Trigonometry in CDS sits at a comfortable intersection of pattern-recognition and formula recall. The exam does not throw you curveballs from higher mathematics — it tests whether you know the core identities cold enough to simplify an ugly expression in under 90 seconds.
Here is the mental model that unifies everything: every trigonometric ratio is a ratio of two sides of a right triangle. Once you fix one ratio, all six ratios are determined. That is it. The identities are not arbitrary rules — they are algebraic consequences of the Pythagorean theorem applied to that right triangle.
Think of it this way. You have a right triangle with hypotenuse h, opposite side p, and adjacent side b. Then:
sin θ = p/h, cos θ = b/h, tan θ = p/bcosec θ = h/p, sec θ = h/b, cot θ = b/pEvery identity you will ever use in CDS flows from p² + b² = h². Divide both sides by h² and you get sin²θ + cos²θ = 1. Divide by b² and you get tan²θ + 1 = sec²θ. Divide by p² and you get 1 + cot²θ = cosec²θ.
The complementary-angle relations come from a different geometric observation: in a right triangle, the two acute angles add up to 90°. So the sine of one is the cosine of the other. sin(90° - θ) = cos θ, tan(90° - θ) = cot θ, and so on.
CDS frequently uses this to collapse expressions like sin 68°/cos 22° into 1, because 68° + 22° = 90°. Spot these pairs in the exam hall — they are almost always a deliberate setup.
The angle sum formulae (sin(A+B), tan(A+B)) appear less often in CDS than in JEE, but they show up in equations where you are given one ratio and asked to find another. Know them, do not derive them in the hall.
These are the backbone. Learn them as a family, not as three separate facts.
| Base identity | Rearrangement 1 | Rearrangement 2 |
|---|---|---|
| sin²θ + cos²θ = 1 | sin²θ = 1 - cos²θ | cos²θ = 1 - sin²θ |
| 1 + tan²θ = sec²θ | tan²θ = sec²θ - 1 | sec²θ - tan²θ = 1 |
| 1 + cot²θ = cosec²θ | cot²θ = cosec²θ - 1 | cosec²θ - cot²θ = 1 |
The factored forms (sec θ - tan θ)(sec θ + tan θ) = 1 and (cosec θ - cot θ)(cosec θ + cot θ) = 1 appear in CDS expressions that look complicated but collapse in one step.
Do not go into the exam without these absolutely automatic:
| θ | 0° | 30° | 45° | 60° | 90° | |---|---|---|---|---|---| | sin | 0 | 1/2 | 1/√2 | √3/2 | 1 | | cos | 1 | √3/2 | 1/√2 | 1/2 | 0 | | tan | 0 | 1/√3 | 1 | √3 | undefined |
Memory ladder for sin: 0, 1/2, 1/√2, √3/2, 1 — numerators under √ go 0, 1, 2, 3, 4, then divide by 2. Cos is sin read backwards.
When two angles add to 90°:
sin(90° - θ) = cos θ
cos(90° - θ) = sin θ
tan(90° - θ) = cot θ
sec(90° - θ) = cosec θ
In CDS, expressions are routinely built with complementary pairs so that they collapse to 1 or cancel out. Your first move on any multi-term expression: scan for pairs that sum to 90°.
Given one ratio — say tan A = 3/4 — draw the triangle immediately. Opposite = 3, adjacent = 4, hypotenuse = 5 (it is a 3-4-5 triangle). Now read off any ratio you need directly. This is faster than the algebraic identity path and less error-prone under time pressure.
The identity path: sin²A + cos²A = 1 requires two algebraic steps and a square root. The triangle method requires one mental picture. Use the triangle.
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)
For CDS, the tan(A+B) formula is the most tested because it lets you work backwards: given tan(A+B) and tan A, find tan B. Set up the equation and solve algebraically.
tan(A - B) = (tan A - tan B) / (1 + tan A tan B)
sin 2A = 2 sin A cos A
cos 2A = cos²A - sin²A = 1 - 2sin²A = 2cos²A - 1
tan 2A = 2 tan A / (1 - tan²A)
CDS uses these to simplify product expressions. If you see sin θ cos θ, think sin 2θ / 2 immediately.
tan θ · cot θ = 1 (always — this eliminates pairs in multi-term products)
sin θ · cosec θ = 1
cos θ · sec θ = 1
In any long product like tan 20° · tan 40° · tan 45° · tan 50° · tan 70°, pair up complementary angles first: tan 20° · tan 70° = 1, tan 40° · tan 50° = 1, leaving only tan 45° = 1. The whole product is 1.
Before simplifying any expression, scan all angle values for pairs summing to 90°. Replace each pair: sin/cos of complementary angles collapse to 1 when divided, and tan/cot pairs also collapse to 1 in products. In the PYQ with 68° and 22°: spotting that 68 + 22 = 90 takes 3 seconds; computing sin 68° numerically takes 30 seconds and is impossible without a calculator anyway. Standard approach: unavailable. Pattern approach: 3 seconds.
When given one ratio (tan, sin, cos, sec, cosec, cot), immediately sketch the 3 sides of a right triangle and label them. For cos A = 4/5: adjacent = 4, hypotenuse = 5, opposite = √(25-16) = 3. Then any other ratio is a direct read — no further algebra needed. Triangle method: 2 steps. Identity algebra method: 4-5 steps with a square root. Use the triangle every time.
When you see (1 + tan²θ) or (sec²θ - 1) in any expression, replace instantly: 1 + tan²θ = sec²θ and 1 - sin²θ = cos²θ. The product sec²θ · cos²θ = 1 in one step. Standard expansion: 6 algebraic steps. Pattern substitution: 2 steps. This specific simplification appears verbatim in CDS PYQs.
In any product of tangents at multiple angles, pair complementary angles first. tan α · tan(90°-α) = tan α · cot α = 1. Eliminate pairs one by one, then handle what remains. For five-angle products in CDS, this reduces the problem to evaluating a single standard angle. Standard multiplication through values: 5 steps + computation. Complementary pairing: 2 steps, all mental.
When an equation gives you tan θ = 1 or sin θ = 1/2, substitute the standard angle directly (θ = 45° or θ = 30°). Then evaluate any expression at that angle using the standard value table. This avoids identity manipulation entirely. Equation tan θ = 1 → θ = 45° takes 3 seconds; solving (sin θ + cos θ)² = ? via identity algebra takes 45 seconds. Substitution wins every time for standard angles.
When a trigonometry question appears in CDS:
Step 1 — Identify the question type.
Step 2 — Scan for complementary pairs. Look for any two angles that sum to 90°. Replace sin/cos pairs and tan/cot pairs immediately. Most CDS simplification problems are 80% solved at this step.
Step 3 — Apply Pythagorean identity substitutions.
Look for 1 + tan²θ, 1 - sin²θ, sec²θ - 1, cosec²θ - 1. Replace with their one-term equivalents.
Step 4 — Handle products.
Use tan θ · cot θ = 1, sin θ · cosec θ = 1, cos θ · sec θ = 1 to cancel reciprocal pairs.
Step 5 — Evaluate residual standard angles. Whatever survives, substitute from the standard value table.
If an expression is still complex after all five steps, check whether it factors as (sec θ - tan θ)(sec θ + tan θ) or a similar Pythagorean factored form.
Why this question: Tests whether you can chain three different complementary-angle and product collapses in one expression — the exact skill CDS rewards.
Solving path: Break into three terms. Term 1: 2 sin 68° / cos 22°. Since sin 68° = sin(90° - 22°) = cos 22°, this equals 2 cos 22° / cos 22° = 2. Term 2: 2 cot 15° / (5 tan 75°). Since cot 15° = tan 75°, this equals 2 tan 75° / (5 tan 75°) = 2/5. Term 3: the product tan 20° · tan 40° · tan 45° · tan 50° · tan 70°. Pair: tan 20° · tan 70° = 1, tan 40° · tan 50° = 1, tan 45° = 1. Product = 1. So Term 3 = 3(1)/5 = 3/5. Final: 2 - 2/5 - 3/5 = 2 - 5/5 = 2 - 1 = 1.
Why this question: Standard angle recall test — the foundation of all CDS trigonometry.
Solving path: sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1. Sum: 1/2 + 1/2 - 1 = 0.
Why this question: Shows the substitution shortcut — recognise a standard angle from a given ratio, then evaluate directly.
Solving path: tan θ = 1 means θ = 45°. At 45°: sin 45° = cos 45° = 1/√2. So (sin θ + cos θ)² = (1/√2 + 1/√2)² = (2/√2)² = (√2)² = 2.
Why this question: Classic right-triangle-method setup — given one ratio, find another using the triangle.
Solving path: cos A = 4/5. Draw the triangle: adjacent = 4, hypotenuse = 5, opposite = √(25 - 16) = 3. Therefore sin A = 3/5 and tan A = sin A / cos A = (3/5)/(4/5) = 3/4.
Why this question: Tests the Pythagorean identity substitution in a compound expression — the most frequent identity trap in CDS.
Solving path: (1 + tan²θ)(1 - sin²θ). Apply identities: 1 + tan²θ = sec²θ and 1 - sin²θ = cos²θ. Product becomes sec²θ · cos²θ = (1/cos²θ) · cos²θ = 1.
Forgetting the triangle for ratio problems. Students who try to find tan A from cos A purely algebraically waste 45 seconds and make sign errors. Draw the right triangle — it takes 5 seconds and is infallible.
Missing complementary pairs in multi-angle expressions. The question is designed to collapse. If your working is getting messy, you have almost certainly missed a complementary pair. Scan angles again before expanding anything.
Confusing sin²θ + cos²θ = 1 with sin θ + cos θ = 1. These are not the same. The linear sum sin θ + cos θ ranges from −√2 to √2 and is not constant. The squared identity is always 1.
Applying the angle sum formula in the wrong direction. In tan(A + B) = (tan A + tan B)/(1 - tan A tan B), if you are solving for tan B, rearrange algebraically step by step. Students often make a sign error by guessing the rearrangement.
Assuming tan(A + B) = tan A + tan B. This is wrong. The product term (1 - tan A tan B) in the denominator is always present and always matters. In CDS problems it is precisely the term that distinguishes correct from incorrect options.
Treating sec θ as 1/sin θ. sec θ = 1/cos θ, not 1/sin θ. The reciprocal of sin is cosec θ. This confusion under time pressure consistently causes wrong answers on otherwise straightforward questions.