Trigonometric ratios are the six fundamental ratios — sine, cosine, tangent, cotangent, secant, cosecant — that relate the angles of a right triangle to its side lengths. But here's what the NDA actually tests: your ability to manipulate these ratios through identities, transform compound-angle expressions quickly, and recognise when a seemingly complicated expression collapses to a simple value.
Think of trigonometric identities as algebraic identities but written in a different language. Just as (a + b)² = a² + 2ab + b² is always true, sin²θ + cos²θ = 1 is always true — for every angle, every time. The entire game of trigonometry at NDA level is recognising which identity to apply and in which direction (sometimes you expand, sometimes you compress).
Here is the analogy that sticks in practice: imagine each identity as a gear in a clock. A question gives you the clock face in one configuration, and you need to reach another configuration by engaging the right gear. The Pythagorean identities let you swap between sin/cos and 1. The compound-angle formulas let you split or merge angles. The double-angle formulas are compression tools — they convert 2θ expressions into θ expressions or vice versa.
For NDA, trigonometry questions fall into a few recurring families: complementary-angle manipulation (like the α + β = π/2 type), evaluating expressions at special angles (30°, 45°, 60°, 18°, 36°), LCM-period questions for combined trigonometric functions, and compound-angle evaluation where you find sin/cos from given values and substitute. Recognising the family immediately is half the battle. The other half is clean arithmetic — fraction errors kill more NDA candidates than conceptual errors.
The primary identity is:
sin²θ + cos²θ = 1
Divide through by cos²θ: tan²θ + 1 = sec²θ
Divide through by sin²θ: 1 + cot²θ = cosec²θ
These three are not separate facts — they are the same fact written three ways. When you see sec²θ - tan²θ, stop — it equals 1 immediately. Similarly, cosec²θ - cot²θ = 1.
You must know these without hesitation:
| Angle | sin | cos | tan | |-------|-----|-----|-----| | 0° | 0 | 1 | 0 | | 30° | 1/2 | √3/2 | 1/√3 | | 45° | 1/√2 | 1/√2 | 1 | | 60° | √3/2 | 1/2 | √3 | | 90° | 1 | 0 | undefined | | 18° | (√5−1)/4 | √(10+2√5)/4 | — | | 36° | √(10−2√5)/4 | (√5+1)/4 | — |
For 18° and 36°, you will not reconstruct these in the exam — memorise sin 18° = (√5 − 1)/4 and cos 36° = (√5 + 1)/4. They appear in NDA with enough frequency to justify rote learning.
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B
tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
Look — the most useful derived form is: (1 + tan θ)/(1 − tan θ) = tan(45° + θ). This identity converts an algebraic fraction into a single trigonometric ratio instantly. Any time you see a fraction of the form (1 ± tan x)/(1 ∓ tan x), fire this pattern.
sin 2θ = 2 sin θ cos θ
cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
tan 2θ = 2 tan θ / (1 − tan²θ)
The three forms of cos 2θ each have a preferred use: use 1 − 2sin²θ when you want to eliminate cos, use 2cos²θ − 1 when you want to eliminate sin, and use cos²θ − sin²θ when both are present and you want to factorise.
Half-angle versions (set θ → θ/2):
sin²(θ/2) = (1 − cos θ)/2
cos²(θ/2) = (1 + cos θ)/2
tan(θ/2) = sin θ / (1 + cos θ) = (1 − cos θ)/sin θ
sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ
sin(180° − θ) = sin θ, cos(180° − θ) = −cos θ, tan(180° − θ) = −tan θ
When a question states "α and β are complementary", immediately write α + β = π/2, which means β = π/2 − α. This lets you replace every trigonometric ratio of β with the co-ratio of α. This single substitution reduces most complementary-angle questions to a one-variable problem.
sin(nx) and cos(nx) have period 2π/n. tan(nx) and cot(nx) have period π/n.
For a sum f(x) = g(x) + h(x), the period of f is LCM(period of g, period of h). To find LCM of two fractions p/q and r/s: LCM = LCM(p, r) / GCD(q, s).
When A + B + C = π (angles of a triangle), many identities simplify because (A + B)/2 = π/2 − C/2, which means tan((A + B)/2) = cot(C/2). This is the key that unlocks expressions like tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1.
When α + β = π/2, lock in α by solving the linear system with the second condition (like α − β = π/6). Add the two equations: 2α = π/2 + π/6 = 2π/3, so α = π/3. All ratios of β become co-ratios of π/3. This eliminates the need to track β at all. Standard method (substituting throughout): 6 steps. This approach: 2 steps to find α, then direct substitution.
Whenever you see (1 + tan x)/(1 − tan x), recognise it as tan(45° + x) using the compound-angle formula for tan with one angle fixed at 45°. Similarly (tan x − 1)/(tan x + 1) = −tan(45° − x). For the question (1 + tan 15°)/(1 − tan 15°): apply pattern directly — = tan(60°) = √3. Standard method (expanding tan 15°): 5-6 steps, ~45 seconds. Pattern recognition: 1 step, ~5 seconds.
Period of sin 3x is 2π/3. Period of cos 4x is π/2. To find LCM(2π/3, π/2): factor out π — find LCM(2/3, 1/2). LCM of fractions = LCM(numerators)/GCD(denominators) = LCM(2,1)/GCD(3,2) = 2/1 = 2. Multiply back: 2π. Standard method (guessing and checking multiples): 60+ seconds. Fraction-LCM formula: 15 seconds.
When tan α and tan β are unit fractions and their sum times denominator product checks out, test if tan(α + β) = 1 — that is, check whether tan α + tan β = 1 − tan α tan β. For tan α = 1/2, tan β = 1/3: LHS = 1/2 + 1/3 = 5/6, RHS = 1 − 1/6 = 5/6. Equal — so tan(α + β) = 1 immediately, without any division. Saves the final fraction division step entirely.
Do not rederive sin 18° — memorise it as (√5 − 1)/4 and its mnemonic: "five minus one, over four" (the number 5 appears because 5 × 18° = 90°). Cross-check: sin 18° ≈ 0.309, and (√5 − 1)/4 ≈ (2.236 − 1)/4 ≈ 1.236/4 ≈ 0.309. Similarly cos 36° = (√5 + 1)/4 ≈ 0.809. These two cover all NDA special-angle questions involving 18° and 36°. Deriving from scratch: 3+ minutes. From memory with verification: 10 seconds.
Read the question and immediately classify it:
Is it a complementary/supplementary angle question? If α + β = 90° or 180° appears, write the substitution for β in terms of α first. Solve the linear system to pin down exact values of α and β before doing anything else.
Is it a compound angle evaluation (cos(A−B), sin(A+B))? Extract the missing ratio using Pythagorean identity (e.g., cos A given → sin A from sin A = √(1 − cos²A)), then plug directly into the formula. No other approach is faster.
Is it a "tan fraction" form? Check if the structure matches (1 ± tan x)/(1 ∓ tan x). If yes, fire the tan(45° ± x) pattern immediately.
Is it a period question? Write the individual periods, compute LCM using the fraction-LCM formula, done.
Is it a special value (18°, 36°, 15°, 75°)? Recall from memory — do not derive.
Is it a triangle identity (A + B + C = 180°)? Use (A+B)/2 = 90° − C/2 to convert and then apply half-angle identities.
Default rule: when in doubt, try expressing everything in terms of sin and cos. Most exotic expressions simplify once you remove sec, cosec, cot.
Why this question: Tests whether you can combine two conditions (complementary + ratio condition) and use componendo-dividendo.
Solving path: Write α + β = π/2 (complementary) and α − β = π/6. Add: 2α = 2π/3, so α = π/3, β = π/6. Since β = π/2 − α, tan β = cot α. So m cot α = n tan α, giving m/n = tan²α = tan²(π/3) = 3. Now (m+n)/(m-n): divide numerator and denominator by n to get (m/n + 1)/(m/n − 1) = (3+1)/(3−1) = 4/2 = 2.
Why this question: Period questions require LCM of fractional periods — a mechanical skill that many candidates mishandle.
Solving path: Period of sin 3x = 2π/3. Period of cos 4x = 2π/4 = π/2. LCM of 2π/3 and π/2: factor out π, find LCM of 2/3 and 1/2 = LCM(2,1)/GCD(3,2) = 2/1 = 2. Multiply π back: period = 2π.
Why this question: The cleanest example of compound-angle evaluation — extract sin from cos using Pythagoras, then substitute.
Solving path: cos A = 3/5 → sin A = 4/5 (3-4-5 triple). cos B = 5/13 → sin B = 12/13 (5-12-13 triple). cos(A − B) = cos A cos B + sin A sin B = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65.
Why this question: Tests whether you recognise that the tan addition formula can simplify to a whole number.
Solving path: tan(α + β) = (1/2 + 1/3)/(1 − 1/2 × 1/3) = (5/6)/(5/6) = 1. Note that numerator and denominator are identical — this is the shortcut check described in the Tricks section.
Why this question: sin 18° is a recurring NDA special value and tests exact value knowledge.
Solving path: Recall directly: sin 18° = (√5 − 1)/4. Verification: (√5 − 1)/4 ≈ (2.236 − 1)/4 ≈ 0.309 ≈ sin 18°. Option A matches.
Why this question: Tests the tan-fraction pattern — (1 + tan θ)/(1 − tan θ) = tan(45° + θ).
Solving path: Apply the identity (1 + tan θ)/(1 − tan θ) = tan(45° + θ) with θ = 15°. Result: tan(60°) = √3.
Why this question: Triangle identities with half-angles appear repeatedly in NDA and require knowing the A + B + C = π reduction.
Solving path: Since A + B + C = 180°, we have (A + B)/2 = 90° − C/2, so tan((A+B)/2) = cot(C/2). Use this to write tan(A/2 + B/2) = 1/tan(C/2), which expands using the tan-addition formula to give (tan(A/2) + tan(B/2))/(1 − tan(A/2)tan(B/2)) = 1/tan(C/2). Cross-multiplying and rearranging yields tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1.
Dropping the negative sign in cos(A − B). The formula is cos A cos B + sin A sin B — note the positive sign, not minus. Students confuse it with cos(A + B) = cos A cos B − sin A sin B. Write the formulas once before each exam session until they are automatic.
Taking sin A = √(1 − cos²A) as positive without checking the quadrant. In NDA questions, both angles are usually stated as acute, so this is safe — but if the question says "A is in the second quadrant", sin A is positive while cos A is negative. Always read the quadrant specification.
Computing LCM of periods by taking the larger period. LCM of 2π/3 and π/2 is 2π, not 2π/3. The LCM of two fractions is not simply the larger one. Use the formula: LCM(a/b, c/d) = LCM(a,c)/GCD(b,d).
Confusing sin 18° = (√5 − 1)/4 with cos 36° = (√5 + 1)/4. The minus sign goes with sin 18°, the plus sign goes with cos 36°. A quick numerical check: sin 18° ≈ 0.309 < 0.5, and (√5 − 1)/4 ≈ 0.309. If you get a value greater than 0.5, you have used the wrong formula.
In the tan-fraction pattern, reversing the sign of the identity. (1 + tan θ)/(1 − tan θ) = tan(45° + θ), not tan(45° − θ). If the question has (1 − tan θ)/(1 + tan θ), that equals tan(45° − θ).
Applying the period formula to |sin x| or sin²x without adjusting. The period of |sin x| is π, not 2π, and the period of sin²x is also π (use sin²x = (1 − cos 2x)/2 — the resulting cosine has period π). Standard period rules apply only to sin(nx) directly — modulus and even powers halve the period.