \tan^{-1}\frac14 + \tan^{-1}\frac29.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q137Moderate

Example 2 · Trigonometric Functions · Inverse Trigonometric Identities — Sums, Substitution and Telescoping

The value of 2tan⁡−1 ⁣12+tan⁡−1 ⁣172\tan^{-1}\!\frac{1}{2}+\tan^{-1}\!\frac{1}{7}

Forgetting the xy < 1 condition

When xy>1xy > 1 and x,y>0x, y > 0, the sum is π+tan⁡−1x+y1−xy\pi + \tan^{-1}\frac{x + y}{1 - xy}. The formula without the π\pi gives a negative angle for a sum of two positive ones — a sign that the condition was ignored.
Drill 13 more on the addition formula — tan⁻¹x ± tan⁻¹y, 2 tan⁻¹x and three-term sums

Concept 3 of 4: Simplifying by Substitution — x = tan θ, cos θ or cos 2θ

An argument like 1−x21+x2\frac{1 - x^2}{1 + x^2} or 1+x−1−x1+x+1−x\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} is a trigonometric ratio in disguise. Choose the substitution that turns it into one ratio of one angle; the inverse then cancels, provided the angle is inside the principal range.

Definition

  • Signals and substitutions: 1+x21 + x^2 → x=tan⁡θx = \tan\theta; 1−x21 - x^2 → x=sin⁡θx = \sin\theta or cos⁡θ\cos\theta; 1±x\sqrt{1 \pm x} → x=cos⁡2θx = \cos 2\theta (then 1+x=2cos⁡θ\sqrt{1 + x} = \sqrt2\cos\theta, 1−x=2sin⁡θ\sqrt{1 - x} = \sqrt2\sin\theta).
  • With x=tan⁡θx = \tan\theta: 2x1+x2=sin⁡2θ\frac{2x}{1 + x^2} = \sin 2\theta, 1−x21+x2=cos⁡2θ\frac{1 - x^2}{1 + x^2} = \cos 2\theta, 2x1−x2=tan⁡2θ\frac{2x}{1 - x^2} = \tan 2\theta, 1−x22x=cot⁡2θ\frac{1 - x^2}{2x} = \cot 2\theta.
  • cos⁡θ+sin⁡θcos⁡θ−sin⁡θ=tan⁡(π4+θ)\dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \tan\left(\frac{\pi}{4} + \theta\right); sec⁡x+tan⁡x=tan⁡(π4+x2)\sec x + \tan x = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right).
  • tan⁡(π4+θ)+tan⁡(π4−θ)=2cos⁡2θ\tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = \frac{2}{\cos 2\theta}.
  • Check the range: cos⁡−1(cos⁡2θ)=2θ\cos^{-1}(\cos 2\theta) = 2\theta only for 0≤2θ≤π0 \le 2\theta \le \pi; the stem's condition on xx (such as 0<x<10 < x < 1) is what guarantees it.

The tan θ substitution

x=tan⁡θ:2x1+x2=sin⁡2θ,1−x21+x2=cos⁡2θ,2x1−x2=tan⁡2θx=\tan\theta:\quad \frac{2x}{1+x^2}=\sin2\theta,\quad \frac{1-x^2}{1+x^2}=\cos2\theta,\quad \frac{2x}{1-x^2}=\tan2\theta

Worked example

Simplify tan⁡−1cos⁡x1−sin⁡x\tan^{-1}\dfrac{\cos x}{1 - \sin x} for −π2<x<π2-\frac{\pi}{2} < x < \frac{\pi}{2}.
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q130Hard

Example 3 · Trigonometric Functions · Inverse Trigonometric Identities — Sums, Substitution and Telescoping

tan⁡ ⁣(π4+12cos⁡−1ab)+tan⁡ ⁣(π4−12cos⁡−1ab)\tan\!\left(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\frac{a}{b}\right)+\tan\!\left(\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\frac{a}{b}\right) is

Cancelling an inverse outside its range

tan⁡−1(tan⁡2θ)=2θ\tan^{-1}(\tan 2\theta) = 2\theta needs −π4<θ<π4-\frac{\pi}{4} < \theta < \frac{\pi}{4}. If x=tan⁡θ>1x = \tan\theta > 1, the cancellation gives the wrong branch; read the stated range of xx before cancelling.
Drill 7 more on simplifying by substitution — x = tan θ, cos θ or cos 2θ

Concept 4 of 4: Telescoping Sums of Inverse Tangents

A long sum of inverse tangents always hides the subtraction formula backwards: each term is tan⁡−1A−tan⁡−1B\tan^{-1}A - \tan^{-1}B where A−BA - B is the numerator and 1+AB1 + AB the denominator. Write each term that way and everything but the first and last cancels.

Definition

  • Split: tan⁡−1A−B1+AB=tan⁡−1A−tan⁡−1B\tan^{-1}\dfrac{A - B}{1 + AB} = \tan^{-1}A - \tan^{-1}B. Find AA and BB whose difference is the numerator and whose product is the denominator minus 1.
  • cot⁡−1(n2+n+1)=tan⁡−111+n(n+1)=tan⁡−1(n+1)−tan⁡−1n\cot^{-1}(n^2 + n + 1) = \tan^{-1}\dfrac{1}{1 + n(n + 1)} = \tan^{-1}(n + 1) - \tan^{-1}n.
  • tan⁡−112r2=tan⁡−124r2=tan⁡−1(2r+1)−tan⁡−1(2r−1)\tan^{-1}\dfrac{1}{2r^2} = \tan^{-1}\dfrac{2}{4r^2} = \tan^{-1}(2r + 1) - \tan^{-1}(2r - 1).
  • tan⁡−12n−11+22n−1=tan⁡−12n−tan⁡−12n−1\tan^{-1}\dfrac{2^{n-1}}{1 + 2^{2n-1}} = \tan^{-1}2^n - \tan^{-1}2^{n-1}; the infinite sum is π2−π4\frac{\pi}{2} - \frac{\pi}{4}.
  • The last step is usually tan⁡(tan⁡−1P−tan⁡−1Q)=P−Q1+PQ\tan(\tan^{-1}P - \tan^{-1}Q) = \frac{P - Q}{1 + PQ} or its cotangent.

The split

tan⁡−1A−B1+AB=tan⁡−1A−tan⁡−1B\tan^{-1}\frac{A-B}{1+AB}=\tan^{-1}A-\tan^{-1}B

Worked example

Evaluate ∑n=110tan⁡−11n2+n+1\sum_{n=1}^{10}\tan^{-1}\dfrac{1}{n^2 + n + 1}.
Practice this conceptself-check

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q134Hard

Example 4 · Trigonometric Functions · Inverse Trigonometric Identities — Sums, Substitution and Telescoping

If ∑r=150tan⁡−112r2=p\sum_{r=1}^{50} \tan^{-1}\frac{1}{2r^2} = p, then tan⁡p\tan p is
Drill 3 more on telescoping sums of inverse tangents

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)

  • Complementary Pairs — sin⁻¹x + cos⁻¹x = π/2

    The three complementary pairs

    sin⁡−1x+cos⁡−1x=tan⁡−1x+cot⁡−1x=sec⁡−1x+csc⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\sec^{-1}x+\csc^{-1}x=\frac{\pi}{2}
  • The Addition Formula — tan⁻¹x ± tan⁻¹y, 2 tan⁻¹x and Three-Term Sums

    Addition formula

    tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy (xy<1)2tan⁡−1x=tan⁡−12x1−x2\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\ (xy<1) \qquad 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}
  • Simplifying by Substitution — x = tan θ, cos θ or cos 2θ

    The tan θ substitution

    x=tan⁡θ:2x1+x2=sin⁡2θ,1−x21+x2=cos⁡2θ,2x1−x2=tan⁡2θx=\tan\theta:\quad \frac{2x}{1+x^2}=\sin2\theta,\quad \frac{1-x^2}{1+x^2}=\cos2\theta,\quad \frac{2x}{1-x^2}=\tan2\theta
  • Telescoping Sums of Inverse Tangents

    The split

    tan⁡−1A−B1+AB=tan⁡−1A−tan⁡−1B\tan^{-1}\frac{A-B}{1+AB}=\tan^{-1}A-\tan^{-1}B

Watch out for (2)

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