Trigonometric Ratios and Identities

What This Topic Covers

The NDA syllabus groups trigonometry into angles and their measurement, trigonometric ratios, identities and their use in problem-solving, and graphs of trigonometric functions. The NCERT Class 11 Chapter 3 treatment covers all of this and is the correct preparation base. The sub-areas examined are:

• Angle measurement — degrees, radians, and conversion between them
• Definition of the six trigonometric ratios from a right triangle and the unit circle
• Standard values at 0°, 30°, 45°, 60°, 90°, 180°, 270°, 360°
• Reciprocal and quotient relations among the six ratios
• Pythagorean identities and their algebraic rewriting
• Sum, difference, double-angle, and half-angle formulas
• Allied angles and quadrant-based sign rules (ASTC)
• Maximum and minimum values of linear combinations \((a\sin\theta + b\cos\theta)\)

Exam Pattern & Weightage

The table below is drawn from the topic-wise PYQ source material covering NDA papers from 2010 to 2025. Both Paper-I sittings per year are counted separately where data is available.

Core Concepts

Trigonometric Ratios

For a right-angled triangle with angle \(\theta\), hypotenuse \(H\), opposite side \(O\), and adjacent side \(A\), the six ratios are defined as \(\sin\theta = O/H\), \(\cos\theta = A/H\), and \(\tan\theta = O/A\), with their reciprocals cosec, sec, and cot respectively. The unit-circle definition extends these to all real angles: a point \((\cos\theta, \sin\theta)\) traces the unit circle as \(\theta\) increases from \(0\) to \(2\pi\). This is the definition that makes allied angles and quadrant rules work.

Trigonometric Identities

An identity holds for all valid values of the angle, unlike an equation which holds only for specific solutions. The three Pythagorean identities are derived from \(\sin^2\theta + \cos^2\theta = 1\) by dividing both sides by \(\cos^2\theta\) and \(\sin^2\theta\) respectively — so memorising one equation gives you all three.

Allied Angles and Quadrant Signs

Allied angles are angles of the form \((n\cdot 90^\circ \pm \theta)\) where \(n\) is a positive integer. The ASTC rule — All, Sin, Tan, Cos — identifies which ratios are positive in quadrants I through IV going anti-clockwise: all ratios positive in Q1, only sin (and cosec) in Q2, only tan (and cot) in Q3, only cos (and sec) in Q4.

Inverse Trigonometric Functions (brief)

Inverse trig functions (\(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) and their principal-value ranges) are a separate NDA topic. They build directly on the ratio definitions covered here. See the dedicated page on Inverse Trigonometric Functions for NDA-specific treatment of that chapter.

Worked Examples — Solved NDA PYQs

Every question below is drawn from NDA previous year papers as classified in the PYQ source material.

Example 1 — Large Angle Reduction: \(\sin(1920^\circ)\) [NDA 2012-I]

Question: What is the value of \(\sin(1920^\circ)\)?

• Divide 1920 by 360: $$1920 = 5 \times 360 + 120 \implies \sin(1920^\circ) = \sin(120^\circ).$$
• 120° lies in the second quadrant (90° to 180°). Use the rule: \(\sin(180^\circ - \theta) = \sin\theta\).
• $$\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}.$$

Answer: \(\dfrac{\sqrt{3}}{2}\)

Example 2 — Compound Angle Expression [NDA 2012-I]

Question: What is the value of \(\sin 420^\circ \cdot \cos 390^\circ + \cos(-300^\circ)\cdot \sin(-330^\circ)\)?

• Reduce \(\sin 420^\circ\): \(420 - 360 = 60^\circ\), so $$\sin 420^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}.$$
• Reduce \(\cos 390^\circ\): \(390 - 360 = 30^\circ\), so $$\cos 390^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}.$$
• Reduce \(\cos(-300^\circ)\): cos is even, so \(\cos(-300^\circ) = \cos(300^\circ)\). \(300^\circ = 360^\circ - 60^\circ\), so $$\cos 300^\circ = \cos 60^\circ = \frac{1}{2}.$$
• Reduce \(\sin(-330^\circ)\): sin is odd, so \(\sin(-330^\circ) = -\sin(330^\circ)\). \(330^\circ = 360^\circ - 30^\circ\), so \(\sin 330^\circ = -\sin 30^\circ = -\tfrac{1}{2}\). Therefore $$\sin(-330^\circ) = \frac{1}{2}.$$
• Substitute: $$\left(\frac{\sqrt{3}}{2}\right)\!\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\!\left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1.$$

Answer: 1

Example 3 — Identity Manipulation [NDA 2013-I]

Question: What is the value of \((\sin^4\theta - \cos^4\theta + 1)\,\csc^2\theta\)?

• Factor \(\sin^4\theta - \cos^4\theta\) using difference of squares: $$\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta).$$
• Since \(\sin^2\theta + \cos^2\theta = 1\), this simplifies to \((\sin^2\theta - \cos^2\theta)\).
• So the bracket becomes: \(\sin^2\theta - \cos^2\theta + 1\).
• Substitute \(1 = \sin^2\theta + \cos^2\theta\): $$\sin^2\theta - \cos^2\theta + \sin^2\theta + \cos^2\theta = 2\sin^2\theta.$$
• Multiply by \(\csc^2\theta = \tfrac{1}{\sin^2\theta}\): $$2\sin^2\theta \cdot \frac{1}{\sin^2\theta} = 2.$$

Answer: 2

Example 4 — \((1+\tan A)(1+\tan B) = 2\) when \(A+B = 45^\circ\) [NDA 2011-I]

Question: If \(\alpha\) and \(\beta\) are positive angles such that \(\alpha + \beta = \pi/4\), then what is \((1 + \tan\alpha)(1 + \tan\beta)\) equal to?

• Since \(\alpha + \beta = 45^\circ\), take tan of both sides: $$\tan(\alpha + \beta) = \tan 45^\circ = 1.$$
• Apply the addition formula: $$\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = 1.$$
• So \(\tan\alpha + \tan\beta = 1 - \tan\alpha\tan\beta\), which gives $$\tan\alpha + \tan\beta + \tan\alpha\tan\beta = 1.$$
• Now expand $$(1 + \tan\alpha)(1 + \tan\beta) = 1 + \tan\alpha + \tan\beta + \tan\alpha\tan\beta.$$
• Substitute the result from step 3: \(1 + 1 = 2\).

Answer: 2

Example 5 — Maximum Value [NDA 2011-I]

Question: What is the maximum value of \(3\cos x + 4\sin x + 5\)?

• Identify the expression \(a\cos x + b\sin x\) where \(a = 3\), \(b = 4\).
• Maximum value of \(a\cos x + b\sin x\): $$\sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$
• The full expression is \((3\cos x + 4\sin x) + 5\). Its maximum is \(5 + 5 = 10\).

Answer: 10

Example 6 — Max-Min of \(a\sin x + b\cos x + c\)

Question: Find the maximum and minimum values of \(3\sin x + 4\cos x - 2\).

• Isolate the trigonometric part: \(3\sin x + 4\cos x\). Here \(a = 3\), \(b = 4\).
• Apply the standard form: $$\text{max of } a\sin x + b\cos x = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = 5; \quad \text{min} = -5.$$
• Apply the constant: max of the full expression \(= 5 + (-2) = 3\); min \(= -5 + (-2) = -7\).

Answer: Max = 3, Min = −7

Exam Shortcuts (Pro-Tips)

Trigonometry is a clock-killer if you solve every question algebraically. The three hacks below convert 2–3 minutes of work into under 20 seconds. Each one targets a specific NDA pattern.

Shortcut 1 — The "Value Putting" Method

If the expression carries a variable like \(\theta\), \(A\), or \(B\) but every answer option is a pure number (0, 1, 2, −1), the expression is independent of the angle. Pick a safe angle (30°, 45°, or 60°) and substitute — the answer falls out in one line.

Example: to evaluate \((1 + \tan\alpha)(1 + \tan\beta)\) when \(\alpha + \beta = 45^\circ\), set \(\alpha = \beta = 22.5^\circ\)? No — just set \(\alpha = 45^\circ\), \(\beta = 0^\circ\): $$(1 + 1)(1 + 0) = 2.$$ Done.

Shortcut 2 — Doubling-Angle Cosine Product

Whenever you see a chain of cosines whose angles double successively (\(A, 2A, 4A, \ldots\)), the entire product collapses to a single fraction. This recurs in NDA papers under the disguise of "evaluate \(\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ\)" — that pattern matches with \(A = 20^\circ\) and \(n = 3\).

Apply directly: $$\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ = \frac{\sin 160^\circ}{8\sin 20^\circ} = \frac{\sin 20^\circ}{8\sin 20^\circ} = \frac{1}{8}.$$

Shortcut 3 — 60° Symmetry Identities

Products of three trig ratios at angles \(\theta\), \((60^\circ - \theta)\), \((60^\circ + \theta)\) compress into a single ratio of \(3\theta\). NDA 2015-I tested this exact pattern. Memorise the three forms — they convert a four-line solution into one substitution.

Example: $$\sin 20^\circ \cdot \sin 40^\circ \cdot \sin 80^\circ = \sin 20^\circ \cdot \sin(60^\circ - 20^\circ) \cdot \sin(60^\circ + 20^\circ) = \tfrac{1}{4}\sin 60^\circ = \frac{\sqrt{3}}{8}.$$

Common Question Patterns

Six patterns cover approximately 85% of PYQs in this cluster. Recognising the pattern before solving is the fastest route to the correct option.

Preparation Strategy

Three phases cover the chapter systematically. Each phase builds on the previous one and ends with a PYQ check.

Phase 1 — Standard Values and Base Identities (Days 1–5): Write out the full standard values table for 0° to 360° from memory every day until it is instantaneous. Derive (do not just memorise) the three Pythagorean identities from \(\sin^2\theta + \cos^2\theta = 1\). Practice 10 simplification problems per day using only these identities. By day 5, identity manipulation problems from 2010–2015 papers should be solvable in under 90 seconds.

Phase 2 — Allied Angles and PYQ Drilling (Days 6–11): Master the ASTC rule and the allied angle transformation table. Practice every "\(\sin\)(large angle)" PYQ from 2010–2024 until the reduction is reflexive. Work through sum-and-difference and double-angle formula PYQs. Aim for zero errors on the compound-angle question type. Use the \((1+\tan A)(1+\tan B)=2\) pattern as a benchmark — if you can reproduce it cold, you have the sum-difference formula properly loaded.

Phase 3 — Mock Integration (Days 12–14): Attempt 2–3 full NDA Maths mocks under timed conditions. In the debrief, categorise every trig question by pattern. Confirm that you are solving each pattern within 60 seconds. After this phase, move to Properties of Triangles (which uses sine and cosine rules built on these identities) and Height and Distance (pure application of standard values and tan ratios).

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