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MHT-CET Maths · Trigonometric Functions

Inverse Trigonometric Equations — Solve, Then Check Every Root

An equation in inverse trigonometric functions is solved by combining terms with a complementary pair or the addition formula, or by converting both sides to one ratio, and every root found must then be checked against the domains and principal ranges it passed through.

All Trigonometric Functions notesMHT-CET Maths strategy

Why this matters

29 PYQs. Twelve combine two or three arctangents with the addition formula, nine convert both sides to one function or substitute x = tan θ, and eight use a complementary pair. The algebra is short. The HARD rows are the ones where a root the algebra produced must be thrown out — a negative root when x ≥ 0, a root that makes xy > 1, a root outside a domain — and the answer is a COUNT of roots.

Concept 1 of 3: Equations Solved by a Complementary Pair

When an equation mixes sin⁡−1x\sin^{-1}x with cos⁡−1x\cos^{-1}x (or tan⁡−1\tan^{-1} with cot⁡−1\cot^{-1}) of the SAME argument, replace one by π2\frac{\pi}{2} minus the other. The equation becomes linear, or quadratic, in a single inverse value.

Definition

  • Replace cos⁡−1x\cos^{-1}x by π2−sin⁡−1x\frac{\pi}{2} - \sin^{-1}x: 4sin⁡−1x+cos⁡−1x=π4\sin^{-1}x + \cos^{-1}x = \pi becomes 3sin⁡−1x=π23\sin^{-1}x = \frac{\pi}{2}.
  • sin⁡−1x13+csc⁡−11312=π2\sin^{-1}\frac{x}{13} + \csc^{-1}\frac{13}{12} = \frac{\pi}{2}: csc⁡−11312=sin⁡−11213\csc^{-1}\frac{13}{12} = \sin^{-1}\frac{12}{13}, so sin⁡−1x13=cos⁡−11213=sin⁡−1513\sin^{-1}\frac{x}{13} = \cos^{-1}\frac{12}{13} = \sin^{-1}\frac{5}{13}.
  • Squares: with t=tan⁡−1xt = \tan^{-1}x, (tan⁡−1x)2+(cot⁡−1x)2=k(\tan^{-1}x)^2 + (\cot^{-1}x)^2 = k is t2+(π2−t)2=kt^2 + \left(\frac{\pi}{2} - t\right)^2 = k, a quadratic in tt. Keep only roots in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).
  • tan⁡−1(1+x)+tan⁡−1(1−x)=π2\tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2} says the two are complementary, so (1+x)(1−x)=1(1 + x)(1 - x) = 1.

Replace one of the pair

cos⁡−1x=π2−sin⁡−1xcot⁡−1x=π2−tan⁡−1x\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x \qquad \cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x

Worked example

Solve 2sin⁡−1x+3cos⁡−1x=4π32\sin^{-1}x + 3\cos^{-1}x = \frac{4\pi}{3}.
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q123Easy

Example 1 · Trigonometric Functions · Inverse Trigonometric Equations

If 4sin⁡−1x+cos⁡−1x=π4\sin^{- 1}x+\cos^{- 1}x=\pi then x=x=

Keeping the root outside the range

2t2−πt−3π28=02t^2 - \pi t - \frac{3\pi^2}{8} = 0 gives t=−π4t = -\frac{\pi}{4} or 3π4\frac{3\pi}{4}, but t=tan⁡−1xt = \tan^{-1}x lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Only −π4-\frac{\pi}{4} is allowed, so x=−1x = -1.
Drill 7 more on equations solved by a complementary pair

Concept 2 of 3: Equations Solved by the Addition Formula — and the Roots It Adds

Combine the inverse tangents on one side into a single tan⁡−1\tan^{-1}, take the tangent of both sides, and solve the algebraic equation. The formula was only valid while xy<1xy < 1, so substitute every root back; the questions that ask how many elements a solution set has are testing exactly this check.

Definition

  • tan⁡−1ax+tan⁡−1bx=π4\tan^{-1}ax + \tan^{-1}bx = \frac{\pi}{4} gives (a+b)x1−abx2=1\dfrac{(a + b)x}{1 - abx^2} = 1, a quadratic. With x≥0x \ge 0 required, only the positive root counts — so the set is a SINGLETON even though the quadratic has two roots.
  • Three terms: combine two first, then the third. tan⁡−1(x+1)+tan⁡−1(x−1)+tan⁡−1x=tan⁡−13\tan^{-1}(x + 1) + \tan^{-1}(x - 1) + \tan^{-1}x = \tan^{-1}3 reduces to a cubic; use the stated condition (x<0x < 0) to pick the root.
  • tan⁡−11−x1+x=π4−tan⁡−1x\tan^{-1}\frac{1 - x}{1 + x} = \frac{\pi}{4} - \tan^{-1}x (for x>−1x > -1): the equation tan⁡−11−x1+x=12tan⁡−1x\tan^{-1}\frac{1 - x}{1 + x} = \frac12\tan^{-1}x becomes tan⁡−1x=π6\tan^{-1}x = \frac{\pi}{6}.
  • Sine forms: sin⁡−113+sin⁡−135+sin⁡−1x=π2\sin^{-1}\frac13 + \sin^{-1}\frac35 + \sin^{-1}x = \frac{\pi}{2} means x=cos⁡(sin⁡−113+sin⁡−135)x = \cos\left(\sin^{-1}\frac13 + \sin^{-1}\frac35\right), a compound-angle expansion.
  • Counting solutions: an equation PQ=P\frac{P}{Q} = P has roots from P=0P = 0 AND from Q=1Q = 1; check each for existence.

Combine, then take tangents

tan⁡−1u+tan⁡−1v=θ  ⇒  u+v1−uv=tan⁡θ(then check uv<1)\tan^{-1}u+\tan^{-1}v=\theta \;\Rightarrow\; \frac{u+v}{1-uv}=\tan\theta\quad(\text{then check }uv<1)

Worked example

How many non-negative solutions does tan⁡−1x+tan⁡−12x=π4\tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4} have?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q106Moderate

Example 2 · Trigonometric Functions · Inverse Trigonometric Equations

Considering only the principal values of inverse function, the set A={x≥0 | tan⁡−1(2x)+tan⁡−1(3x)=π4}A = \left\{x \ge 0\,\middle|\, \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\right\}

Answering with the number of roots of the quadratic

6x2+5x−1=06x^2 + 5x - 1 = 0 has two roots, 16\frac16 and −1-1, but the set was defined with x≥0x \ge 0. 'Contains two elements' is the planted option; the set is a singleton.
Drill 11 more on equations solved by the addition formula — and the roots it adds

Concept 3 of 3: Converting Both Sides to One Function, Substituting, and Checking the Domain

When the two sides use different functions — sin⁡(cot⁡−1x)\sin(\cot^{-1}x) against cos⁡(tan⁡−1(1+x))\cos(\tan^{-1}(1 + x)) — convert each to an algebraic expression with a right triangle and equate. When the arguments are 2x1+x2\frac{2x}{1 + x^2} and its relatives, substitute x=tan⁡θx = \tan\theta. Either way, the final step is to check the domain the square roots and inverse functions impose.

Definition

  • sin⁡(cot⁡−1u)=cos⁡(tan⁡−1u)=11+u2\sin(\cot^{-1}u) = \cos(\tan^{-1}u) = \dfrac{1}{\sqrt{1 + u^2}}, so sin⁡(cot⁡−1x)=cos⁡(tan⁡−1(1+x))\sin(\cot^{-1}x) = \cos(\tan^{-1}(1 + x)) gives 1+x2=1+(1+x)21 + x^2 = 1 + (1 + x)^2, x=−12x = -\frac12.
  • Substitute x=tan⁡θx = \tan\theta: sin⁡−12x1+x2=2θ\sin^{-1}\frac{2x}{1 + x^2} = 2\theta, cos⁡−11−x21+x2=2θ\cos^{-1}\frac{1 - x^2}{1 + x^2} = 2\theta, tan⁡−12x1−x2=2θ\tan^{-1}\frac{2x}{1 - x^2} = 2\theta (for ∣x∣<1|x| < 1), which turns a three-term equation into one in θ\theta.
  • Domain first: tan⁡−1x(x+1)+sin⁡−1x2+x+1=π2\tan^{-1}\sqrt{x(x + 1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{2} needs x(x+1)≥0x(x + 1) \ge 0 and x2+x+1≤1x^2 + x + 1 \le 1. Together they force x(x+1)=0x(x + 1) = 0: two solutions, 00 and −1-1.
  • Complementary square roots: cos⁡−1p+cos⁡−11−p=π2\cos^{-1}\sqrt{p} + \cos^{-1}\sqrt{1 - p} = \frac{\pi}{2} for 0≤p≤10 \le p \le 1.
  • Squaring can admit a root with the wrong sign: substitute back. In sin⁡−14x+sin⁡−143x=−π2\sin^{-1}4x + \sin^{-1}4\sqrt3 x = -\frac{\pi}{2}, squaring gives x=±18x = \pm\frac18, but only −18-\frac18 satisfies the original.

Two conversions and a substitution

sin⁡(cot⁡−1u)=cos⁡(tan⁡−1u)=11+u2x=tan⁡θ: sin⁡−12x1+x2=2θ (∣x∣≤1)\sin(\cot^{-1}u)=\cos(\tan^{-1}u)=\frac{1}{\sqrt{1+u^2}} \qquad x=\tan\theta:\ \sin^{-1}\frac{2x}{1+x^2}=2\theta\ (|x|\le1)

Worked example

Solve cos⁡(tan⁡−1x)=sin⁡(cot⁡−1(x−2))\cos(\tan^{-1}x) = \sin(\cot^{-1}(x - 2)).
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q114Moderate

Example 3 · Trigonometric Functions · Inverse Trigonometric Equations

If sin⁡(cot⁡−1(x+1))=cos⁡(tan⁡−1x)\sin(\cot^{-1}(x+1))=\cos(\tan^{-1}x), then considering positive square roots, xx has the value

Reporting both signs after squaring

Squaring sin⁡−14x+sin⁡−143x=−π2\sin^{-1}4x + \sin^{-1}4\sqrt3 x = -\frac{\pi}{2} gives x=±18x = \pm\frac18, but x=18x = \frac18 makes the left side +π2+\frac{\pi}{2}. Substitute each candidate. (The paper's options are all written with ±\pm, and its key is ±18\pm\frac18.)
Drill 8 more on converting both sides to one function, substituting, and checking the domain

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 (3)

  • Equations Solved by a Complementary Pair

    Replace one of the pair

    cos⁡−1x=π2−sin⁡−1xcot⁡−1x=π2−tan⁡−1x\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x \qquad \cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x
  • Equations Solved by the Addition Formula — and the Roots It Adds

    Combine, then take tangents

    tan⁡−1u+tan⁡−1v=θ  ⇒  u+v1−uv=tan⁡θ(then check uv<1)\tan^{-1}u+\tan^{-1}v=\theta \;\Rightarrow\; \frac{u+v}{1-uv}=\tan\theta\quad(\text{then check }uv<1)
  • Converting Both Sides to One Function, Substituting, and Checking the Domain

    Two conversions and a substitution

    sin⁡(cot⁡−1u)=cos⁡(tan⁡−1u)=11+u2x=tan⁡θ: sin⁡−12x1+x2=2θ (∣x∣≤1)\sin(\cot^{-1}u)=\cos(\tan^{-1}u)=\frac{1}{\sqrt{1+u^2}} \qquad x=\tan\theta:\ \sin^{-1}\frac{2x}{1+x^2}=2\theta\ (|x|\le1)

Watch out for (3)

Test yourself on Trigonometric Functions

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