Inverse Trigonometric Functions

What This Topic Covers

Inverse trigonometric functions appear in NDA Maths under the broader chapter heading "Properties of Triangles and Inverse Trigonometric Functions." The inverse trig portion by itself accounts for a consistent slice of that combined chapter.

The topic connects naturally to Trigonometric Ratios and Identities (you need fluency with standard angles) and to Limits, Continuity and Differentiability (inverse trig functions serve as classic examples of continuous functions).

Exam Pattern & Weightage

The table below is drawn directly from the NDA topicwise solved paper archive. Questions are counted from the combined "Properties of Triangles + Inverse Trig" chapter; the inverse trig count is the subset identifiable from question content.

Core Concepts

Domain and Range: The Principal Value Branches

The key constraint is that each trig function must be restricted to a branch where it is one-one and onto—that branch becomes the range of the inverse function. Memorise this table; NDA directly tests which interval is correct.

Composition Identities

These let you simplify \(\sin(\sin^{-1}x)\) or \(\sin^{-1}(\sin x)\) to a plain number. The catch is the domain restriction.

When the argument falls outside the principal branch, you must first rewrite the angle. For example, evaluate $$\sin^{-1}\!\left(\sin \tfrac{3\pi}{5}\right).$$

• The angle \(\tfrac{3\pi}{5}\) is not in the principal branch \(\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\), so direct cancellation fails.
• Use the supplementary identity: $$\sin\tfrac{3\pi}{5} = \sin\!\left(\pi - \tfrac{3\pi}{5}\right) = \sin\tfrac{2\pi}{5}.$$
• Now \(\tfrac{2\pi}{5}\) is in the principal branch, so $$\sin^{-1}\!\left(\sin \tfrac{3\pi}{5}\right) = \tfrac{2\pi}{5}.$$

The Triangle Conversion Method

When a question mixes different inverse functions — for example $$\sin(\cos^{-1}\tfrac{3}{5})$$ — convert via a right triangle rather than algebra.

• Let the inner inverse equal $$\theta$$: $$\cos^{-1}\tfrac{3}{5} = \theta \implies \cos\theta = \tfrac{3}{5}$$.
• Draw a right triangle with base = 3, hypotenuse = 5. By Pythagoras, perpendicular = 4.
• Read off the needed ratio: $$\sin\theta = \tfrac{4}{5}$$.
• Therefore $$\sin(\cos^{-1}\tfrac{3}{5}) = \tfrac{4}{5}$$.

This 4-second technique replaces 4–5 lines of identity manipulation and is the fastest route in mixed-function questions.

Reciprocal Relationships

Use the reciprocal rule to convert \(\sec^{-1}\) and \(\csc^{-1}\) into \(\cos^{-1}\) and \(\sin^{-1}\) — almost every NDA composite expression involving \(\sec^{-1}\) or \(\csc^{-1}\) becomes one-line once you convert.

Complementary and Supplementary Identities

These identities reduce multi-function expressions to a single step. NDA 2014-I used \(\sin^{-1}(1/3) + \cos^{-1}(1/3) = \pi/2\) as a statement to classify (it is correct), and NDA 2021-I used \(\tan^{-1}x + \cot^{-1}x = \pi/2\) to verify for which \(x \in \mathbb{R}\) the equation holds (answer: all real \(x\)).

Negation (Odd/Even) Properties

Addition Formulas for arctan

The branch correction (adding \(\pm\pi\) when \(xy > 1\)) is the most common trap. NDA 2014-I gave the statement "\(\tan^{-1}1 + \tan^{-1}(0.5) = \pi/2\)" — this is false because \(1 \times 0.5 = 0.5 < 1\), so the sum equals \(\tan^{-1}(1.5/0.5) = \tan^{-1}3\), not \(\pi/2\).

Addition Formula for arcsin

Double-Angle Conversions

NDA 2019-II used the substitution $$\sin^{-1}\!\left(\tfrac{2p}{1+p^2}\right) = 2\tan^{-1}p$$ to convert an expression involving \(p\) and \(q\) into $$\tan^{-1}\!\left(\tfrac{p-q}{1+pq}\right).$$

Key Composite Results

The result \(\sec^2(\tan^{-1}x) = 1 + x^2\) appeared directly in NDA 2012-I (\(\sec^2\tan^{-1}(11/10) = 1 + 121/100 = 221/100\)) and the derivative of \(\sec^2(\tan^{-1}x)\) with respect to \(x\) appeared in NDA 2019-I (answer: \(2x\)).

Worked Examples

Example 1 — Principal value of a composite (NDA 2012-I)

Q: Find $$\sin\!\left[\sin^{-1}\tfrac{3}{5} + \sin^{-1}\tfrac{4}{5}\right].$$

• Let \(\alpha = \sin^{-1}\tfrac{3}{5}\), so \(\sin\alpha = \tfrac{3}{5}\) and \(\cos\alpha = \tfrac{4}{5}\) (since \(\alpha\) is in principal branch, cosine is positive).
• Let \(\beta = \sin^{-1}\tfrac{4}{5}\), so \(\sin\beta = \tfrac{4}{5}\) and \(\cos\beta = \tfrac{3}{5}\).
• Check: $$\left(\tfrac{3}{5}\right)^2 + \left(\tfrac{4}{5}\right)^2 = \tfrac{9}{25} + \tfrac{16}{25} = 1,$$ so \(x^2 + y^2 = 1 \le 1\), use the standard formula.
• $$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \tfrac{3}{5}\cdot\tfrac{3}{5} + \tfrac{4}{5}\cdot\tfrac{4}{5} = \tfrac{9}{25} + \tfrac{16}{25} = 1.$$
• Answer: \(1\).

Example 2 — Solving an equation (NDA 2018-II)

Q: Find \(x\) if \(\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4\).

• Use the addition formula: $$\tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\!\left(\tfrac{2x + 3x}{1 - 2x \cdot 3x}\right) \text{ when } 6x^2 < 1.$$
• Set $$\tan^{-1}\!\left(\tfrac{5x}{1 - 6x^2}\right) = \tfrac{\pi}{4},$$ so $$\tfrac{5x}{1 - 6x^2} = \tan\tfrac{\pi}{4} = 1.$$
• $$5x = 1 - 6x^2 \implies 6x^2 + 5x - 1 = 0 \implies (6x - 1)(x + 1) = 0.$$
• \(x = \tfrac{1}{6}\) or \(x = -1\). Check \(x = -1\): \(6(-1)^2 = 6 > 1\), the formula does not apply directly; and the original equation gives \(\tan^{-1}(-2) + \tan^{-1}(-3)\) which is negative, not \(\pi/4\). So \(x = -1\) is rejected.
• Answer: \(x = \tfrac{1}{6}\) (option (a) "3 only" in the original set, corresponding to \(\tfrac{1}{6}\) being listed as value 3).

Example 3 — Triangle with arctan angles (NDA 2011-I)

Q: In triangle \(ABC\), \(A = \tan^{-1}2\) and \(B = \tan^{-1}3\). Find \(C\).

• Since \(xy = 2 \times 3 = 6 > 1\) and both \(x, y > 0\), use the modified formula: $$\tan^{-1}2 + \tan^{-1}3 = \pi + \tan^{-1}\!\left(\tfrac{2+3}{1-6}\right) = \pi + \tan^{-1}(-1).$$
• \(\tan^{-1}(-1) = -\tfrac{\pi}{4}\), so \(A + B = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4}\).
• \(C = \pi - (A + B) = \pi - \tfrac{3\pi}{4} = \tfrac{\pi}{4}\).
• Answer: \(C = \tfrac{\pi}{4}\).

Example 4 — Using the complementary identity (NDA 2022-I)

Q: If \(3\sin^{-1}x + \cos^{-1}x = \pi\), find \(x\).

• Write \(\cos^{-1}x = \tfrac{\pi}{2} - \sin^{-1}x\) (complementary identity).
• Substitute: $$3\sin^{-1}x + \left(\tfrac{\pi}{2} - \sin^{-1}x\right) = \pi.$$
• \(2\sin^{-1}x = \pi - \tfrac{\pi}{2} = \tfrac{\pi}{2}\), so \(\sin^{-1}x = \tfrac{\pi}{4}\).
• \(x = \sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}\).
• Answer: \(x = \tfrac{1}{\sqrt{2}}\).

Example 5 — Principal value with allied angle reduction

Q: Find $$\sin^{-1}\!\left(\sin \tfrac{2\pi}{3}\right)$$.

• Tempting wrong answer: cancel $$\sin^{-1}$$ and $$\sin$$ to get $$\tfrac{2\pi}{3}$$. But $$\tfrac{2\pi}{3} = 120°$$ is NOT in $$\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$$, so the identity does not apply directly.
• Reduce using the allied-angle identity $$\sin(\pi - \theta) = \sin\theta$$: $$\sin \tfrac{2\pi}{3} = \sin\!\left(\pi - \tfrac{\pi}{3}\right) = \sin \tfrac{\pi}{3}$$.
• Now $$\tfrac{\pi}{3}$$ is inside the principal branch, so $$\sin^{-1}\!\left(\sin \tfrac{\pi}{3}\right) = \tfrac{\pi}{3}$$.
• Answer: $$\tfrac{\pi}{3}$$.

Example 6 — Substitution to simplify a radical

Q: Simplify $$\tan^{-1}\!\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right)$$.

• Spot the $$\sqrt{1 + x^2}$$ pattern; substitute $$x = \tan\theta$$, so $$\theta = \tan^{-1}x$$.
• $$1 + \tan^2\theta = \sec^2\theta$$, so the expression becomes $$\tan^{-1}\!\left(\dfrac{\sec\theta - 1}{\tan\theta}\right)$$.
• Convert to sine/cosine: $$\dfrac{\sec\theta - 1}{\tan\theta} = \dfrac{1 - \cos\theta}{\sin\theta}$$.
• Apply half-angle identities: $$1 - \cos\theta = 2\sin^2\!\tfrac{\theta}{2}$$ and $$\sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$$, giving $$\tan\tfrac{\theta}{2}$$.
• So the whole expression equals $$\tan^{-1}\!\left(\tan\tfrac{\theta}{2}\right) = \tfrac{\theta}{2} = \tfrac{1}{2}\tan^{-1}x$$.
• Answer: $$\tfrac{1}{2}\tan^{-1}x$$.

Example 7 — cos(2cos⁻¹x) evaluation (NDA 2016-I)

Q: Find \(\cos(2\cos^{-1}(0.8))\).

• Let \(\theta = \cos^{-1}(0.8)\), so \(\cos\theta = 0.8\).
• Use double-angle formula: $$\cos(2\theta) = 2\cos^2\theta - 1 = 2(0.8)^2 - 1 = 2(0.64) - 1 = 0.28.$$
• Alternatively, using the identity \(\cos(\cos^{-1}x) = x\): $$\cos(2\cos^{-1}x) = 2x^2 - 1 = 2(0.64) - 1 = 0.28.$$
• Answer: \(0.28\).

Exam Shortcuts (Pro-Tips)

Inverse trig questions reward pattern-recognition over algebra. The four hacks below cut multi-step calculation into a single observation — each one has solved at least one real NDA question with no scratch work.

Shortcut 1 — Memorise the Branch Table Cold

Every wrong principal-value answer in NDA comes from a misremembered range. Lock these six in by writing them daily until they are automatic — especially the asymmetric cases.

Mnemonic: sin/tan/cosec are the "symmetric about origin" trio; cos/cot/sec are the "strictly non-negative" trio. NDA 2017-II tested this verbatim.

Shortcut 2 — The "Maximum-Sum" Identity Hack

When an equation forces an inverse trig expression to its extreme value, skip the algebra and use logic.

Reason: each term is bounded above by $$\frac{\pi}{2}$$, so the only way three of them can hit $$\frac{3\pi}{2}$$ is if every one of them equals $$\frac{\pi}{2}$$. Same logic applied to $$\cos^{-1}$$ sums forces each variable to $$-1$$.

Shortcut 3 — Sum Formula Branch Correction (xy < 1)

The arctan addition formula has a condition that NDA setters love to plant traps around.

Shortcut 4 — Substitution to Collapse Radicals

When you see a radical inside an inverse trig function, swap the variable for a trig ratio. The expression usually collapses to a half-angle or single angle.

After substitution, use $$1 + \tan^2\theta = \sec^2\theta$$ and half-angle identities to simplify, then resubstitute $$\theta = \tan^{-1}(x/a)$$ or equivalent.

Common Question Patterns

Pattern 1 — Principal value finder

Given an expression like \(\sec^{-1}(2/\sqrt{3})\) or \(\csc^{-1}(-2)\), find the exact angle. Strategy: convert to a known standard angle, check it lies in the principal branch, and write the answer. NDA 2011-II asked for the principal value of \(\sec^{-1}(2/\sqrt{3})\); the answer is \(\pi/6\) because \(\sec(\pi/6) = 2/\sqrt{3}\) and \(\pi/6 \in [0, \pi] - \{\pi/2\}\).

Pattern 2 — Composite evaluation

Expressions like \(\cos(2\cos^{-1}(0.8))\) or \(\sin(\sin^{-1}(3/5) + \sin^{-1}(4/5))\) require you to unwrap the inverse and apply a standard trig identity. The answer is always a clean decimal or simple fraction from the original input numbers.

Pattern 3 — Statement correct/incorrect

NDA loves "Consider the following statements" items. Common traps: (a) \(\tan^{-1}1 + \tan^{-1}(0.5) \ne \pi/2\) (it equals \(\tan^{-1}3\)); (b) there exists \(\theta \in (-\pi/4, \pi/4)\) for which \(\tan^{-1}(\tan\theta) \ne \theta\) — this is false, the identity holds throughout that interval; (c) \(\sin^{-1}x + \cos^{-1}x = \pi/2\) holds for \(x \in [-1, 1]\) — this is always true.

Pattern 4 — Equation solving

You are given an equation like \(\tan^{-1}(2x) + \tan^{-1}(3x) = \pi/4\) or \(4\sin^{-1}x + \cos^{-1}x = \pi\) and must find \(x\). Steps: apply the appropriate addition formula, convert to an algebraic equation, solve, and check that the solution lies within the domain.

Pattern 5 — Triangle angle from arctan

When two angles of a triangle are given as arctan values (e.g., \(A = \tan^{-1}2\), \(B = \tan^{-1}3\)), find the third angle. Use \(\tan^{-1}x + \tan^{-1}y\) with the branch correction, then subtract from \(\pi\). This pattern appeared in NDA 2011-I, 2012-I, and 2013-I.

Pattern 6 — Substitution using 2tan⁻¹

Expressions of the form $$\sin^{-1}\!\left(\tfrac{2p}{1+p^2}\right) \quad \text{or} \quad \cos^{-1}\!\left(\tfrac{1-p^2}{1+p^2}\right)$$ reduce to \(2\tan^{-1}p\). NDA 2013-I used this to convert $$\sin^{-1}\!\left(\tfrac{2a}{1+a^2}\right) + \sin^{-1}\!\left(\tfrac{2b}{1+b^2}\right) = 2\tan^{-1}x,$$ arriving at \(x = \tfrac{a+b}{1-ab}\).

Preparation Strategy

Week 1 — Domain and range cold-recall

Write out the six domain-range pairs from memory every day until they are automatic. Focus especially on the asymmetric ranges: \(\cos^{-1}\) uses \([0, \pi]\), not \([-\pi/2, \pi/2]\); \(\sec^{-1}\) excludes \(\pi/2\) from \([0, \pi]\); \(\csc^{-1}\) excludes \(0\) from \([-\pi/2, \pi/2]\). These distinctions generate wrong answers if you mix them up.

Week 2 — Identity drill

Practice the complementary identity (\(\sin^{-1}x + \cos^{-1}x = \pi/2\)) and the arctan addition formula with 20 varied inputs. Include cases where \(xy > 1\) so you are comfortable with the branch-correction. Then practice the \(2\tan^{-1}x \leftrightarrow \sin^{-1}\) and \(\cos^{-1}\) conversions.

Week 3 — PYQ simulation

Solve the entire "Properties of Triangles + Inverse Trig" chapter from the NDA topicwise book under timed conditions (roughly 3 minutes per question). Flag every question where you used a formula wrongly or ran out of time. Those are your revision points.

Linking topics

Connect this topic with Properties of Triangles — NDA frequently bundles both into the same direction-based set. Also review Trigonometric Ratios and Identities for the standard angle values you need when evaluating principal values, and glance at Derivatives and Their Applications since \(\tfrac{d}{dx}(\sin^{-1}x) = \tfrac{1}{\sqrt{1-x^2}}\) and similar derivatives appear in calculus questions that also reference inverse trig.

High-yield shortcuts

• Memorise: \(\tan^{-1}\tfrac{1}{2} + \tan^{-1}\tfrac{1}{3} = \tfrac{\pi}{4}\). This appears frequently (NDA 2013-I).
• Memorise: \(\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi\). Useful when angles-of-a-triangle questions involve these values.
• For \(\cot^{-1}(\csc^{-1}\) or \(\sec^{-1}\) expressions), convert everything to sin/cos first using definitions, then simplify.
• NDA 2025-I tested: if \(k\) is a root of \(x^2 - 4x + 1 = 0\), find \(\tan^{-1}k + \tan^{-1}(1/k)\). Since \(x^2 - 4x + 1 = 0\) gives \(k + 1/k = 4\) (both roots positive and \(k \cdot (1/k) = 1\)), the sum is \(\pi/2\) by \(\tan^{-1}x + \tan^{-1}(1/x) = \pi/2\) for \(x > 0\).

Use NDA mock tests to practise this topic under exam conditions. Past mocks consistently show that students who know the domain-range table and the arctan addition formula with its branch correction score full marks on inverse trig in under 5 minutes total.

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