Trigonometry studies the relationship between the angles and sides of a right-angled triangle. In the CDS / OTA paper it delivers several almost-guaranteed marks because the questions follow a small, fixed set of ratios and identities. This Cavalier guide builds the chapter from the right triangle upward, fixes the standard-angle table, and drills the exact identity patterns the exam repeats.
Why trigonometry matters in CDS
Trigonometry is one of the most dependable scoring areas in CDS / OTA Maths. Every paper carries a cluster of questions on ratios, identities and standard-angle values, and they reward memory more than long calculation. Unlike geometry, where a single problem can swallow several minutes, a well-prepared candidate clears most trigonometry items in under a minute each.
The topic also powers the Height and Distance chapter, where angles of elevation and depression are solved using the same ratios. So the few hours you spend here pay off twice on the paper. The word itself comes from the Greek for “measuring triangles”, and that is exactly what every question reduces to once you strip away the wording.
For the OTA paper in particular, the cut-off rewards speed and accuracy on these guaranteed questions, so leaving trigonometry half-prepared is a costly gap. Aim to treat this chapter as a free block of marks rather than an optional extra.
In CDS, angles are almost always the standard ones — 0°, 30°, 45°, 60° and 90°. If you know their ratio values cold, most questions become single-step substitutions.
The right triangle and six ratios
Take a right-angled triangle and fix attention on one acute angle θ. Relative to θ, the three sides are named: the side facing θ is the perpendicular (opposite), the side touching θ that is not the hypotenuse is the base (adjacent), and the longest side facing the right angle is the hypotenuse.
The six ratios are built from these three sides:
- sin θ = perpendicular ÷ hypotenuse
- cos θ = base ÷ hypotenuse
- tan θ = perpendicular ÷ base
- cosec θ = 1 ÷ sin θ
- sec θ = 1 ÷ cos θ
- cot θ = 1 ÷ tan θ
tan θ = sin θ ÷ cos θ and cot θ = cos θ ÷ sin θ. Every ratio can be rebuilt from sin and cos alone.
The mnemonic “Pandit Badri Prasad / Har Har Bole / Sona Chandi Tolo” gives Perp/Hyp, Base/Hyp, Perp/Base for sin, cos, tan respectively. Use whatever phrase helps you place the sides correctly.
The standard-angle table
This single table answers the majority of CDS trig questions. Memorise the sin row and you can build the rest.
- sin: 0° = 0, 30° = ½, 45° = 1÷√2, 60° = √3÷2, 90° = 1
- cos: 0° = 1, 30° = √3÷2, 45° = 1÷√2, 60° = ½, 90° = 0
- tan: 0° = 0, 30° = 1÷√3, 45° = 1, 60° = √3, 90° = not defined
Write 0, 1, 2, 3, 4 under 0°, 30°, 45°, 60°, 90°. Divide each by 4, take the square root, and you get the sin row (0, ½, 1÷√2, √3÷2, 1). Reverse it for cos. This rebuilds the whole table if your memory blanks.
tan 90° and sec 90° are not defined (cos 90° = 0 makes the denominator zero). Likewise cot 0° and cosec 0° are not defined. Writing “0” or “∞” as a final option here loses easy marks.
The three Pythagorean identities
These three identities are the engine of every “prove that” and “simplify” question. They hold for every angle for which the ratios are defined.
sin2θ + cos2θ = 1
1 + tan2θ = sec2θ
1 + cot2θ = cosec2θ
Each one is just the first identity in disguise. Divide sin2θ + cos2θ = 1 by cos2θ to get the second, and by sin2θ to get the third. The useful rearrangements are worth memorising too:
- sin2θ = 1 − cos2θ and cos2θ = 1 − sin2θ
- sec2θ − tan2θ = 1
- cosec2θ − cot2θ = 1
sin2θ means (sin θ)2, not sin(θ2). The superscript 2 squares the whole ratio, not the angle.
Complementary-angle relations
Two angles are complementary when they add up to 90°. In a right triangle the two acute angles are always complementary, because the three angles total 180° and one of them is already the 90° right angle. This single fact gives a neat set of conversions that CDS examiners love to test.
sin(90° − θ) = cos θ, cos(90° − θ) = sin θ
tan(90° − θ) = cot θ, cot(90° − θ) = tan θ
sec(90° − θ) = cosec θ, cosec(90° − θ) = sec θ
The pattern is simple: each ratio turns into its co-ratio (sin↔cos, tan↔cot, sec↔cosec) when the angle is replaced by its complement. This is why a question like sin 60° can also be read as cos 30°. The prefix “co” in cosine, cotangent and cosecant literally records this link — cosine is the sine of the complementary angle.
A typical exam line is sin 25° sec 65°. Since 25° and 65° are complementary, sec 65° = cosec 25° = 1÷sin 25°, so the whole product collapses to 1. Spotting the complement saves you from ever touching a table.
When you see angles that add to 90° — like sin 35° and cos 55° — convert one so both match. Often the whole expression collapses to 1 or 0 instantly.
Range and basic limits of ratios
CDS sometimes tests whether a given value of a ratio is even possible, so the natural limits are worth knowing.
- sin θ and cos θ always lie between −1 and 1, that is −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1.
- cosec θ and sec θ are never between −1 and 1; their magnitude is always ≥ 1.
- tan θ and cot θ can take any real value.
If a question states sin θ = 3÷2 or cos θ = 1.4, the answer is “not possible”, because these exceed the maximum value of 1. Watch for this trap in the options.
For the acute angles met in CDS (0° to 90°), every ratio is non-negative, so you rarely worry about signs unless the question explicitly moves beyond a right triangle.
It also helps to remember how each ratio behaves as the angle grows from 0° to 90°. Sine rises steadily from 0 to 1, cosine falls from 1 to 0, and tangent climbs from 0 without any upper bound. Knowing these trends lets you reject impossible options at a glance, even before you reach for an exact value.
Worked example: evaluating an expression
A classic CDS pattern asks you to plug standard-angle values into a fraction and simplify. Take the values straight from the table.
Evaluate (sin 30° + cos 60°) ÷ (tan 45°).
The expression equals 1. No triangle is drawn — the standard-angle table alone finishes the sum in three lines.
Worked example: proving an identity
Identity questions look hard but almost always reduce to sin2θ + cos2θ = 1. Convert everything to sin and cos, then simplify.
Prove that (1 − sin2θ) × sec2θ = 1.
By replacing 1 − sin2θ with cos2θ and writing sec2θ as its reciprocal, the whole left side cancels to 1.
The winning strategy for almost every CDS identity: convert each ratio to sin and cos, look for sin2θ + cos2θ, and cancel. It rarely fails.
Finding other ratios from one
A frequent question gives one ratio and asks for another. Two methods work; pick whichever is faster.
Method 1: the triangle method
If sin θ = 3÷5, treat 3 as the perpendicular and 5 as the hypotenuse. By Pythagoras the base = √(52 − 32) = 4. Now every ratio is readable: cos θ = 4÷5, tan θ = 3÷4, and so on.
Method 2: the identity method
Use cos θ = √(1 − sin2θ). With sin θ = 3÷5, cos θ = √(1 − 9÷25) = √(16÷25) = 4÷5. Same answer, no figure needed.
The triangle method is faster when the numbers form a known Pythagorean triple (3-4-5, 5-12-13, 8-15-17). Spot the triple and you skip the square-root step entirely.
Keep the three common triples on the tip of your tongue. The moment a ratio shows 5 and 12, or 8 and 15, you already know the missing side is 13 or 17 respectively, and every other ratio follows in seconds. This recognition is what separates a fast attempt from a slow one in the exam hall.
Previous-year style question
This question mirrors the difficulty and phrasing of recent CDS papers, where one ratio is given and an expression is asked.
Q. If tan θ = 4÷3, then the value of (sin θ + cos θ) is:
(a) 7÷5 (b) 1 (c) 5÷7 (d) 7÷3
Answer: (a) 7÷5. With tan θ = 4÷3, take perpendicular = 4, base = 3, so hypotenuse = √(42 + 32) = 5. Then sin θ = 4÷5 and cos θ = 3÷5, giving sin θ + cos θ = 4÷5 + 3÷5 = 7÷5.
Recognise the 3-4-5 triple the moment you see tan θ = 4÷3. The hypotenuse is 5 without any calculation, and the rest follows instantly.
Common traps to avoid
A handful of slips cost most of the lost marks in this chapter. Train yourself to spot them.
- Treating sin2θ as sin(θ2) — it means (sin θ)2.
- Forgetting that tan 90°, sec 90°, cot 0° and cosec 0° are not defined.
- Accepting sin θ or cos θ values outside the range −1 to 1.
- Mixing up the complementary rule — sin(90° − θ) is cos θ, not −sin θ.
- Adding ratios wrongly: sin(A + B) is not sin A + sin B.
cosec θ is the reciprocal of sin θ, but it is not sin−1θ. The notation sin−1 means the inverse function (the angle), a completely different thing.
Quick recap and revision
- sin = perp÷hyp, cos = base÷hyp, tan = perp÷base; the other three are their reciprocals.
- Standard angles: sin row is 0, ½, 1÷√2, √3÷2, 1 for 0° to 90°; reverse for cos.
- Identities: sin2θ + cos2θ = 1, 1 + tan2θ = sec2θ, 1 + cot2θ = cosec2θ.
- Complement rule: each ratio → its co-ratio when the angle becomes 90° − θ.
- sin and cos stay within −1 and 1; sec and cosec stay outside.
- For identities, convert all to sin and cos and look for sin2 + cos2.
Revise these six lines the night before the exam and practise five mixed sums. That alone secures the trigonometry marks on the CDS / OTA paper.
Frequently asked questions
Which angles appear most in CDS trigonometry questions?
The standard angles 0°, 30°, 45°, 60° and 90° dominate the paper. If you memorise the sin, cos and tan values for these five angles, most questions become a one-step substitution.
How can I rebuild the standard-angle table if I forget it?
Write 0, 1, 2, 3, 4 under the five angles, divide each by 4, then take the square root. This gives the sin row (0, 1/2, 1/√2, √3/2, 1). Reverse it for cos, and tan equals sin divided by cos.
What is the single most useful trigonometric identity?
sin²θ + cos²θ = 1. Nearly every CDS identity or simplification question reduces to it once you convert all ratios to sine and cosine, so it is the first tool to reach for.
Why are tan 90° and sec 90° undefined?
Because both have cos 90° in the denominator, and cos 90° = 0. Division by zero is undefined, so tan 90° and sec 90° have no value. The same reasoning makes cot 0° and cosec 0° undefined.
How do I find one ratio when another is given?
Use the right-triangle method: place the given ratio's numerator and denominator as two sides, find the third side by Pythagoras, then read off the ratio you need. Recognising triples like 3-4-5 makes this almost instant.
Is cosec θ the same as sin⁻¹θ?
No. cosec θ is the reciprocal 1/sin θ, while sin⁻¹θ denotes the inverse function that returns an angle. They are different concepts and mixing them up is a common exam error.
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