Class 10 Mathematics Notes

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Trigonometric Identities Class 10 Maths Notes

After the six ratios come one relation that ties them together. In the rationalised 2024-25 syllabus exactly one trigonometric identity is retained: sin²A + cos²A = 1. This page proves it from a right triangle, rewrites it in every form the questions need, and shows how to use it to evaluate expressions and to find a ratio you were not given.

Class:10Subject:MathematicsUnit:VCovers:CBSE 2024-25

6 Key Formulas

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

State the only trigonometric identity retained in the 2024-25 syllabus and prove it.

The identity is sin²A + cos²A = 1. In a right triangle ABC with ∠B = 90° and ∠A = A, sin A = BC / AC and cos A = AB / AC, so sin²A + cos²A = (BC² + AB²) / AC² = AC² / AC² = 1, because the legs of a right triangle satisfy opposite² + adjacent² = hypotenuse².

What an Identity Is

Definition

Trigonometric identity

An identity is an equation that is true for every value of the angle for which the ratios are defined, not just for one special angle. That is the difference between an identity and an equation: sin²30° + cos²30° = 1 is true, but only because 30° happens to be one of the three standard angles, so it is an evaluation and not a proof.

For CBSE 2024-25 the retained content of this chapter is one identity and its proof, its rearrangements, and its use in simplifying and evaluating expressions. Nothing else is expected, so a short chapter is not a lightly marked one — the identity is the base of every later trigonometry calculation.

Why the letter A and not θ

Textbooks use A for the angle in an identity and θ for the angle in a ratio definition. Both mean the same thing. Do not read anything extra into the letter; just keep the symbol consistent within one piece of working.

The One Retained Identity

The Pythagorean identity

sin^{2} A + cos^{2} A = 1

It is called the Pythagorean identity because it carries the Pythagoras relation of a right triangle into trigonometric form. For every acute angle A the two numbers sin²A and cos²A are the fractions of the hypotenuse contributed by the two legs, and those two fractions always add up to the whole, which is 1.

sin²A means (sin A)², that is the ratio squared, not sin(A²). The square applies to the value of the ratio.
The identity holds for every acute angle, and also at 0° and 90° wherever the ratios exist, giving 0 + 1 = 1 and 1 + 0 = 1.
It says nothing on its own about the individual values of sin A and cos A; it fixes only their squares.
Everything else in this chapter is a rearrangement of this one line or a direct application of it.

The three lines the board wants

Write the identity, substitute the two ratios from the triangle, and replace the sum of the squares of the legs by the square of the hypotenuse. Those three lines are the whole proof and they carry every mark available for it.

The Proof from the Right Triangle

Take a right-angled triangle ABC in which the angle at B is 90° and the angle at A is A. Then the side opposite A is BC, the side adjacent to A is AB, and the hypotenuse is AC. Read the two ratios off the triangle and square them.

Substitute the two ratios

sin^{2} A + cos^{2} A = (\frac{B C}{A C})^{2} + (\frac{A B}{A C})^{2} = \frac{B C ^{2} + A B ^{2}}{A C ^{2}}

Use the right-triangle relation and finish

\frac{B C ^{2} + A B ^{2}}{A C ^{2}} = \frac{A C ^{2}}{A C ^{2}} = 1 ⟹ sin^{2} A + cos^{2} A = 1

The step that carries the proof is BC² + AB² = AC², the defining property of the hypotenuse of a right triangle, since BC and AB are exactly the two legs.
Put the two fractions over a single denominator before simplifying; each one has the same denominator AC².
Every quantity in the proof is a length, so squaring removes any sign issue and no absolute values are needed for an acute angle.
The proof is identical for any acute A, which is exactly what makes it an identity rather than a numerical result.

Rearrangements and What Each Is For

Most examination questions are the identity moved slightly to one side. Learn the rearrangements by seeing what each one gives you, because that tells you which one a given question needs.

The two rearrangements

1 - sin^{2} A = cos^{2} A, 1 - cos^{2} A = sin^{2} A

To find a cos² value when sin A is known, use 1 − sin²A = cos²A. This is the form used in every 'if sin A = 3 / 5' question.
To find a sin² value when cos A is known, use 1 − cos²A = sin²A, which is the same identity rearranged the other way.
To remove a squared ratio from under a square root, use √(1 − sin²A) = cos A for an acute A, since cos A is positive there.
To simplify a fraction that is one squared ratio over the other, divide the identity by that other ratio.
Note that 1 − sin A is NOT cos²A. The square belongs on the sine, and dropping it is the single most common slip in this chapter.

The square cannot be dropped

sin²A + cos²A = 1 gives 1 − sin²A = cos²A. It does not give 1 − sin A = cos²A, and it does not give 1 − sin²A = cos A either — the right-hand side is the square, so take the square root deliberately and separately, and only for an acute angle.

Evaluating Expressions

In an evaluation question you usually need both the identity and the standard values, so do the squares and the surds carefully and keep the numbers exact until the last line.

Example 1

2 sin 3 0^{\circ} + 3 cos 3 0^{\circ} = 2 (\frac{1}{2}) + 3 (\frac{3}{2}) = 1 + \frac{3}{2} = \frac{5}{2}

Example 2

4 sin^{2} 6 0^{\circ} + 3 cos^{2} 6 0^{\circ} = 4 (\frac{3}{4}) + 3 (\frac{1}{4}) = 3 + \frac{3}{4} = \frac{15}{4}

Example 3, simplify for any acute A

\frac{1 - cos ^{2} A}{sin ^{2} A} = \frac{sin ^{2} A}{sin ^{2} A} = 1

A whole expression in one step

If every term in an expression is k sin²A + k cos²A for the same k, the answer is simply k. So 3 sin²60° + 3 cos²60° = 3, with no substitution at all, and a five-mark simplification often reduces to exactly this.

Finding a Missing Ratio

The second classic use is being given one ratio and asked for the rest. There are two starting points, and both are common.

Starting from a sine value

sin A = \frac{3}{5} ⟹ cos^{2} A = 1 - \frac{9}{25} = \frac{16}{25} ⟹ cos A = \frac{4}{5}

From sin A = 3 / 5 you then get cosec A = 5 / 3, tan A = 3 / 4, cot A = 4 / 3 and sec A = 5 / 4, all by the reciprocal and quotient relationships.
From tan A = 3 / 4, take opposite = 3 and adjacent = 4, so the hypotenuse is √(9 + 16) = 5 and sin A = 3 / 5, cos A = 4 / 5, cosec A = 5 / 3, sec A = 5 / 4, cot A = 4 / 3.
The 3-4-5 right triangle is the workhorse of this chapter, and the 5-12-13 triangle is the next most useful.
Take the positive square root only, because A is acute and every ratio of an acute angle is positive.
Write the answer as a simplified fraction, not as √(16 / 25).

Example, a two-term expansion

(sin A + cos A)^{2} = sin^{2} A + 2 sin A cos A + cos^{2} A = 1 + 2 sin A cos A

Out of Scope for 2024-25

Only simple identities survive in the rationalised syllabus, and the deletions here are wide enough that a book written for the old syllabus will mislead you. Treat anything beyond the next callout as out of play.

Only sin²A + cos²A = 1 is retained, together with its rearrangements and its applications.
No double-angle or triple-angle identity is retained, so nothing in the form of sin 2A, cos 2A or sin 3A will appear.
No identity for the sum or difference of two angles is retained.
An identity such as sec²A − tan²A = 1 does follow from the retained one, but it is not on the syllabus list and no question will require it.
The entire Constructions chapter is also deleted this year, so no construction of tangents or perpendiculars will be asked.

Complementary-angle ratios are deleted — do not use them

The relations sin(90° − A) = cos A, cos(90° − A) = sin A, tan(90° − A) = cot A, sec(90° − A) = cosec A and cosec(90° − A) = sec A were removed from the 2024-25 rationalised syllabus. Do not use any of them as a working step and do not quote them as an identity in an answer. Work from the standard values at 30°, 45° and 60° instead. If a question appears to need a complementary-angle relation, it is out of scope and will not be set this year.

How the Questions Are Asked

This chapter supports the short questions of Sections B and C and supplies the technique used in the three-mark items of the trigonometry unit. Nothing here needs a long proof beyond the single identity.

Prove that sin²A + cos²A = 1. (A three-mark short answer; the three lines above earn all three marks.)
Evaluate 2 sin 30° + √3 cos 30°. (A two or three mark question.)
Simplify 1 − sin²45° and hence find cos 45°. (Two or three marks.)
If tan A = 3 / 4, find sin A and cos A. (A three-mark short answer.)
Show that (1 − cos²A) / sin²A = 1. (A three-mark short answer.)
A case-study question may set this identity inside a practical context, for instance a cable-tension or an inclined-plane figure, where the second sub-part usually needs a trigonometric ratio instead.

State the identity before using it

In any evaluation or simplification, open by writing sin²A + cos²A = 1 or the rearrangement you need. The mark for choosing the right rearrangement is separate from the mark for the arithmetic, and it is the one most often lost by jumping straight to numbers.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

The Pythagorean identity

sin^{2} A + cos^{2} A = 1

The only trigonometric identity retained in the 2024-25 syllabus.

Sine rearrangement

1 - sin^{2} A = cos^{2} A

Use it when sin A is given and cos A is wanted.

Cosine rearrangement

1 - cos^{2} A = sin^{2} A

Use it when cos A is given and sin A is wanted.

The proof, step by step

(\frac{B C}{A C})^{2} + (\frac{A B}{A C})^{2} = \frac{B C ^{2} + A B ^{2}}{A C ^{2}} = \frac{A C ^{2}}{A C ^{2}} = 1

In right triangle ABC with ∠B = 90° and ∠A = A, BC is opposite and AB adjacent, so BC² + AB² = AC².

Removing a squared ratio from a root

1 - sin^{2} A = cos A (A acute)

Only valid for an acute angle, when cos A is positive.

The reciprocal and quotient pair

sec A = \frac{1}{cos A}, cosec A = \frac{1}{sin A}, cot A = \frac{1}{tan A}, tan A = \frac{sin A}{cos A}

Once one ratio is known, these give every other ratio without another triangle.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

The only identity retained for 2024-25 is sin²A + cos²A = 1; no double-angle, sum or difference identity will be asked.
sin²A means (sin A)² — the square is on the ratio, never on the angle.
The proof needs three lines: substitute both ratios, put them over the common denominator AC², then use BC² + AB² = AC².
1 − sin²A = cos²A, not 1 − sin A = cos²A. Keeping the square is the whole difficulty of the rearrangements.
The complementary-angle relations sin(90° − A) = cos A, cos(90° − A) = sin A and tan(90° − A) = cot A were deleted and must not be used or quoted.
Any expression of the form k sin²A + k cos²A has the value k, with no substitution needed.
When a sine or cosine value is given, the 3-4-5 triangle produces the remaining ratios; when a tangent is given, build the triangle first.
Take positive square roots only, since every ratio of an acute angle is positive, and write √(16/25) as 4/5.
State which rearrangement you are using before substituting, because the choice of identity carries its own mark.
Trigonometry carries 12 of the 80 theory marks and internal assessment is 20 marks — 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical — with no chapter-wise split.

FAQ

Frequently asked questions

Prove that sin²A + cos²A = 1.

In a right triangle ABC with ∠B = 90° and ∠A = A, sin A = BC / AC and cos A = AB / AC. Squaring and adding gives sin²A + cos²A = (BC / AC)² + (AB / AC)² = (BC² + AB²) / AC². Since BC and AB are the two legs of the triangle, BC² + AB² = AC², so the expression is AC² / AC² = 1. This holds for every acute angle A, which is what makes it an identity.

What other forms of the identity should I memorise?

Two rearrangements cover the questions: 1 − sin²A = cos²A and 1 − cos²A = sin²A. Two consequences are also worth having, namely √(1 − sin²A) = cos A for an acute angle, and (1 − cos²A) / sin²A = 1. Note that sin²A means (sin A)², and that 1 − sin A is not cos²A.

If sin A = 3 / 5, how do I find the other ratios?

Use the rearrangement 1 − sin²A = cos²A, which gives cos²A = 1 − 9/25 = 16/25, so cos A = 4/5 for an acute angle. Then cosec A = 1 / sin A = 5/3, sec A = 1 / cos A = 5/4, tan A = sin A / cos A = 3/4 and cot A = 1 / tan A = 4/3. The same six numbers come from a 3-4-5 right triangle with the right angle between the two legs.

Can I use sin(90° − A) = cos A in my answers?

No. The complementary-angle relations, including sin(90° − A) = cos A, cos(90° − A) = sin A and tan(90° − A) = cot A, were deleted in the rationalised 2024-25 syllabus. Using one of them as a working step or quoting it as an identity is not correct for this paper. Use the standard values at 30°, 45° and 60° instead, and note that a question requiring a complementary angle will not be set this year.

How do I evaluate an expression such as 4 sin²60° + 3 cos²60°?

Substitute the squared standard values: sin²60° = (√3 / 2)² = 3 / 4 and cos²60° = (1 / 2)² = 1 / 4. So the expression is 4 × 3 / 4 + 3 × 1 / 4 = 3 + 3 / 4 = 15 / 4. As a shortcut, check whether the coefficients are equal, since any expression k sin²A + k cos²A is simply k.

Next Chapters

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Class 10 Maths Notes Introduction to Trigonometry Class 10 Maths Notes Some Applications of Trigonometry Class 10 Maths Notes All Class 11 & 12 Notes

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