MathsISC Class 12 · Inverse Trigonometric Functions

Solving Inverse Trigonometric Equations: Method and Root Checks

Updated 15 October 2026 · Fact-checked

To solve an inverse trigonometric equation, combine the terms into one inverse function using an identity, or move one term across and take sin, cos or tan of both sides. This gives an algebraic equation. Solve it, then substitute every root back, because squaring and the sum formulas can create roots that do not satisfy the original.

Pick the route from the look of the equation, solve the algebra, and always substitute every root back into the original.

Illustration drawn for QuestionBuddy.

Understand Solving Inverse Trigonometric Equations

An inverse trigonometric equation has the unknown inside $sin^{- 1}$ , $cos^{- 1}$ , $tan^{- 1}$ and so on, such as $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{4}$ . The aim is to remove the inverse functions and reach ordinary algebra. There are three routes, and choosing the right one is half the work.

Route 1: combine, then take the function. If every term is $tan^{- 1}$ , use the sum formula to combine the left side into a single $tan^{- 1}$ , then take $tan$ of both sides. For $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{4}$ this gives $\frac{10 x}{1 - 24 x ^{2}} = 1$ , a quadratic.

Route 2: use a complementary pair. If the equation mixes $sin^{- 1} x$ and $cos^{- 1} x$ (or $tan^{- 1} x$ and $cot^{- 1} x$ ) of the same argument, replace one by $\frac{π}{2}$ minus the other. The equation $sin^{- 1} x = \frac{π}{6} + cos^{- 1} x$ becomes $2 sin^{- 1} x = \frac{2 π}{3}$ at once.

Route 3: isolate one term, then take sin or cos. For mixed arguments, such as $sin^{- 1} 2 x + sin^{- 1} x = \frac{π}{3}$ , write $sin^{- 1} 2 x = \frac{π}{3} - sin^{- 1} x$ and take $sin$ of both sides. Expanding $sin (A - B)$ needs $cos (sin^{- 1} x) = 1 - x^{2}$ , and the root sign usually forces one squaring. This is the method expected in the ISC 2024 question $sin^{- 1} \frac{x}{2} + cos^{- 1} x = \frac{π}{6}$ , where the examiners commented that some candidates chose lengthy calculations and could not finish.

Why roots must be checked. Taking $tan$ of both sides loses information: $tan θ = 1$ is true for $θ = \frac{π}{4}$ and also for $θ = - \frac{3 π}{4}$ . Squaring loses signs. And the sum formula $tan^{- 1} x + tan^{- 1} y = tan^{- 1} \frac{x + y}{1 - x y}$ needs $x y < 1$ . So the algebra can give values that do not satisfy the original equation. Every candidate root must pass three checks:

Domain: arguments of $sin^{- 1}$ and $cos^{- 1}$ must lie in $[- 1, 1]$ .
Sign and size: the left side, evaluated at the root, must have the same sign and size as the right side. For example, if $x < 0$ then $tan^{- 1} 4 x + tan^{- 1} 6 x < 0$ , so it cannot equal $\frac{π}{4}$ .
Substitution: put the root back and confirm, ideally with exact values.

Equations with algebraic arguments. When an argument looks like $\frac{2 x}{1 - x ^{2}}$ , $\frac{1 - x ^{2}}{1 + x ^{2}}$ or $2 x 1 - x^{2}$ , substitute $x = tan θ$ or $x = sin θ$ to collapse it to a multiple of $θ$ . The domain of $x$ then decides which branch you land in, and that often changes the sign or adds $π$ .

Equations with no solution. Ranges set limits. $tan^{- 1} x + cot^{- 1} x$ is always $\frac{π}{2}$ , so it can never equal $\frac{π}{3}$ . Recognising such a case quickly earns the mark without any algebra.

Solving Inverse Trigonometric Equations: formulas to remember

Complementary pairs: $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}, tan^{- 1} x + cot^{- 1} x = \frac{π}{2}$; Use when the same argument appears in both functions.
Sum of tan inverses: $tan^{- 1} x + tan^{- 1} y = tan^{- 1} \frac{x + y}{1 - x y}, x y < 1$; Check xy < 1 for every root you keep.
Cos of a sin inverse: $cos (sin^{- 1} x) = 1 - x^{2}, sin (cos^{- 1} x) = 1 - x^{2}$; Both are non-negative on [-1, 1].
Sine of a difference: $sin (A - B) = sin A cos B - cos A sin B$; Used after isolating one inverse term (Route 3).
Double angle substitution: $x = tan θ \Rightarrow \frac{2 x}{1 - x ^{2}} = tan 2 θ, \frac{1 - x ^{2}}{1 + x ^{2}} = cos 2 θ$; Track the range of θ from the domain of x.
Principal ranges: $sin^{- 1} : [- \frac{π}{2}, \frac{π}{2}], cos^{- 1} : [0, π], tan^{- 1} : (- \frac{π}{2}, \frac{π}{2})$; A right-hand side outside the possible range means no solution.

How to solve Solving Inverse Trigonometric Equations questions

Follow these steps for any "Solve for x" question on inverse trigonometric functions.

1Write down the domain restrictions: every argument of $sin^{- 1}$ or $cos^{- 1}$ must lie in $[- 1, 1]$ .
2Choose the route: same argument in a complementary pair (Route 2); all $tan^{- 1}$ (Route 1); mixed arguments (Route 3, isolate one term).
3Take $tan$ , $sin$ or $cos$ of both sides and expand, using $cos (sin^{- 1} x) = 1 - x^{2}$ where needed.
4If a square root remains, isolate it and square once. Simplify to a polynomial equation.
5Solve the polynomial, factorising where possible.
6Check each root: domain, the sign of the left side, and the condition $x y < 1$ if you used the $tan^{- 1}$ sum formula. Reject any root that fails, and say why.
7State the final answer clearly, for example " $x = \frac{1}{12}$ ( $x = - \frac{1}{2}$ is rejected)".

Solving Inverse Trigonometric Equations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Sum of inverse tangents equal to π/2

Suppose a drone hovers at a height of $h$ m directly above a point P on flat ground. Riya stands 30 m from P and Kabir stands 10 m from P on the same side. The angles of elevation of the drone from Riya and Kabir add up to a right angle.

Form an inverse trigonometric equation and find $h$ .

Try it first, then show the solutionSolution

Elevation from Riya $= tan^{- 1} \frac{h}{30}$ ; from Kabir $= tan^{- 1} \frac{h}{10}$ .
Equation: $tan^{- 1} \frac{h}{30} + tan^{- 1} \frac{h}{10} = \frac{π}{2}$ .
Since $tan^{- 1} a + tan^{- 1} \frac{1}{a} = \frac{π}{2}$ for $a > 0$ , we need $\frac{h}{10} = \frac{30}{h}$ , that is $\frac{h}{30} \cdot \frac{h}{10} = 1$ .
$h^{2} = 300 \Rightarrow h = 103 \approx 17.3$ m (height is positive).
Check: $tan^{- 1} \frac{1}{3} + tan^{- 1} 3 = \frac{π}{6} + \frac{π}{3} = \frac{π}{2}$ .

Answer: The drone is at $103 \approx 17.3$ m.

Memory hook: Two elevation angles adding to π/2 means the two tangents multiply to 1.

Equation with 2tan⁻¹x

Imagine you look at the top of a lamp post from a point A on level ground, and the angle of elevation is $tan^{- 1} \frac{1}{2}$ . You walk 6 m straight towards the post and the angle of elevation becomes exactly twice as large.

Find the height of the lamp post and your starting distance from it (ignore your own height).

Try it first, then show the solutionSolution

Let the height be $h$ and the starting distance $d$ . From A: $\frac{h}{d} = \frac{1}{2}$ , so $d = 2 h$ .
The new angle is $2 tan^{- 1} \frac{1}{2} = tan^{- 1} \frac{1}{1 - \frac{1}{4}} = tan^{- 1} \frac{4}{3}$ .
From the new point: $\frac{h}{d - 6} = \frac{4}{3} \Rightarrow 3 h = 4 (2 h - 6) = 8 h - 24$ .
$5 h = 24 \Rightarrow h = 4.8$ m, and $d = 9.6$ m.
Check: $\frac{4.8}{3.6} = \frac{4}{3}$ .

Answer: The lamp post is 4.8 m tall and you started 9.6 m from it.

Memory hook: Doubling an angle means using 2tan⁻¹x = tan⁻¹(2x/(1 − x²)).

Tricks to remember Solving Inverse Trigonometric Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Same argument? Pair it.
If sin⁻¹x and cos⁻¹x (or tan⁻¹x and cot⁻¹x) appear with the same x, replace one by π/2 minus the other before anything else.
Use it when: Spotting one-line solutions.
Isolate, then take the function
Move one term across so that each side becomes one angle; then take sin, cos or tan.
Use it when: Mixed equations like sin⁻¹2x + sin⁻¹x = π/3.
Every root on trial
Domain, sign, substitute: a root must pass all three before you keep it.
Use it when: Avoiding extra roots after squaring or the tan⁻¹ sum formula.
Negative x, negative angles
sin⁻¹ and tan⁻¹ of negative numbers are negative, so they cannot add up to a positive right-hand side.
Use it when: Quick rejection of negative roots.

Solving Inverse Trigonometric Equations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two inverse tangents equal to π/4 · 3 marks · Short answerBoard level

Solve for $x$ : $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{4}$ .

Try it first, then show the solutionSolution

Assuming $24 x^{2} < 1$ : $tan^{- 1} \frac{4 x + 6 x}{1 - 24 x ^{2}} = \frac{π}{4}$ , so $\frac{10 x}{1 - 24 x ^{2}} = 1$ . (1 mark)
$24 x^{2} + 10 x - 1 = 0 \Rightarrow (12 x - 1) (2 x + 1) = 0 \Rightarrow x = \frac{1}{12}$ or $x = - \frac{1}{2}$ . (1 mark)
For $x = - \frac{1}{2}$ , the left side is $tan^{- 1} (- 2) + tan^{- 1} (- 3)$ : both terms are negative, so it cannot equal $\frac{π}{4}$ (also $24 x^{2} = 6 > 1$ ). Reject. (½ mark)
For $x = \frac{1}{12}$ : $24 x^{2} = \frac{1}{6} < 1$ and $tan^{- 1} \frac{1}{3} + tan^{- 1} \frac{1}{2} = \frac{π}{4}$ . (½ mark)

Answer: $x = \frac{1}{12}$ .

Example 2: A complementary pair in one line · 1 mark · MCQBoard level

If $sin^{- 1} x = \frac{π}{6} + cos^{- 1} x$ , then $x$ equals:
(a) $\frac{1}{2}$
(b) $\frac{1}{2}$
(c) $\frac{3}{2}$
(d) $1$

Try it first, then show the solutionSolution

Put $cos^{- 1} x = \frac{π}{2} - sin^{- 1} x$ : $2 sin^{- 1} x = \frac{2 π}{3}$ , so $sin^{- 1} x = \frac{π}{3}$ and $x = \frac{3}{2}$ .

Answer: (c) $\frac{3}{2}$ .

Example 3: An equation with three roots · 3 marks · Short answerHarder board level

Solve: $cos^{- 1} x - sin^{- 1} x = cos^{- 1} (3 x)$ .

Try it first, then show the solutionSolution

Domain: $x \in [- 1, 1]$ and $3 x \in [- 1, 1]$ , so $∣ x ∣ \leq \frac{1}{3}$ . (½ mark)
Put $cos^{- 1} x = \frac{π}{2} - sin^{- 1} x$ : the left side becomes $\frac{π}{2} - 2 sin^{- 1} x$ . For $∣ x ∣ \leq \frac{1}{3}$ this lies in $[0, π]$ , where cosine is one-one, so taking $cos$ of both sides gives an equivalent equation. (1 mark)
$cos (\frac{π}{2} - 2 sin^{- 1} x) = sin (2 sin^{- 1} x) = 2 x 1 - x^{2}$ , so $2 x 1 - x^{2} = 3 x$ . (½ mark)
$x (2 1 - x^{2} - 3) = 0 \Rightarrow x = 0$ or $1 - x^{2} = \frac{3}{4}$ , that is $x = \pm \frac{1}{2}$ . All three lie in the domain. Do not cancel $x$ , or the root $x = 0$ is lost. (½ mark)
Check: $x = 0$ gives $\frac{π}{2} - 0 = cos^{- 1} 0$ ; $x = \frac{1}{2}$ gives $\frac{π}{3} - \frac{π}{6} = \frac{π}{6} = cos^{- 1} \frac{3}{2}$ ; $x = - \frac{1}{2}$ gives $\frac{2 π}{3} + \frac{π}{6} = \frac{5 π}{6} = cos^{- 1} (- \frac{3}{2})$ . (½ mark)

Answer: $x = 0$ or $x = \pm \frac{1}{2}$ .

Example 4: Mixed arguments with one squaring · 3 marks · Short answerHarder board level

Solve: $sin^{- 1} 2 x + sin^{- 1} x = \frac{π}{3}$ .

Try it first, then show the solutionSolution

Write $sin^{- 1} 2 x = \frac{π}{3} - sin^{- 1} x$ and take $sin$ : $2 x = \frac{3}{2} 1 - x^{2} - \frac{1}{2} x$ . (1 mark)
$\frac{5 x}{2} = \frac{3}{2} 1 - x^{2} \Rightarrow 5 x = 3 1 - x^{2}$ . The right side is non-negative, so $x \geq 0$ . (½ mark)
Square: $25 x^{2} = 3 - 3 x^{2} \Rightarrow x^{2} = \frac{3}{28}$ . With $x \geq 0$ : $x = \frac{3}{28} = \frac{1}{2} \frac{3}{7}$ . (1 mark)
Domain check: $2 x = \frac{3}{7} < 1$ , so both terms are defined. (½ mark)

Answer: $x = \frac{1}{2} \frac{3}{7}$ (about 0.327).

Example 5: Two roots that both survive · 3 marks · Short answerHarder board level

Solve: $tan^{- 1} \frac{x - 1}{x - 2} + tan^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}$ .

Try it first, then show the solutionSolution

Combine: $\frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \cdot \frac{x + 1}{x + 2}} = 1$ . Multiply top and bottom by $(x - 2) (x + 2)$ . (1 mark)
Numerator: $(x - 1) (x + 2) + (x + 1) (x - 2) = 2 x^{2} - 4$ . Denominator: $(x^{2} - 4) - (x^{2} - 1) = - 3$ . (1 mark)
$\frac{2 x ^{2} - 4}{- 3} = 1 \Rightarrow 2 x^{2} = 1 \Rightarrow x = \pm \frac{1}{2}$ . (½ mark)
Check: for $x = \frac{1}{2}$ the two arguments are about $0.23$ and $0.63$ , both positive with product less than 1; for $x = - \frac{1}{2}$ they swap. Both roots satisfy. (½ mark)

Answer: $x = \pm \frac{1}{2}$ .

Example 6: Branch reasoning with a substitution · 5 marks · Long answerHOTS / competency

Solve for $x > 1$ : $cos^{- 1} \frac{x ^{2} - 1}{x ^{2} + 1} + tan^{- 1} \frac{2 x}{x ^{2} - 1} = \frac{2 π}{3}$ .

Try it first, then show the solutionSolution

Put $x = tan θ$ . Since $x > 1$ , $θ \in (\frac{π}{4}, \frac{π}{2})$ , so $2 θ \in (\frac{π}{2}, π)$ . (1 mark)
$\frac{x ^{2} - 1}{x ^{2} + 1} = - cos 2 θ = cos (π - 2 θ)$ , and $π - 2 θ \in (0, \frac{π}{2})$ , so $cos^{- 1} \frac{x ^{2} - 1}{x ^{2} + 1} = π - 2 θ$ . (1 mark)
$\frac{2 x}{x ^{2} - 1} = - tan 2 θ = tan (π - 2 θ)$ , and $π - 2 θ \in (0, \frac{π}{2})$ , so $tan^{- 1} \frac{2 x}{x ^{2} - 1} = π - 2 θ$ . (1 mark)
The equation becomes $2 (π - 2 θ) = \frac{2 π}{3} \Rightarrow π - 2 θ = \frac{π}{3} \Rightarrow θ = \frac{π}{3}$ . (1 mark)
$x = tan \frac{π}{3} = 3$ , which satisfies $x > 1$ . Check: $cos^{- 1} \frac{2}{4} + tan^{- 1} \frac{2 3}{2} = \frac{π}{3} + \frac{π}{3}$ . (1 mark)

Answer: $x = 3$ . Writing $cos^{- 1} \frac{x ^{2} - 1}{x ^{2} + 1} = 2 tan^{- 1} x$ without the branch check would give the wrong equation.

Where students lose marks in Solving Inverse Trigonometric Equations

The slips examiners see most often, and what to write instead.

Mistake: Keeping a root that makes the left side the wrong sign
Taking tan of both sides cannot tell π/4 from −3π/4, so the algebra produces extra roots.
Wrong working
$(12 x - 1) (2 x + 1) = 0 \Rightarrow x = \frac{1}{12}, - \frac{1}{2}$ .
Right working
$x = - \frac{1}{2}$ makes the left side $tan^{- 1} (- 2) + tan^{- 1} (- 3) < 0 \neq = \frac{π}{4}$ , so only $x = \frac{1}{12}$ .
Do this: Test each root in the original equation and write the reason for any rejection.
Usually costs: 1 mark in a 3-mark question.
Mistake: Not finishing a long calculation
Choosing a clumsy combination of terms makes the algebra long; the ISC 2024 examiners reported candidates who knew the identities but could not finish.
Do this: Isolate the term with the simplest argument on one side before taking sin or cos, so that only one square root appears.
Usually costs: Up to the whole question.
Mistake: Forgetting the domain of sin⁻¹ and cos⁻¹
A root such as x = 2 in sin⁻¹(x/2) is fine, but in sin⁻¹x it is outside the domain.
Do this: Write the domain at the start and check roots against it at the end.
Usually costs: ½ to 1 mark.
Mistake: Squaring without fixing the sign first
Squaring 5x = √3√(1 − x²) also accepts the negative root, which does not satisfy the unsquared equation.
Do this: Before squaring, note the sign forced by the square root (here x ≥ 0) and use it to choose the root.
Usually costs: 1 mark for an extra wrong answer.

Exam tips

Before any algebra, look for a complementary pair with the same argument; many 1- and 2-mark questions end there.
Show the rejection step in words, such as "x = −1/2 is rejected as the left side is negative". It is often worth its own half-mark.
Keep the exact form of the answer ( $\frac{1}{2} \frac{3}{7}$ , $33 \pm 11$ ) and add a decimal only in brackets.
For substitution questions, write the interval of $θ$ explicitly; examiners give credit for the branch argument.
If the algebra leads to a quadratic with no real roots, say so: "no real solution" is a valid final answer.

Practice questions on Solving Inverse Trigonometric Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $sin^{- 1} x = \frac{π}{5}$ for some $x \in (- 1, 1)$ , then $cos^{- 1} x$ equals:
(a) $\frac{3 π}{10}$
(b) $\frac{5 π}{10}$
(c) $\frac{7 π}{10}$
(d) $\frac{9 π}{10}$
Hint
Use the complementary pair.
Answer
(a) $\frac{π}{2} - \frac{π}{5} = \frac{3 π}{10}$ .
Q2 · 2 marks · Short answerBoard level
Solve: $tan^{- 1} x + 2 cot^{- 1} x = \frac{2 π}{3}$ .
Hint
Split $2 cot^{- 1} x$ as $cot^{- 1} x + cot^{- 1} x$ .
Answer
$(tan^{- 1} x + cot^{- 1} x) + cot^{- 1} x = \frac{2 π}{3} \Rightarrow cot^{- 1} x = \frac{2 π}{3} - \frac{π}{2} = \frac{π}{6}$ , so $x = cot \frac{π}{6} = 3$ .
Q3 · 3 marks · Short answerHarder board level
Solve: $tan^{- 1} (x + 1) + tan^{- 1} (x - 1) = tan^{- 1} \frac{8}{31}$ .
Hint
You will get two roots; test the sign of the left side.
Answer
$\frac{2 x}{1 - ( x ^{2} - 1 )} = \frac{8}{31} \Rightarrow 62 x = 16 - 8 x^{2} \Rightarrow 4 x^{2} + 31 x - 8 = 0 \Rightarrow (4 x - 1) (x + 8) = 0$ . $x = - 8$ makes both terms negative (and $(x + 1) (x - 1) = 63 > 1$ ), so reject. $x = \frac{1}{4}$ .
Q4 · 3 marks · Short answerHarder board level
Solve: $sin^{- 1} 6 x + sin^{- 1} 63 x = - \frac{π}{2}$ .
Hint
Isolate: $sin^{- 1} 6 x = - \frac{π}{2} - sin^{- 1} 63 x$ , then take $sin$ .
Answer
$6 x = - cos (sin^{- 1} 63 x) = - 1 - 108 x^{2}$ , so $x < 0$ . Squaring: $36 x^{2} = 1 - 108 x^{2} \Rightarrow x^{2} = \frac{1}{144}$ , and with $x < 0$ , $x = - \frac{1}{12}$ . Check: $- \frac{π}{6} - \frac{π}{3} = - \frac{π}{2}$ .
Q5 · 3 marks · Short answerHarder board level
For $a, b \geq 0$ , solve $2 tan^{- 1} x = cos^{- 1} \frac{1 - a ^{2}}{1 + a ^{2}} - cos^{- 1} \frac{1 - b ^{2}}{1 + b ^{2}}$ .
Hint
For $t \geq 0$ , $cos^{- 1} \frac{1 - t ^{2}}{1 + t ^{2}} = 2 tan^{- 1} t$ .
Answer
The right side is $2 tan^{- 1} a - 2 tan^{- 1} b$ , so $tan^{- 1} x = tan^{- 1} a - tan^{- 1} b = tan^{- 1} \frac{a - b}{1 + ab}$ (since $ab \geq 0 > - 1$ ). Hence $x = \frac{a - b}{1 + ab}$ .
Q6 · 1 mark · MCQHOTS / competency
The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{2}$ is:
(a) 0
(b) 1
(c) 2
(d) infinitely many
Hint
Both square roots must be real, and $x^{2} + x + 1 \leq 1$ .
Answer
(c) 2. Let $u = x^{2} + x$ . We need $u \geq 0$ and $u + 1 \leq 1$ , so $u = 0$ , giving $x = 0$ or $x = - 1$ . Both give $tan^{- 1} 0 + sin^{- 1} 1 = \frac{π}{2}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a hoarding on a building has its lower edge 2 m and its upper edge 8 m above a pedestrian's eye level. The pedestrian stands $x$ m from the building, and $θ$ is the angle the hoarding subtends at the eye.
(i) Show that $θ = tan^{- 1} \frac{6 x}{x ^{2} + 16}$ . (1 mark)
(ii) Find the distances at which $θ = \frac{π}{6}$ . (2 marks)
(iii) Show that no position gives $θ = \frac{π}{4}$ , and find the largest possible $θ$ . (1 mark)
Hint
In (iii), look at the discriminant, then at x + 16/x.
Answer
(i) $tan^{- 1} \frac{8}{x} - tan^{- 1} \frac{2}{x} = tan^{- 1} \frac{6/ x}{1 + 16/ x ^{2}} = tan^{- 1} \frac{6 x}{x ^{2} + 16}$ . (ii) $\frac{6 x}{x ^{2} + 16} = \frac{1}{3} \Rightarrow x^{2} - 63 x + 16 = 0 \Rightarrow x = 33 \pm 11$ , about 1.88 m or 8.51 m. (iii) $θ = \frac{π}{4}$ needs $x^{2} - 6 x + 16 = 0$ , whose discriminant is $36 - 64 < 0$ : impossible. Since $x + \frac{16}{x} \geq 8$ , $tan θ \leq \frac{6}{8}$ ; the largest angle is $tan^{- 1} \frac{3}{4}$ , at $x = 4$ m.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Solve $sin^{- 1} x + cos^{- 1} x = \frac{π}{3}$ .Show answer
No solution, since the left side is always π/2.
2.What is $cos (sin^{- 1} x)$ ?Show answer
$1 - x^{2}$ .
3.Solve $tan^{- 1} x = cot^{- 1} x$ .Show answer
$x = 1$ (both equal $\frac{π}{4}$ ).
4.Name the three checks for a root of an inverse trigonometric equation.Show answer
Domain of the arguments, sign and size of the left side, and direct substitution.
5.Solve $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{4}$ .Show answer
$x = \frac{1}{12}$ ( $x = - \frac{1}{2}$ rejected).
6.For $x > 0$ , simplify $tan^{- 1} x + tan^{- 1} \frac{1}{x}$ .Show answer
$\frac{π}{2}$ .

Solving Inverse Trigonometric Equations: frequently asked questions

Why do I get extra roots when solving inverse trigonometric equations?

Taking tan, sin or cos of both sides keeps the trigonometric value but forgets which angle it came from, and squaring forgets signs. Both steps can let in roots that the original equation does not allow, so every root has to be checked.

How do I decide whether to take sin, cos or tan of both sides?

Match the function of the isolated term. If one side is sin⁻¹(something), take sin; if it is cos⁻¹, take cos; if every term is tan⁻¹, combine them and take tan. This keeps the number of square roots to a minimum.

Can an inverse trigonometric equation have no solution?

Yes. The ranges limit what each side can equal. For example, sin⁻¹x + cos⁻¹x is always π/2, and tan⁻¹ of anything is strictly between −π/2 and π/2, so equations asking for other values have no solution.

Do I need to show the domain in an ISC answer?

It is good practice and examiners credit it, especially when you reject a root. A one-line statement such as "x must lie in [0, 1]" is enough.

How many marks do inverse trigonometric equations carry in the ISC paper?

In recent ISC papers they have appeared as longer questions worth several marks (for example a 4-mark "solve for x" in 2024), as well as in 1-mark items in Section A. Unit 1, Relations and Functions with inverse trigonometric functions, carries 10 marks in the 2027 paper.