Trigonometric Identities (Class 10): 3 Formulas & Proofs

In one line: a trigonometric identity is an equation involving trigonometric ratios that is true for every value of the angle, and Class 10 has three fundamental ones — all derived from the Pythagoras theorem.

The three identities are sin²A + cos²A = 1, 1 + tan²A = sec²A, and 1 + cot²A = cosec²A. Each is obtained by taking the Pythagoras relation for a right triangle and dividing it by a different side. They let you rewrite any one ratio in terms of others, which is the key to solving and proving nearly every identity-based question in the chapter.

An identity differs from an ordinary equation: an equation like x + 2 = 5 is true for only one value of x, but an identity is true for all allowed values of the angle. So sin²A + cos²A = 1 holds at 30°, at 45°, at 89° — everywhere in its valid range. This guide states all three identities with their proofs and conditions, then gives you a dependable step-by-step method for proving exam identities. If you first need the ratios these identities are built from, review Trigonometric Ratios (Class 10).

What is a trigonometric identity?

A trigonometric identity is an equation involving the trigonometric ratios of an angle that is true for all values of the angle(s) for which both sides are defined. The word “identity” is what matters here: unlike a normal equation that you solve for a specific value, an identity is always true, so your job is to prove it or use it, never to solve it.

The three Class 10 identities are sometimes called the Pythagorean identities, because each one is really the Pythagoras theorem in disguise. Once you see where they come from, you never have to memorise them blindly.

The three trigonometric identities

Here are all three, each with its meaning, the variables involved, and the exact condition under which it is true. A short verification follows each one.

Identity 1 — sin²A + cos²A = 1. Meaning: for any acute angle A, the square of its sine plus the square of its cosine always equals 1. Condition: valid for 0° ≤ A ≤ 90°. Check with a 3-4-5 triangle where sin A = 3/5 and cos A = 4/5: (3/5)² + (4/5)² = 9/25 + 16/25 = 25/25 = 1. ✓

Identity 2 — 1 + tan²A = sec²A. Meaning: one plus the square of the tangent equals the square of the secant. Condition: valid for 0° ≤ A < 90° (excluded at 90°, where tan and sec are undefined). Check with tan A = 3/4 and sec A = 5/4: 1 + (3/4)² = 1 + 9/16 = 25/16 = (5/4)² = sec²A. ✓

Identity 3 — 1 + cot²A = cosec²A. Meaning: one plus the square of the cotangent equals the square of the cosecant. Condition: valid for 0° < A ≤ 90° (excluded at 0°, where cot and cosec are undefined). Check with cot A = 4/3 and cosec A = 5/3: 1 + (4/3)² = 1 + 16/9 = 25/9 = (5/3)² = cosec²A. ✓

Notation note: sin²A means (sin A)² — the ratio squared — written that way just for convenience. It is not sin(A²).

🎬 Watch — 30-sec Short: Can you recall this essential trig identity? 🔑

Proof: where sin²A + cos²A = 1 comes from

All three identities start from one place — the Pythagoras theorem in a right triangle ABC, right-angled at B:

AB² + BC² = AC² … (the Pythagoras relation)

Identity 1 comes from dividing every term by AC² (the hypotenuse squared):

(AB/AC)² + (BC/AC)² = (AC/AC)²

Now AB/AC = cos A and BC/AC = sin A, so this becomes cos²A + sin²A = 1, i.e. sin²A + cos²A = 1. Because the Pythagoras theorem holds for the whole right triangle, this is true for every A from 0° to 90°.

Seeing a formula built from something you already trust, rather than handed to you, is what makes it stay in memory — and it is how examfront’s lessons frame every identity so you can rebuild it under exam pressure instead of drawing a blank.

Proof: the other two identities

The remaining two come from dividing the same Pythagoras relation by a different side.

Identity 2 — divide AB² + BC² = AC² by AB²:

(AB/AB)² + (BC/AB)² = (AC/AB)² → 1 + tan²A = sec²A

Here BC/AB = tan A and AC/AB = sec A. Since sec and tan need a non-zero adjacent side AB, this fails at A = 90° (where AB shrinks to zero), so the condition is 0° ≤ A < 90°.

Identity 3 — divide AB² + BC² = AC² by BC²:

(AB/BC)² + (BC/BC)² = (AC/BC)² → cot²A + 1 = cosec²A

Here AB/BC = cot A and AC/BC = cosec A. Since cot and cosec need a non-zero opposite side BC, this fails at A = 0°, so the condition is 0° < A ≤ 90°.

One theorem, three identities — divide by the hypotenuse, the adjacent side, or the opposite side, and each division gives one identity.

🎬 Watch — 30-sec Short: Can you complete this essential sec identity? 🤔

How to prove any trigonometric identity (method)

Most exam questions ask you to prove an identity — show that a messy left-hand side (LHS) equals a simpler right-hand side (RHS). Use this reliable routine:

1. Pick the more complicated side (usually the LHS) and work on it alone. Never shift terms across the equals sign as if solving — you are proving both sides are already equal.
2. Convert everything to sin and cos. Rewriting tan, cot, sec and cosec in terms of sine and cosine turns the problem into ordinary algebra.
3. Simplify using algebra and the identity sin²A + cos²A = 1 (and its rearrangements, 1 − sin²A = cos²A and 1 − cos²A = sin²A).
4. Stop when your side matches the other side. That completes the proof.

Exam tip: If one side has a single term and the other is a sum or fraction, start from the bigger, messier side and simplify down to the small one. Simplifying is far easier than trying to build complexity up.

Worked example 1 — find every ratio from one identity

Problem: Given tan A = 1/√3, find sec A, cos A, sin A, cosec A and cot A using the identities.

• Step 1 — reciprocal: cot A = 1 ÷ tan A = √3.
• Step 2 — use Identity 2: sec²A = 1 + tan²A = 1 + (1/√3)² = 1 + 1/3 = 4/3, so sec A = 2/√3 and therefore cos A = √3/2.
• Step 3 — use Identity 1: sin²A = 1 − cos²A = 1 − 3/4 = 1/4, so sin A = 1/2, and cosec A = 2.

From a single ratio, the identities deliver all the rest — no triangle drawing needed. (These are exactly the values of a 30° angle, a handy self-check.)

Worked example 2 — a proof

Problem: Prove that sec A (1 − sin A)(sec A + tan A) = 1.

• Step 1 — convert to sin and cos. Write sec A = 1/cos A and tan A = sin A/cos A: LHS = (1/cos A)(1 − sin A)(1/cos A + sin A/cos A).
• Step 2 — combine. The last bracket is (1 + sin A)/cos A, so LHS = (1 − sin A)(1 + sin A) ÷ cos²A = (1 − sin²A) ÷ cos²A.
• Step 3 — apply Identity 1. Since 1 − sin²A = cos²A, LHS = cos²A ÷ cos²A = 1 = RHS. ✓

The whole proof turned on one substitution, 1 − sin²A = cos²A. Spotting where sin²A + cos²A = 1 hides inside an expression is the skill these questions test.

🎬 Watch — 30-sec Short: Can you simplify this trig expression? 🧠

Going further

This guide covers the core of trigonometric identities: what an identity is, the three Pythagorean identities with full proofs and conditions, and the standard method for proving them, with a worked “find all ratios” example and a worked proof. The harder variations — multi-step proofs like (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ), fractions that need rationalising, expressions mixing all six ratios, and the trickier “prove and justify” questions — are where guided practice makes the biggest difference. Work through the full identity-proof set with step-by-step solutions and personalised help inside examfront, where Chapter Quizzes, Strength & Weakness Analysis and Progress Tracking pinpoint the proof types you still need to drill. To revisit the ratios underneath every identity, keep Trigonometric Ratios (Class 10) open alongside.

Key Takeaways

• A trigonometric identity is true for all allowed angle values — you prove or use it, never solve it.
• The three identities are sin²A + cos²A = 1, 1 + tan²A = sec²A, and 1 + cot²A = cosec²A.
• All three come from the Pythagoras theorem, divided by the hypotenuse, adjacent, or opposite side.
• Conditions: Identity 1 holds for 0°–90°; Identity 2 excludes 90°; Identity 3 excludes 0°.
• To prove an identity, take the messier side, convert to sin and cos, and simplify using sin²A + cos²A = 1.
• Rearrangements 1 − sin²A = cos²A and 1 − cos²A = sin²A unlock most proofs.

Quick Facts

• Trigonometric identity: an equation of ratios true for all valid angles.
• Identity 1: sin²A + cos²A = 1 (0° ≤ A ≤ 90°).
• Identity 2: 1 + tan²A = sec²A (0° ≤ A < 90°).
• Identity 3: 1 + cot²A = cosec²A (0° < A ≤ 90°).
• Source: all three are the Pythagoras theorem, divided by a side.
• sin²A means (sin A)², not sin(A²).
• Chapter: Introduction to Trigonometry · Class: 10 · Subject: Maths · Board: CBSE.

Common Mistakes

1. Solving an identity like an equation. Moving terms across the equals sign. In a proof you work on one side only until it matches the other.
2. Writing sin²A as sin(A²). sin²A means (sin A)² — the ratio squared, not the angle squared.
3. Forgetting the conditions. Using 1 + tan²A = sec²A at A = 90°, where tan and sec are undefined. Identity 2 excludes 90°; Identity 3 excludes 0°.
4. Not converting to sin and cos. Getting stuck by leaving tan, sec, cosec in place. Rewriting everything in sine and cosine almost always clears the path.
5. Missing the hidden identity. Failing to see that 1 − sin²A is cos²A (or 1 − cos²A is sin²A). Watch for “1 − square” patterns and substitute at once.

FAQ

Q. What are the three trigonometric identities in Class 10? sin²A + cos²A = 1 (for 0° ≤ A ≤ 90°), 1 + tan²A = sec²A (for 0° ≤ A < 90°), and 1 + cot²A = cosec²A (for 0° < A ≤ 90°). All three are derived from the Pythagoras theorem.

Q. How do you prove sin²A + cos²A = 1? Start from AB² + BC² = AC² in a right triangle right-angled at B. Divide every term by AC² to get (AB/AC)² + (BC/AC)² = 1, which is cos²A + sin²A = 1, i.e. sin²A + cos²A = 1.

Q. What does sin²A mean? sin²A is shorthand for (sin A)² — the sine of A, squared. It is not sin(A²). This convention is used only to make the identities easier to write.

Q. How do you prove a trigonometric identity? Take the more complicated side (usually the LHS), convert every ratio to sin and cos, then simplify using sin²A + cos²A = 1 and algebra until it equals the other side. Work on one side at a time.

Q. Why is 1 + tan²A = sec²A not valid at A = 90°? Because tan 90° and sec 90° are undefined — at 90° the adjacent side becomes zero, and tan and sec require dividing by it. So the identity holds only for 0° ≤ A < 90°.

Related Concepts

• Trigonometric Ratios (Class 10): the six ratios (sin, cos, tan, cosec, sec, cot) that every identity is built from.
• Introduction to Trigonometry (Class 10): the full Chapter 8 roadmap, including standard-angle values used to check identities.
• Heights and Distances (Class 10): where ratios and identities are applied to real problems.
• Angle of Elevation and Angle of Depression (Class 10): the angle set-ups behind applied trigonometry.

Ready to master trigonometric identities? Practise proofs step by step on examfront’s Topic Practice, and let the Practice Companion, Chapter Quizzes and Strength & Weakness Analysis guide you from the three formulas to full exam-level proofs. Start on examfront →