Using Identities to solve/prove simple algebraic trigonometric expressions | ICSE Class 10 Maths Notes

This note covers trigonometric notation, the three basic identities, their angle restrictions, rearranging identities, finding unknown ratios, simplifying expressions, and proving equalities by substitution, factorisation, common denominators and expansion.

What do the symbols in a trigonometric identity mean?

Trigonometry studies relationships between the sides and angles of a triangle. A trigonometric ratio compares two side lengths in a right-angled triangle. A right angle measures 90°, where the symbol ° means degrees. An acute angle measures more than 0° but less than 90°.

Let A name the angle under consideration. Relative to A, the opposite side faces the angle, the adjacent side touches it and is not the hypotenuse, and the hypotenuse is the side opposite the right angle.

How are reciprocals and squares written?

A reciprocal is one divided by a non-zero quantity. The sign = means “equals”. Thus cosec A = 1/sin A, sec A = 1/cos A and cot A = 1/tan A wherever these expressions are defined. The slash indicates division, with the expression below a fraction called its denominator.

The useful quotient relations are tan A = sin A/cos A and cot A = cos A/sin A. These let us replace ratios by expressions involving sine and cosine. A numerator is the expression above the fraction line.

The superscript ² means squaring. In particular, sin² A means (sin A)², the square of the complete ratio. It does not mean the sine of a squared angle. Similarly, cos² A, tan² A and the other squared ratios refer to squares of their values.

Definition: A trigonometric identity is an equation involving trigonometric ratios that is true for every permitted value of the angle for which its expressions are defined.

A variable is a symbol whose value may vary. Here the angle is the variable. The notation sin A names one ratio; it is not multiplication of something called “sin” by A. A ratio must keep its angle attached throughout a calculation.

An expression is a mathematical combination of quantities and operations. An equation states that two expressions are equal. To simplify an expression is to replace it by an equal, more manageable form; to prove an identity is to justify the equality throughout its permitted range.

How are the three basic identities obtained?

Take a triangle with vertices, or corner points, labelled A, B and C, right-angled at B. The labels AB, BC and AC denote the corresponding side lengths. With respect to angle A, BC is opposite, AB is adjacent and AC is the hypotenuse.

The Pythagoras theorem says that the square of the hypotenuse equals the sum of the squares of the other two sides. Thus AB² + BC² = AC², where + means addition. Dividing this equation by different squared side lengths produces the three basic identities.

A is above B, C is to the left of B, and the sloping side joins C to A. A small square marks the right angle at B, and an arc marks angle A.

See Fig. 8.21 in your NCERT textbook

Identity: sin² A + cos² A = 1

1. Start with AB² + BC² = AC² in the right-angled triangle.
2. Divide every term by AC² to obtain AB²/AC² + BC²/AC² = 1.
3. Replace AB/AC by cos A and BC/AC by sin A, giving cos² A + sin² A = 1.

sin² A + cos² A = 1 also holds at the endpoints A = 0° and A = 90°. The full range here is 0° ≤ A ≤ 90°, where ≤ means “less than or equal to”. The order of the two added terms does not change their sum.

Identity: 1 + tan² A = sec² A

1. Begin again with AB² + BC² = AC².
2. Divide every term by AB², giving 1 + BC²/AB² = AC²/AB².
3. Use BC/AB = tan A and AC/AB = sec A to obtain the required identity.

sec² A = 1 + tan² A holds for 0° ≤ A < 90°, where < means “less than”. At 90°, cos A is zero, so tan A and sec A are not defined. At 0°, tan A = 0 and sec A = 1, satisfying the equality.

Identity: 1 + cot² A = cosec² A

1. Return to AB² + BC² = AC².
2. Divide every term by BC², giving AB²/BC² + 1 = AC²/BC².
3. Replace AB/BC by cot A and AC/BC by cosec A to obtain cot² A + 1 = cosec² A.

cosec² A = 1 + cot² A holds for 0° < A ≤ 90°. At 0°, sin A is zero, making cot A and cosec A undefined. At 90°, cot A = 0 and cosec A = 1, so this identity remains valid.

Note: An identity does not give a value to an undefined ratio. Check the angle restrictions before applying a formula, especially when the calculation involves a reciprocal or division.

The triangle derivations use acute angles, where the side lengths are positive. The endpoint statements use the defined values of the ratios at 0° and 90°. This keeps the geometrical proof separate from checking which endpoint substitutions are valid.

How do rearranged identities simplify expressions?

A rearrangement changes the form of an equation by performing the same operation on both sides. The basic identities therefore provide several equivalent replacement rules. Matching the whole expression to one of these rules often avoids a longer calculation.

The symbol − means subtraction. Pay attention to its position: subtracting tan² A from sec² A gives 1. Reversing the order reverses the sign. Writing the original identity first is a useful way to check an uncertain rearrangement.

How can a common numerical factor help?

Worked example 1. For an acute angle A, simplify 9 sec² A − 9 tan² A.

Answer: Both terms contain the common factor 9, a quantity multiplying each term. Factor it out: 9 sec² A − 9 tan² A = 9(sec² A − tan² A) = 9 × 1 = 9. Here × means multiplication.

The calculation does not require a numerical angle. The bracket equals 1 throughout the permitted range. Finding separate values of sec A and tan A would introduce unnecessary work because their squared difference is already known.

How can both parts of a fraction be simplified?

Worked example 2. For an acute angle A, simplify (1 + tan² A)/(1 + cot² A).

Answer: Apply the matching identities to obtain sec² A/cosec² A. Replace the reciprocals to give (1/cos² A)/(1/sin² A) = sin² A/cos² A = tan² A.

In this example, numerator and denominator each match a different identity. Dividing by a non-zero fraction means multiplying by its reciprocal. Since A is acute, sin A and cos A are positive, so the displayed divisions are valid.

Do not cancel the two occurrences of 1 across the addition signs. Cancellation removes a shared non-zero factor from the whole numerator and denominator; it does not remove matching terms from unrelated sums. The identities turn those sums into useful single squared ratios.

How can one ratio determine other ratios?

An identity can turn information about one ratio into information about another. First choose the identity containing the given ratio. Substitute its value, isolate the required square, and then take its square root. A square root is a number whose square gives the original quantity.

The symbol √ denotes the non-negative square root. For acute angles, all six ratios are positive. Consequently, when a squared ratio is known, use its positive square root. The angle condition supplies the reason for choosing that sign.

How does a tangent value give the remaining ratios?

Worked example 3. Given tan A = 1/√3 for an acute angle A, determine cot A, sec A, cos A, sin A and cosec A.

Answer: The reciprocal gives cot A = √3. Next, sec² A = 1 + tan² A = 1 + 1/3 = 4/3, so sec A = 2/√3.

Taking the reciprocal gives cos A = √3/2. Then sin² A = 1 − cos² A = 1 − 3/4 = 1/4, so sin A = 1/2. Finally, cosec A = 2.

This calculation uses two different basic identities and three reciprocal relations. Each step answers a particular need: the tangent identity finds secant, the reciprocal finds cosine, and the sine-cosine identity finds sine. No measurement of a triangle is required.

How can answers remain in terms of a ratio?

Worked example 4. For an acute angle A, express cos A, tan A and sec A in terms of sin A.

Answer: From cos² A = 1 − sin² A, obtain cos A = √(1 − sin² A). Therefore tan A = sin A/√(1 − sin² A), and sec A = 1/√(1 − sin² A).

“In terms of sin A” means that the final expressions use sin A instead of other trigonometric ratios. The answer is a formula, not a numerical value. Its square-root denominators are positive because the acute-angle condition gives cos A greater than zero.

Keep the entire difference 1 − sin² A inside the square-root sign. Taking the square root applies to the result of the subtraction, not separately to its two terms. Brackets make the intended order of operations visible in a written fraction.

A final check can run backwards: squaring the expression for cos A gives 1 − sin² A, while multiplying the expression for sec A by cos A gives 1. Both checks recover the relations used to obtain the answers.

How should a proof of an identity begin?

The left-hand side, abbreviated LHS, is the expression before an equality sign. The right-hand side, abbreviated RHS, is the expression after it. In a proof, start with one side and replace it by equal expressions until the other side appears.

Choose the side that offers a useful operation: a recognisable identity, a factor, a common denominator or a conversion into sine and cosine. The required result is the destination. It must not be assumed as the reason why the starting expression is equal to it.

What is a useful sequence of checks?

1. Read the angle condition. Establish where the original expressions and their denominators are defined.
2. Inspect the structure. Look for squared ratios, reciprocal pairs, sums or differences that can be factored.
3. Choose a justified replacement. Use a definition, a basic identity or an ordinary algebraic operation.
4. Compare with the target. Continue until the transformed expression has the same form as the other side.

Worked example 5. Prove sec A(1 − sin A)(sec A + tan A) = 1 for an acute angle A.

Answer: Start with the LHS. Replace sec A by 1/cos A and tan A by sin A/cos A, giving (1/cos A)(1 − sin A)(1 + sin A)/cos A.

Combine the fractions: (1 − sin A)(1 + sin A)/cos² A = (1 − sin² A)/cos² A. Using 1 − sin² A = cos² A, this becomes cos² A/cos² A = 1, which is the RHS.

The important move is converting the sum sec A + tan A into (1 + sin A)/cos A. That exposes the product (1 − sin A)(1 + sin A). Expanding that product produces exactly the difference required by the sine-cosine identity.

The final cancellation is valid because cos A is non-zero for an acute angle. At A = 90°, the original expression contains undefined ratios, so the simplified answer 1 does not make that substitution valid.

Verification at one angle and proof of an identity serve different purposes. A numerical substitution checks one permitted case. The proof above keeps A unspecified and justifies every transformation, establishing the result throughout the stated range.

How do factorisation and differences of squares help?

Factorisation rewrites an expression as a product. Two useful operations are taking out a common factor and applying a difference of squares. Let x and y denote real numbers, meaning numbers representable on a number line. The algebraic identity (x − y)(x + y) = x² − y² explains the second operation.

The same rule applies when x or y is a trigonometric ratio. Thus (1 − sin A)(1 + sin A) = 1 − sin² A, and (1 − cos A)(1 + cos A) = 1 − cos² A. The basic identities then replace these differences by squares.

How does factoring make cancellation valid?

Worked example 6. For an acute angle A, prove (cot A − cos A)/(cot A + cos A) = (cosec A − 1)/(cosec A + 1).

Answer: Replace cot A by cos A/sin A. The numerator becomes cos A(1/sin A − 1), and the denominator becomes cos A(1/sin A + 1).

Cancel the non-zero factor cos A from the complete numerator and denominator. The result is (1/sin A − 1)/(1/sin A + 1) = (cosec A − 1)/(cosec A + 1), the RHS.

The original fraction contains additions and subtractions. Factoring first shows that the whole numerator and the whole denominator each contain cos A as a multiplier. This explains why the later cancellation is legitimate even though crossing out the original cos A terms would not be.

How can a squared difference become a fraction?

Worked example 7. For an acute angle A, prove (cosec A − cot A)² = (1 − cos A)/(1 + cos A).

Answer: Convert the bracket to sine and cosine: cosec A − cot A = (1 − cos A)/sin A. Squaring gives (1 − cos A)²/sin² A.

Replace sin² A by 1 − cos² A and factor the denominator: (1 − cos A)²/[(1 − cos A)(1 + cos A)]. Cancelling the non-zero factor 1 − cos A gives the required RHS.

For an acute angle, cos A is less than 1, so the factor 1 − cos A is non-zero. The condition matters because the cancelled factor would vanish at 0°, where the original cosecant and cotangent expressions are undefined.

A factor multiplies the rest of an expression; a term is separated from other terms by addition or subtraction. Distinguishing factors from terms is essential when simplifying a trigonometric fraction. The identities supply equal replacements, but ordinary algebra still controls each cancellation.

How can common denominators and separate simplification prove identities?

A common denominator is a denominator used to express fractions on the same basis before adding them. To obtain it, multiply the numerator and denominator of a fraction by the same non-zero quantity. Its value is unchanged because that multiplier divided by itself equals 1.

How does adding two fractions expose an identity?

Worked example 8. For an acute angle A, prove cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A.

Answer: Use the common denominator cos A(1 + sin A). The combined numerator is cos² A + (1 + sin A)².

Expand the square to obtain cos² A + 1 + 2 sin A + sin² A. Since cos² A + sin² A = 1, this becomes 2 + 2 sin A = 2(1 + sin A).

The fraction is therefore 2(1 + sin A)/[cos A(1 + sin A)] = 2/cos A = 2 sec A, proving the result.

Expansion multiplies out brackets. In this example, the square contributes the middle term 2 sin A. Omitting that term would prevent the numerator from becoming 2(1 + sin A), so the cancellation and the final result would both be wrong.

When is simplifying both sides useful?

Some identities have two sides that look quite different, but both simplify to the same expression. In that method, keep the two calculations separate. Their common final expression establishes the equality without assuming it at the beginning.

Worked example 9. For an acute angle A, prove (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A).

Answer: On the LHS, cosec A − sin A = (1 − sin² A)/sin A = cos² A/sin A. Similarly, sec A − cos A = sin² A/cos A.

Multiplying gives (cos² A/sin A)(sin² A/cos A) = sin A cos A. On the RHS, tan A + cot A = sin A/cos A + cos A/sin A = (sin² A + cos² A)/(sin A cos A) = 1/(sin A cos A).

Thus the RHS, its reciprocal, also equals sin A cos A. Both sides have reached the same expression, proving the identity.

Here the outer reciprocal on the RHS must remain visible. Simplifying tan A + cot A gives its denominator expression, not the entire RHS. Taking the reciprocal once more is the step that gives the required common result.

How are squares and square roots handled in longer identities?

The algebraic expansion (x + y)² = x² + 2xy + y² applies to real numbers x and y, and therefore to defined trigonometric ratios. The middle term is twice the product. A square of a sum must not be replaced by a sum of squares.

How do reciprocal products simplify an expansion?

Worked example 10. For an acute angle A, prove (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

Answer: Expand the LHS to give sin² A + 2 sin A cosec A + cosec² A + cos² A + 2 cos A sec A + sec² A.

The reciprocal products sin A cosec A and cos A sec A each equal 1. Also, sin² A + cos² A = 1. The expression is therefore 5 + cosec² A + sec² A.

Use cosec² A = 1 + cot² A and sec² A = 1 + tan² A. The result is 5 + 1 + cot² A + 1 + tan² A = 7 + tan² A + cot² A.

Grouping the terms by their mathematical role makes the proof easier to follow: the sine and cosine squares form one identity, the reciprocal products become constants, and the remaining squares convert into the ratios appearing in the target.

Why does a square-root proof need a sign check?

Worked example 11. For an acute angle A, prove √[(1 + sin A)/(1 − sin A)] = sec A + tan A.

Answer: Multiply the fraction inside the square root by (1 + sin A)/(1 + sin A). This gives (1 + sin A)²/(1 − sin² A) = (1 + sin A)²/cos² A.

The quotient (1 + sin A)/cos A is positive for an acute angle. The non-negative square root of its square is therefore (1 + sin A)/cos A = 1/cos A + sin A/cos A = sec A + tan A.

The multiplication inside the root is justified because 1 + sin A is non-zero. The denominator 1 − sin A is also positive for an acute angle. These conditions keep the original fraction defined and support the positive sign selected when taking the root.

Squaring alone can lose sign information. A proof involving square roots must therefore explain why the proposed answer has the required non-negative sign. Here the acute-angle condition provides exactly that information, so the final expression agrees with the meaning of √.

For a final review of a longer proof, check every bracket, every squared ratio and every denominator. Then check that the concluding expression is precisely the requested one. The most useful simplification is the one that preserves equality at each step and makes the target visible.

Glossary

• Trigonometric ratio — A ratio of two side lengths of a right-angled triangle, specified relative to an acute angle.
• Identity — An equation true for every permitted value of its variables for which the expressions are defined.
• Acute angle — An angle measuring more than zero degrees and less than ninety degrees.
• Hypotenuse — The longest side of a right-angled triangle, lying opposite its right angle.
• Reciprocal — The result of dividing one by a specified non-zero quantity.
• Numerator — The expression above a fraction line, divided by the denominator below it.
• Denominator — The expression below a fraction line, which must be non-zero for division to be defined.
• Factorisation — Rewriting an expression as a product of factors that multiply to give it.
• Common factor — A factor shared by every term of an expression or by expressions being compared.
• Difference of squares — A subtraction of two squared quantities, factorised as their difference multiplied by their sum.
• Expansion — Multiplying out brackets to express a product as a sum or difference of terms.
• Square root — A number whose square equals the given quantity; the symbol √ selects its non-negative value.

Common errors and misconceptions

• Misconception: sin² A means the sine of the angle A². Correct: It means (sin A)². Find or manipulate the sine ratio first, then square that value.
• Misconception: 1 + tan² A equals cosec² A. Correct: It equals sec² A. The cotangent identity is 1 + cot² A = cosec² A.
• Misconception: Every basic identity can be evaluated at both 0° and 90°. Correct: Tangent and secant are undefined at 90°; cotangent and cosecant are undefined at 0°.
• Misconception: The 1 terms cancel in (1 + tan² A)/(1 + cot² A). Correct: Cancel common non-zero factors of the complete numerator and denominator, not terms joined by addition.
• Misconception: A square of a sum contains just the two squared terms. Correct: Expansion also includes twice the product, as in (sin A + cosec A)².
• Misconception: Either square-root sign is acceptable when finding an acute angle's ratio. Correct: The ratios are positive for acute angles, so the positive value is required.
• Misconception: Checking an equality for one angle proves it is an identity. Correct: A proof must establish the equality for every permitted angle where the expressions are defined.
• Misconception: Simplifying an expression removes its original restrictions. Correct: A cancelled denominator can still exclude an angle from the original expression, even if the final form looks defined there.

Exam-style questions with model answers

Q1. For an acute angle A, evaluate 9 sec² A − 9 tan² A using an identity. [2 marks]

1. Factor out the common multiplier 9 to write the expression as 9(sec² A − tan² A).
2. Use sec² A − tan² A = 1. The required value is therefore 9 × 1 = 9.

Q2. State the identity connecting tangent and secant, and give its permitted angle range within 0° ≤ A ≤ 90°, explaining the excluded endpoint. [2 marks]

1. The identity is 1 + tan² A = sec² A, valid here for 0° ≤ A < 90°.
2. At 90°, cos A = 0. Hence tan A = sin A/cos A and sec A = 1/cos A are undefined, so that endpoint is excluded.

Q3. Given tan A = 1/√3 and A is acute, find sec A, cos A and sin A using identities. [3 marks]

1. Use sec² A = 1 + tan² A = 1 + 1/3 = 4/3. Since A is acute, take the positive root to obtain sec A = 2/√3.
2. Cosine is the reciprocal of secant, so cos A = 1/sec A = √3/2.
3. Now sin² A = 1 − cos² A = 1 − 3/4 = 1/4. Sine is positive for an acute angle, giving sin A = 1/2.

Q4. For an acute angle A, prove sec A(1 − sin A)(sec A + tan A) = 1. [4 marks]

1. Starting with the LHS, substitute sec A = 1/cos A and tan A = sin A/cos A.
2. Combine the fractions to obtain (1 − sin A)(1 + sin A)/cos² A.
3. Apply the difference of squares, giving (1 − sin² A)/cos² A, and replace 1 − sin² A by cos² A.
4. The result is cos² A/cos² A = 1, the RHS. Cancellation is valid because cos A is non-zero for the given acute angle.

Q5. For an acute angle A, prove (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A. Show the expansion and identity substitutions. [5 marks]

1. Expand both squares on the LHS, retaining the middle terms: sin² A + 2 sin A cosec A + cosec² A + cos² A + 2 cos A sec A + sec² A.
2. Use sin A cosec A = 1 and cos A sec A = 1. Their doubled products therefore contribute 2 + 2 = 4.
3. Group sin² A with cos² A and replace their sum by 1. The whole expression becomes 5 + cosec² A + sec² A.
4. Apply the remaining basic identities: cosec² A = 1 + cot² A and sec² A = 1 + tan² A.
5. Collect the constants to obtain 7 + tan² A + cot² A, which is the RHS. All reciprocal ratios used are defined because A is acute.

Q6. For an acute angle A, prove cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A by combining the fractions. [5 marks]

1. Start with the LHS and take the common denominator cos A(1 + sin A). Both factors are non-zero because A is acute.
2. The combined numerator is cos² A + (1 + sin A)². Expand the square to obtain cos² A + 1 + 2 sin A + sin² A.
3. Apply cos² A + sin² A = 1. The numerator now simplifies to 2 + 2 sin A.
4. Factor the numerator as 2(1 + sin A), then cancel the common non-zero factor 1 + sin A from the fraction.
5. The remaining expression is 2/cos A. Using sec A = 1/cos A gives 2 sec A, exactly the required RHS.

Q7. For an acute angle A, prove (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A) by simplifying both sides separately. [5 marks]

1. In the first LHS bracket, replace cosec A by 1/sin A. Then cosec A − sin A = (1 − sin² A)/sin A = cos² A/sin A.
2. Similarly, sec A − cos A = (1 − cos² A)/cos A = sin² A/cos A. The two LHS brackets therefore multiply to sin A cos A.
3. For the RHS denominator, use tan A + cot A = sin A/cos A + cos A/sin A.
4. Combine these fractions to obtain (sin² A + cos² A)/(sin A cos A) = 1/(sin A cos A).
5. The complete RHS is the reciprocal of that denominator, hence sin A cos A. Both sides are equal, and the acute-angle condition makes the divisions valid.

Q8. For an acute angle A, prove √[(1 + sin A)/(1 − sin A)] = sec A + tan A, explaining the sign of the square root. [4 marks]

1. Inside the square root, multiply numerator and denominator by 1 + sin A. The fraction becomes (1 + sin A)²/(1 − sin² A).
2. Use 1 − sin² A = cos² A, giving (1 + sin A)²/cos² A inside the root.
3. For an acute angle, (1 + sin A)/cos A is positive. Taking the non-negative square root therefore gives this quotient.
4. Separate the quotient into 1/cos A + sin A/cos A = sec A + tan A, completing the proof.

Key takeaways

• The three basic identities follow from the Pythagoras theorem by dividing by different squared side lengths.
• Check which ratios are defined at an angle before substituting it into an identity or a fraction.
• Rearranged identities replace differences such as 1 − sin² A and sec² A − 1 with single squared ratios.
• Convert reciprocal and quotient ratios into sine and cosine when that exposes useful factors or common denominators.
• Cancellation removes common non-zero factors of the complete numerator and denominator, not individual terms within sums.
• For acute angles, trigonometric ratios are positive; this determines the sign when recovering ratios from their squares.
• Prove an identity through justified equalities, either transforming one side into the other or simplifying both separately.
• In square expansions retain the middle product, and in square-root proofs explain why the chosen result is non-negative.

Test yourself

What does sin² A mean?

It means (sin A)², the square of the complete sine ratio of angle A.

Which identity replaces 1 − cos² A?

Since sin² A + cos² A = 1, the expression equals sin² A.

Why is 1 + tan² A = sec² A not evaluated at A = 90°?

At 90°, cos A is zero, so tangent and secant involve division by zero and are undefined.

For an acute angle A, how does (1 − sin A)(1 + sin A) simplify?

The difference of squares gives 1 − sin² A, which equals cos² A.

For an acute angle A, why can cos A be cancelled from a numerator and denominator that both contain it as a factor?

Cos A is non-zero for an acute angle, so division of both expressions by that common factor is valid.

Given tan A = 1/√3 and A acute, what is sec² A?

Use sec² A = 1 + tan² A = 1 + 1/3 = 4/3.

What is the middle term when (sin A + cosec A)² is expanded for acute A?

It is 2 sin A cosec A, which equals 2 because sine and cosecant are reciprocals.

Does checking a proposed identity at one angle prove it?

No. A proof must establish the equality for every permitted angle where both original expressions are defined.