MathsISC Class 12 · Inverse Trigonometric Functions

Inverse Trigonometric Identities: Properties, Formulas and Proofs

Q: Which substitution should I use in an inverse trigonometric proof?

Use x = tan θ when the expression has 1 + x² or 1 − x² over 1 + x², x = sin θ when it has √(1 − x²), and x = cos θ when it has 2x² − 1. The aim is to turn the argument into a single trigonometric function of a multiple of θ.

Q: Are the formulas for 3tan⁻¹x and 2sin⁻¹x in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists the formulae for 2sin⁻¹x, 2cos⁻¹x, 2tan⁻¹x and 3tan⁻¹x, along with the sum and difference formulas for sin⁻¹, cos⁻¹ and tan⁻¹, and their applications.

Q: How are inverse trigonometric identities tested in the ISC paper?

They appear as 1-mark evaluations in Section A, 2-mark short questions using a complementary pair, and longer proofs where three inverse functions add up to π or π/2. Graph-based case questions may also ask you to combine two inverse functions.

Updated 15 October 2026 · Fact-checked

The key inverse trigonometric identities are sin⁻¹x + cos⁻¹x = π/2, tan⁻¹x + cot⁻¹x = π/2, tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)) for xy < 1, and the double-angle forms of 2tan⁻¹x. Each holds only on a stated interval, and proofs work by putting x = sin θ, tan θ or cos θ.

The two acute angles of a right triangle add to π/2, so sin⁻¹x and cos⁻¹x (the angles at the ends of side x) always sum to π/2.

Illustration drawn for QuestionBuddy.

Understand Inverse Trigonometric Identities

Every inverse trigonometric function returns an angle in its principal value branch: $sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}]$ , $cos^{- 1} x \in [0, π]$ , $tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2})$ , $cot^{- 1} x \in (0, π)$ . Every identity in this topic is just an ordinary trigonometric identity, read backwards, with a check that the angle you get lands inside the right branch. That check is why each identity carries a condition.

Complementary pairs. If $sin^{- 1} x = θ$ , then $x = sin θ = cos (\frac{π}{2} - θ)$ , and $\frac{π}{2} - θ$ lies in $[0, π]$ , so $cos^{- 1} x = \frac{π}{2} - θ$ . Hence $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$ for $x \in [- 1, 1]$ . In the same way $tan^{- 1} x + cot^{- 1} x = \frac{π}{2}$ and $sec^{- 1} x + cosec^{- 1} x = \frac{π}{2}$ . These turn a mixed expression into one function, as in ISC 2025, where $(tan^{- 1} x)^{2} - (cot^{- 1} x)^{2}$ factorised into a sum times a difference.

Reciprocal and negative arguments. $sin^{- 1} x = cosec^{- 1} \frac{1}{x}$ and $cos^{- 1} x = sec^{- 1} \frac{1}{x}$ . For negatives, the odd functions keep the sign outside: $sin^{- 1} (- x) = - sin^{- 1} x$ and $tan^{- 1} (- x) = - tan^{- 1} x$ . But $cos^{- 1} (- x) = π - cos^{- 1} x$ and $cot^{- 1} (- x) = π - cot^{- 1} x$ , because their branches lie in $[0, π]$ .

Converting between functions. Draw a right triangle. If $sin^{- 1} x = θ$ with $0 \leq x < 1$ , the sides are $x$ , $1 - x^{2}$ and $1$ , so $sin^{- 1} x = cos^{- 1} 1 - x^{2} = tan^{- 1} \frac{x}{1 - x ^{2}}$ . The tangent form holds for all $∣ x ∣ < 1$ ; the cosine form only for $0 \leq x \leq 1$ .

Sum and difference formulas. From $tan (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ :
$tan^{- 1} x + tan^{- 1} y = tan^{- 1} \frac{x + y}{1 - x y}$ when $x y < 1$ , and $tan^{- 1} x - tan^{- 1} y = tan^{- 1} \frac{x - y}{1 + x y}$ when $x y > - 1$ . If $x, y > 0$ and $x y > 1$ , the sum exceeds $\frac{π}{2}$ and you must add $π$ : $tan^{- 1} 2 + tan^{- 1} 3 = π + tan^{- 1} (- 1) = \frac{3 π}{4}$ . From $sin (A \pm B)$ and $cos (A \pm B)$ : $sin^{- 1} x \pm sin^{- 1} y = sin^{- 1} (x 1 - y^{2} \pm y 1 - x^{2})$ and $cos^{- 1} x \pm cos^{- 1} y = cos^{- 1} (x y \mp 1 - x^{2} 1 - y^{2})$ , under conditions that keep the angle in the branch.

Multiple-angle forms. With $x = tan θ$ : $2 tan^{- 1} x = tan^{- 1} \frac{2 x}{1 - x ^{2}}$ for $∣ x ∣ < 1$ , $= sin^{- 1} \frac{2 x}{1 + x ^{2}}$ for $∣ x ∣ \leq 1$ , $= cos^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}$ for $x \geq 0$ ; and $3 tan^{- 1} x = tan^{- 1} \frac{3 x - x ^{3}}{1 - 3 x ^{2}}$ for $∣ x ∣ < \frac{1}{3}$ . With $x = sin θ$ : $2 sin^{- 1} x = sin^{- 1} (2 x 1 - x^{2})$ for $∣ x ∣ \leq \frac{1}{2}$ . With $x = cos θ$ : $2 cos^{- 1} x = cos^{- 1} (2 x^{2} - 1)$ for $0 \leq x \leq 1$ .

How proofs go. Name the angles ( $A = sin^{- 1} x$ and so on), turn the given relation into one about $A, B, C$ , take a suitable trigonometric function of both sides, and use the ordinary identity. When there are three terms, move one to the other side first.

Inverse Trigonometric Identities: formulas to remember

Complementary pairs: $sin^{- 1} x + cos^{- 1} x = tan^{- 1} x + cot^{- 1} x = \frac{π}{2}$; sin/cos: x in [-1, 1]; tan/cot: all real x. Also sec⁻¹x + cosec⁻¹x = π/2 for |x| ≥ 1.
Negative arguments: $sin^{- 1} (- x) = - sin^{- 1} x, cos^{- 1} (- x) = π - cos^{- 1} x$; tan⁻¹ and cosec⁻¹ behave like sin⁻¹; cot⁻¹ and sec⁻¹ behave like cos⁻¹.
Reciprocals: $sin^{- 1} x = cosec^{- 1} \frac{1}{x}, tan^{- 1} x = cot^{- 1} \frac{1}{x} (x > 0)$; For x < 0, cot⁻¹(1/x) = π + tan⁻¹x, because cot⁻¹ takes values in (0, π).
Sum of tan inverses: $tan^{- 1} x + tan^{- 1} y = tan^{- 1} \frac{x + y}{1 - x y}, x y < 1$; If x, y > 0 and xy > 1, add π.
Difference of tan inverses: $tan^{- 1} x - tan^{- 1} y = tan^{- 1} \frac{x - y}{1 + x y}, x y > - 1$; Used for angles subtended, such as tan⁻¹2 − tan⁻¹1.
Sum of sin inverses: $sin^{- 1} x + sin^{- 1} y = sin^{- 1} (x 1 - y^{2} + y 1 - x^{2})$; Valid when x, y ≥ 0 and x² + y² ≤ 1 (the sum stays at most π/2).
Double angle via tan: $2 tan^{- 1} x = tan^{- 1} \frac{2 x}{1 - x ^{2}} = sin^{- 1} \frac{2 x}{1 + x ^{2}} = cos^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}$; Conditions: |x| < 1, |x| ≤ 1 and x ≥ 0 respectively.
Triple angle via tan: $3 tan^{- 1} x = tan^{- 1} \frac{3 x - x ^{3}}{1 - 3 x ^{2}}, ∣ x ∣ < \frac{1}{3}$; Comes from tan 3θ = (3 tan θ − tan³θ)/(1 − 3 tan²θ).
Double angle via sin and cos: $2 sin^{- 1} x = sin^{- 1} (2 x 1 - x^{2}), 2 cos^{- 1} x = cos^{- 1} (2 x^{2} - 1)$; |x| ≤ 1/√2 for the first; 0 ≤ x ≤ 1 for the second.

How to solve Inverse Trigonometric Identities questions

This approach handles the three common ISC formats: evaluate an expression, prove an identity with numbers, and prove a relation between x, y and z.

1Look for a complementary pair first ( $sin^{- 1} x$ with $cos^{- 1} x$ , $tan^{- 1} x$ with $cot^{- 1} x$ ). Replacing one by $\frac{π}{2}$ minus the other often finishes the question.
2If the arguments are numbers, convert everything to $tan^{- 1}$ using a right triangle (for example $sin^{- 1} \frac{3}{5} = tan^{- 1} \frac{3}{4}$ ), then combine with the sum formula.
3Before using the sum formula, check its condition: is $x y < 1$ ? If not, adjust by $π$ .
4For a double angle such as $2 tan^{- 1} \frac{1}{3}$ , use $2 tan^{- 1} x = tan^{- 1} \frac{2 x}{1 - x ^{2}}$ first, then add the remaining term.
5For algebraic proofs, substitute $x = tan θ$ (when you see $1 + x^{2}$ ), $x = sin θ$ (when you see $1 - x^{2}$ ) or $x = cos θ$ (when you see $2 x^{2} - 1$ ), and simplify to a single angle.
6For a relation such as $tan^{- 1} x + tan^{- 1} y + tan^{- 1} z = π$ , move one term to the right, take $tan$ (or $sin$ , $cos$ ) of both sides, and simplify.
7End by stating the result and, where relevant, the interval on which it holds.

Inverse Trigonometric Identities 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.

Difference of two inverse tangents

Suppose you stand on level ground 6 m from the foot of a 6 m tall building that has a flagpole on its roof. The top of the flagpole is 12 m above the ground. Ignore your own height.

Find the angle the flagpole subtends at your eye, as a single inverse tangent.

Try it first, then show the solutionSolution

Angle of elevation of the top of the pole: $tan^{- 1} \frac{12}{6} = tan^{- 1} 2$ .
Angle of elevation of the foot of the pole (the roof): $tan^{- 1} \frac{6}{6} = tan^{- 1} 1 = \frac{π}{4}$ .
Angle subtended $= tan^{- 1} 2 - tan^{- 1} 1$ . Since $2 \times 1 > - 1$ , use the difference formula.
$= tan^{- 1} \frac{2 - 1}{1 + 2} = tan^{- 1} \frac{1}{3}$ , about $18. 4^{\circ}$ .

Answer: The flagpole subtends $tan^{- 1} \frac{1}{3}$ (about $18. 4^{\circ}$ ).

Memory hook: The angle a pole subtends is a difference of two elevation angles: use the tan⁻¹ difference formula.

Sum of two inverse tangents

Imagine a 6 m pole held upright by two cables from its top, fixed to the ground on opposite sides at 2 m and 3 m from its foot.

Find the angle between the two cables at the top of the pole.

Try it first, then show the solutionSolution

Each cable makes an angle with the pole: $tan^{- 1} \frac{2}{6} = tan^{- 1} \frac{1}{3}$ and $tan^{- 1} \frac{3}{6} = tan^{- 1} \frac{1}{2}$ .
The angle between the cables is the sum: $tan^{- 1} \frac{1}{3} + tan^{- 1} \frac{1}{2}$ .
Here $x y = \frac{1}{6} < 1$ , so the sum formula applies: $tan^{- 1} \frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{6}} = tan^{- 1} \frac{5/6}{5/6} = tan^{- 1} 1$ .
So the angle is $\frac{π}{4}$ .

Answer: The cables meet at exactly $\frac{π}{4}$ ( $4 5^{\circ}$ ).

Memory hook: One half plus one third gives one: tan⁻¹(1/2) + tan⁻¹(1/3) = π/4, the cable angle.

Tricks to remember Inverse Trigonometric Identities

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

Co-functions add to a right angle
sin with cos, tan with cot, sec with cosec: each pair of inverses adds to π/2.
Use it when: Spotting the first step in evaluation questions.
Odd ones keep the minus
sin⁻¹, tan⁻¹, cosec⁻¹ are odd: the minus comes out. cos⁻¹, cot⁻¹, sec⁻¹ give π minus.
Use it when: Negative arguments in MCQs.
Same as tan(A + B)
The tan⁻¹ sum formula is tan(A + B) read backwards: "sum over one minus product".
Use it when: Recalling the formula and its sign in the difference version.
Triangle trick
To change sin⁻¹(3/5) into tan⁻¹, draw a 3-4-5 triangle: opposite 3, adjacent 4, so tan⁻¹(3/4).
Use it when: Converting before adding inverse functions with numerical arguments.

Inverse Trigonometric Identities: solved examples

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

Example 1: Using a complementary pair · 2 marks · Short answerBoard level

If $(sin^{- 1} x)^{2} - (cos^{- 1} x)^{2} = \frac{π ^{2}}{8}$ , find the value of $sin^{- 1} x - cos^{- 1} x$ .

Try it first, then show the solutionSolution

Factorise: $(sin^{- 1} x + cos^{- 1} x) (sin^{- 1} x - cos^{- 1} x) = \frac{π ^{2}}{8}$ . (½ mark)
Use $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$ . (½ mark)
So $\frac{π}{2} (sin^{- 1} x - cos^{- 1} x) = \frac{π ^{2}}{8} \Rightarrow sin^{- 1} x - cos^{- 1} x = \frac{π}{4}$ . (1 mark)

Answer: $sin^{- 1} x - cos^{- 1} x = \frac{π}{4}$ (and then $sin^{- 1} x = \frac{3 π}{8}$ ). ISC 2025 asked the same idea with $tan^{- 1} x$ and $cot^{- 1} x$ ; the examiners noted that candidates failed to use $tan^{- 1} x + cot^{- 1} x = \frac{π}{2}$ .

Example 2: A standard sum · 1 mark · MCQBoard level

$tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3}$ equals:
(a) $\frac{π}{6}$
(b) $\frac{π}{4}$
(c) $\frac{π}{3}$
(d) $\frac{π}{2}$

Try it first, then show the solutionSolution

$x y = \frac{1}{6} < 1$ , so the sum is $tan^{- 1} \frac{\frac{5}{6}}{\frac{5}{6}} = tan^{- 1} 1$ .

Answer: (b) $\frac{π}{4}$ .

Example 3: Double angle then sum · 3 marks · Short answerHarder board level

Prove that $2 tan^{- 1} \frac{1}{4} + tan^{- 1} \frac{7}{23} = \frac{π}{4}$ .

Try it first, then show the solutionSolution

Since $\frac{1}{4} < 1$ : $2 tan^{- 1} \frac{1}{4} = tan^{- 1} \frac{\frac{1}{2}}{1 - \frac{1}{16}} = tan^{- 1} \frac{\frac{1}{2}}{\frac{15}{16}} = tan^{- 1} \frac{8}{15}$ . (1 mark)
Now $\frac{8}{15} \cdot \frac{7}{23} = \frac{56}{345} < 1$ , so $tan^{- 1} \frac{8}{15} + tan^{- 1} \frac{7}{23} = tan^{- 1} \frac{\frac{8}{15} + \frac{7}{23}}{1 - \frac{56}{345}}$ . (1 mark)
$= tan^{- 1} \frac{\frac{289}{345}}{\frac{289}{345}} = tan^{- 1} 1 = \frac{π}{4}$ . (1 mark)

Answer: Proved.

Example 4: Sum of two sin inverses as a cos inverse · 3 marks · Short answerHarder board level

Prove that $sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{5}{13} = cos^{- 1} \frac{16}{65}$ .

Try it first, then show the solutionSolution

Let $A = sin^{- 1} \frac{4}{5}$ and $B = sin^{- 1} \frac{5}{13}$ , both acute. Then $cos A = \frac{3}{5}$ and $cos B = \frac{12}{13}$ . (1 mark)
$cos (A + B) = cos A cos B - sin A sin B = \frac{3}{5} \cdot \frac{12}{13} - \frac{4}{5} \cdot \frac{5}{13} = \frac{36 - 20}{65} = \frac{16}{65}$ . (1 mark)
Since $A, B \in (0, \frac{π}{2})$ , $A + B \in (0, π)$ , the principal branch of $cos^{- 1}$ . So $A + B = cos^{- 1} \frac{16}{65}$ . (1 mark)

Answer: Proved. The branch check in the last step is what justifies writing cos⁻¹.

Example 5: Proving a double-angle form by substitution · 3 marks · Short answerHarder board level

Prove that $sin^{- 1} (2 x 1 - x^{2}) = 2 sin^{- 1} x$ for $- \frac{1}{2} \leq x \leq \frac{1}{2}$ .

Try it first, then show the solutionSolution

Put $x = sin θ$ . Since $∣ x ∣ \leq \frac{1}{2}$ , $θ \in [- \frac{π}{4}, \frac{π}{4}]$ and $cos θ \geq 0$ , so $1 - x^{2} = cos θ$ . (1 mark)
Then $2 x 1 - x^{2} = 2 sin θ cos θ = sin 2 θ$ , with $2 θ \in [- \frac{π}{2}, \frac{π}{2}]$ . (1 mark)
So $sin^{- 1} (sin 2 θ) = 2 θ = 2 sin^{- 1} x$ . (1 mark)

Answer: Proved. Outside this interval the result fails: for $x = 0.9$ , $2 sin^{- 1} x$ exceeds $\frac{π}{2}$ .

Example 6: Three inverse tangents adding to π · 5 marks · Long answerHOTS / competency

(i) If $tan^{- 1} x + tan^{- 1} y + tan^{- 1} z = π$ , show that $x + y + z = x y z$ .
(ii) Verify the result for $x = 1, y = 2, z = 3$ , given that $tan^{- 1} 1 + tan^{- 1} 2 + tan^{- 1} 3 = π$ .
(iii) Hence find $tan^{- 1} 2 + tan^{- 1} 3$ .

Try it first, then show the solutionSolution

(i) Let $A = tan^{- 1} x$ , $B = tan^{- 1} y$ , $C = tan^{- 1} z$ . Then $A + B = π - C$ . (1 mark)
Take $tan$ of both sides: $tan (A + B) = tan (π - C) = - tan C$ . (1 mark)
$\frac{x + y}{1 - x y} = - z \Rightarrow x + y = - z + x y z \Rightarrow x + y + z = x y z$ . (1 mark)
(ii) $1 + 2 + 3 = 6$ and $1 \times 2 \times 3 = 6$ . Verified. (1 mark)
(iii) $tan^{- 1} 2 + tan^{- 1} 3 = π - tan^{- 1} 1 = π - \frac{π}{4} = \frac{3 π}{4}$ . Note that the sum formula alone gives $tan^{- 1} (- 1) = - \frac{π}{4}$ , which is wrong because $2 \times 3 > 1$ . (1 mark)

Answer: (i) Proved; (ii) both sides equal 6; (iii) $\frac{3 π}{4}$ .

Where students lose marks in Inverse Trigonometric Identities

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

Mistake: Using the tan⁻¹ sum formula when xy > 1
The formula gives an angle in (−π/2, π/2), but the true sum of two positive angles with xy > 1 is more than π/2.
Wrong working
$tan^{- 1} 2 + tan^{- 1} 3 = tan^{- 1} \frac{5}{1 - 6} = tan^{- 1} (- 1) = - \frac{π}{4}$ .
Right working
$x y = 6 > 1$ with $x, y > 0$ , so the sum is $π + tan^{- 1} (- 1) = \frac{3 π}{4}$ .
Do this: Check xy against 1 before applying the formula, and write the check.
Usually costs: The final answer mark, and the method mark if no check is shown.
Mistake: Writing cos⁻¹(−x) = −cos⁻¹x
cos⁻¹ takes values in [0, π], so it can never be negative.
Wrong working
$cos^{- 1} (- \frac{1}{2}) = - \frac{π}{3}$ .
Right working
$cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ .
Do this: Learn which functions are odd (sin, tan, cosec) and which give π minus (cos, cot, sec).
Usually costs: 1 mark in Section A.
Mistake: Picking an awkward combination of terms
In three-term proofs, combining the wrong pair leads to long algebra that rarely finishes; the ISC 2024 examiners noted exactly this.
Do this: Move the term whose function matches the target (for example the sin⁻¹x term when x² appears alone) to the other side, then take the matching trigonometric function.
Usually costs: Up to the whole proof.
Mistake: Dropping the condition on x
Identities such as 2tan⁻¹x = sin⁻¹(2x/(1 + x²)) hold only on an interval; outside it they are false.
Do this: When you substitute x = tan θ or sin θ, write the range of θ and check that the final angle is in the principal branch.
Usually costs: Half a mark to a mark in proofs.

Exam tips

Write the principal value branch next to any answer that is an angle; examiners look for it when the angle could be in two places.
For numerical identities, convert everything to $tan^{- 1}$ first. It has the simplest sum formula and the widest domain.
For a three-term relation equal to $π$ or $\frac{π}{2}$ , take $tan$ , $sin$ or $cos$ of both sides after moving one term across; never try to add all three inverse functions at once.
When a proof needs squaring, isolate the square root first so that only one squaring is needed.
Keep exact values ( $\frac{π}{4}$ , $tan^{- 1} \frac{3}{4}$ ). Decimal angles are not accepted in proofs.

Practice questions on Inverse Trigonometric Identities

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
$sin^{- 1} (- \frac{1}{2}) + cos^{- 1} (- \frac{1}{2})$ equals:
(a) $0$
(b) $\frac{π}{2}$
(c) $π$
(d) $- \frac{π}{2}$
Hint
Same argument in both terms.
Answer
(b) $\frac{π}{2}$ : $- \frac{π}{6} + \frac{2 π}{3} = \frac{π}{2}$ , as the complementary identity predicts.
Q2 · 2 marks · Short answerBoard level
Find the value of $tan^{- 1} 3 - cot^{- 1} (- 3)$ .
Hint
$cot^{- 1} (- x) = π - cot^{- 1} x$ .
Answer
$tan^{- 1} 3 = \frac{π}{3}$ and $cot^{- 1} (- 3) = π - \frac{π}{6} = \frac{5 π}{6}$ . Value $= \frac{π}{3} - \frac{5 π}{6} = - \frac{π}{2}$ .
Q3 · 2 marks · Short answerHarder board level
Show that $tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{2}{11} = tan^{- 1} \frac{3}{4}$ .
Hint
Check that the product is less than 1 first.
Answer
$\frac{1}{2} \cdot \frac{2}{11} = \frac{1}{11} < 1$ . Sum $= tan^{- 1} \frac{\frac{1}{2} + \frac{2}{11}}{1 - \frac{1}{11}} = tan^{- 1} \frac{15/22}{10/11} = tan^{- 1} \frac{3}{4}$ .
Q4 · 3 marks · Short answerHarder board level
Prove that $2 tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{7} = sin^{- 1} \frac{31}{25 2}$ .
Hint
First show the left side is $tan^{- 1} \frac{31}{17}$ , then draw a triangle.
Answer
$2 tan^{- 1} \frac{1}{2} = tan^{- 1} \frac{1}{3/4} = tan^{- 1} \frac{4}{3}$ . Then $tan^{- 1} \frac{4}{3} + tan^{- 1} \frac{1}{7} = tan^{- 1} \frac{31/21}{17/21} = tan^{- 1} \frac{31}{17}$ (product $\frac{4}{21} < 1$ ). Hypotenuse $3 1^{2} + 1 7^{2} = 1250 = 252$ , so the angle is $sin^{- 1} \frac{31}{25 2}$ (acute, so in the branch).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $cos^{- 1} (cos \frac{4 π}{3}) = \frac{4 π}{3}$ .
Reason (R): The principal value branch of $cos^{- 1}$ is $[0, π]$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$\frac{4 π}{3}$ is outside $[0, π]$ .
Answer
(d). $cos \frac{4 π}{3} = - \frac{1}{2}$ and $cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ , so A is false; R is true.
Q6 · 3 marks · Short answerHOTS / competency
If $cos^{- 1} x + cos^{- 1} y + cos^{- 1} z = π$ , prove that $x^{2} + y^{2} + z^{2} + 2 x y z = 1$ .
Hint
Write $cos^{- 1} x + cos^{- 1} y = π - cos^{- 1} z$ and take cosines.
Answer
Taking cosines: $x y - 1 - x^{2} 1 - y^{2} = - z$ . So $x y + z = (1 - x^{2}) (1 - y^{2})$ . Squaring: $x^{2} y^{2} + 2 x y z + z^{2} = 1 - x^{2} - y^{2} + x^{2} y^{2}$ , which gives $x^{2} + y^{2} + z^{2} + 2 x y z = 1$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a painting hangs on a gallery wall with its lower edge 1 m and its upper edge 3 m above a visitor's eye level. The visitor stands $x$ m from the wall, and $θ$ is the angle the painting subtends at the eye.
(i) Show that $θ = tan^{- 1} \frac{2 x}{x ^{2} + 3}$ . (2 marks)
(ii) Find $θ$ when $x = 3$ . (1 mark)
(iii) Find $θ$ when $x = 3$ , and say why this is the best viewing distance. (1 mark)
Hint
$θ = tan^{- 1} \frac{3}{x} - tan^{- 1} \frac{1}{x}$ ; for (iii), $\frac{2 x}{x ^{2} + 3} = \frac{2}{x + 3/ x}$ .
Answer
(i) $θ = tan^{- 1} \frac{3}{x} - tan^{- 1} \frac{1}{x} = tan^{- 1} \frac{2/ x}{1 + 3/ x ^{2}} = tan^{- 1} \frac{2 x}{x ^{2} + 3}$ (product $\frac{3}{x ^{2}} > - 1$ ). (ii) $tan^{- 1} \frac{6}{12} = tan^{- 1} \frac{1}{2}$ . (iii) $tan^{- 1} \frac{2 3}{6} = tan^{- 1} \frac{1}{3} = \frac{π}{6}$ . Since $x + \frac{3}{x} \geq 23$ with equality at $x = 3$ , $tan θ$ and hence $θ$ is greatest there.

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.State $sin^{- 1} x + cos^{- 1} x$ and its domain.Show answer
$\frac{π}{2}$ , for $x \in [- 1, 1]$ .
2.Write the formula for $tan^{- 1} x - tan^{- 1} y$ .Show answer
$tan^{- 1} \frac{x - y}{1 + x y}$ , for $x y > - 1$ .
3.Express $2 tan^{- 1} x$ as a $cos^{- 1}$ .Show answer
$cos^{- 1} \frac{1 - x ^{2}}{1 + x ^{2}}$ , for $x \geq 0$ .
4.Find $tan^{- 1} 1 + tan^{- 1} 2 + tan^{- 1} 3$ .Show answer
$π$ .
5.What is $cot^{- 1} (- 1)$ ?Show answer
$π - \frac{π}{4} = \frac{3 π}{4}$ .
6.Write $sin^{- 1} \frac{3}{5}$ as an inverse tangent.Show answer
$tan^{- 1} \frac{3}{4}$ .

Inverse Trigonometric Identities: frequently asked questions

Why do inverse trigonometric identities have conditions?

Each inverse function returns an angle in a fixed principal branch. An identity is true only when the angle produced on both sides lies in that branch. Outside the stated interval the two sides can differ by π or have opposite signs.

What happens to tan⁻¹x + tan⁻¹y when xy is greater than 1?

If x and y are both positive, the sum is π + tan⁻¹((x + y)/(1 − xy)). If both are negative, it is −π + tan⁻¹((x + y)/(1 − xy)). The plain formula only works when xy < 1.

Which substitution should I use in an inverse trigonometric proof?

Use x = tan θ when the expression has 1 + x² or 1 − x² over 1 + x², x = sin θ when it has √(1 − x²), and x = cos θ when it has 2x² − 1. The aim is to turn the argument into a single trigonometric function of a multiple of θ.

Are the formulas for 3tan⁻¹x and 2sin⁻¹x in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists the formulae for 2sin⁻¹x, 2cos⁻¹x, 2tan⁻¹x and 3tan⁻¹x, along with the sum and difference formulas for sin⁻¹, cos⁻¹ and tan⁻¹, and their applications.

How are inverse trigonometric identities tested in the ISC paper?

They appear as 1-mark evaluations in Section A, 2-mark short questions using a complementary pair, and longer proofs where three inverse functions add up to π or π/2. Graph-based case questions may also ask you to combine two inverse functions.