Trigonometric Identities: A Memory Framework Instead of Rote Learning

Three identities, a dozen ratios, and a page of proofs: trigonometric identities have a reputation for needing a strong memory. In fact, they need a small memory and a good method. You only have to remember one identity. The other two can be rebuilt from it in ten seconds, and a short list of strategies handles almost every proof the chapter will ask.

This guide sets out that framework, tests it with a simple right triangle, works through four proofs, and lists the slips to avoid. For the underlying chapter, see these notes on trigonometric identities, trigonometric ratios and the introduction to trigonometry. The same ratios are put to work in heights and distances.

Start From the Triangle

Take a right-angled triangle with an acute angle θ. Call the side opposite θ the opposite, the side next to it the adjacent, and the longest side the hypotenuse.

A small trap sits in the reciprocals: sin pairs with cosec, and cos pairs with sec. The “co” in the name does not mean it goes with cosine.

The One Identity to Remember

In the triangle, opposite² + adjacent² = hypotenuse² by Pythagoras’ theorem. Divide every term by hypotenuse²:

(opposite/hypotenuse)² + (adjacent/hypotenuse)² = 1

sin²θ + cos²θ = 1

That is the only identity you need to remember. It is just Pythagoras, written in ratios.

Two Derivations in Ten Seconds

Divide by cos²θ. Starting from sin²θ + cos²θ = 1, divide every term by cos²θ:

sin²θ/cos²θ + 1 = 1/cos²θ, which is tan²θ + 1 = sec²θ.

Divide by sin²θ. Divide every term by sin²θ instead:

1 + cos²θ/sin²θ = 1/sin²θ, which is 1 + cot²θ = cosec²θ.

The framework is therefore: remember one identity, and the division trick. Everything else is a rearrangement.

A Safety Net: The 3-4-5 Triangle

Before proving any identity, you can test whether it is plausible with the 3-4-5 right triangle. Take θ as the angle opposite the side of length 3. Then sin θ = 3/5, cos θ = 4/5, tan θ = 3/4, cot θ = 4/3, sec θ = 5/4 and cosec θ = 5/3.

Check the three identities.

• sin²θ + cos²θ = 9/25 + 16/25 = 1. ✓
• 1 + tan²θ = 1 + 9/16 = 25/16 = sec²θ. ✓
• 1 + cot²θ = 1 + 16/9 = 25/9 = cosec²θ. ✓

A test with numbers does not prove an identity, because an identity must hold for every angle. But if an expression fails with 3-4-5, it is certainly wrong, so it is a quick way to catch a mis-remembered formula.

Five Strategies for Proving Identities

1. Work on one side only, usually the more complicated one, and try to reach the other side.
2. Convert everything to sin and cos. It turns unfamiliar ratios into familiar ones.
3. Look for the three identities hiding in the expression, such as 1 − cos²θ, which is sin²θ.
4. Combine fractions with a common denominator, or multiply by a conjugate such as (1 − sin θ) when there is a 1 + sin θ.
5. Do not cross-multiply across the equals sign, because that assumes the thing you want to prove.

Four Worked Proofs

Proof one

Prove: (1 − cos²θ)(1 + cot²θ) = 1.

LHS = sin²θ × cosec²θ, using 1 − cos²θ = sin²θ and 1 + cot²θ = cosec²θ. Since cosec²θ = 1/sin²θ, LHS = sin²θ × (1/sin²θ) = 1 = RHS. ✓

Proof two

Prove: (sec θ + tan θ)(1 − sin θ) = cos θ.

LHS = (1/cos θ + sin θ/cos θ)(1 − sin θ) = (1 + sin θ)(1 − sin θ)/cos θ = (1 − sin²θ)/cos θ = cos²θ/cos θ = cos θ = RHS. ✓

Proof three

Prove: sin θ/(1 + cos θ) + (1 + cos θ)/sin θ = 2 cosec θ.

Combine over a common denominator: LHS = [sin²θ + (1 + cos θ)²]/[sin θ (1 + cos θ)] = [sin²θ + 1 + 2cos θ + cos²θ]/[sin θ (1 + cos θ)]. Since sin²θ + cos²θ = 1, the numerator is 2 + 2cos θ = 2(1 + cos θ). So LHS = 2(1 + cos θ)/[sin θ (1 + cos θ)] = 2/sin θ = 2 cosec θ = RHS. ✓

Proof four

Prove: tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ.

Write t for tan θ, so cot θ = 1/t. First term: t/(1 − 1/t) = t²/(t − 1). Second term: (1/t)/(1 − t) = −1/[t(t − 1)]. Sum = [t³ − 1]/[t(t − 1)] = (t² + t + 1)/t = t + 1 + 1/t = 1 + tan θ + cot θ. Now tan θ + cot θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/(sin θ cos θ) = sec θ cosec θ. So the sum = 1 + sec θ cosec θ. ✓

Finding Other Ratios From One

Identities also let you find all the ratios from one of them.

Example. If sin θ = 3/5 and θ is acute, find cos θ and tan θ.

cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25, so cos θ = 4/5 (positive for an acute angle). tan θ = sin θ/cos θ = 3/4.

Slips That Cost Marks

Identities in Your Syllabus

Introduction to Trigonometry is a chapter of the Class 10 Maths book in both the CBSE and the Karnataka SSLC textbooks. Published deleted-syllabus lists differ on how much of the ratios of complementary angles CBSE still expects, so check your own board’s syllabus for that sub-topic. Practise with the chapter MCQs, the important questions and the chapter notes. The identities grow into Class 11 trigonometric functions, as in this overview of trigonometric functions and the notes on sum and difference formulas, which matter for JEE Main and KCET.

Frequently Asked Questions

Q1. What are the three basic trigonometric identities?

sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Q2. How can I remember them without memorising all three?

Remember sin²θ + cos²θ = 1, which is Pythagoras’ theorem written in ratios. Dividing by cos²θ gives the second identity, and dividing by sin²θ gives the third.

Q3. What is the best way to prove an identity?

Work on one side only, usually the more complicated one, convert to sin and cos, use the basic identities, and aim to reach the other side.

Q4. Which ratio is the reciprocal of cos θ?

sec θ. The reciprocal of sin θ is cosec θ, and the reciprocal of tan θ is cot θ.

Q5. Can I prove an identity by putting in a value of θ?

No. A value can check that an identity is plausible, but a proof must hold for every angle.

One Identity, Many Results

An identity is not a thing to memorise three times. It is one fact about right triangles, seen from three angles. Learn the facts, learn the division trick, and use the strategies, and the chapter becomes a short list of patterns.