Starting from sin2 A + cos2 A = 1, derive 1 + cot2 A = cosec2 A and state the condition you need.

Divide every term by sin2 A to get 1 + (cos2 A)/(sin2 A) = (1)/(sin2 A), that is 1 + cot2 A = cosec2 A. The division needs sin A ≠ 0, so A ≠ 0°.

Without tables, find the value of 5sec2 38° - 5tan2 38°.

5(sec2 38° - tan2 38°) = 5 × 1 = 5.

Prove that (1)/(1 + sin A) + (1)/(1 - sin A) = 2sec2 A.

LHS = ((1 - sin A) + (1 + sin A))/(1 - sin2 A) = (2)/(cos2 A) = 2sec2 A = RHS.

Trigonometrical Identities

Q: Define a trigonometric identity and explain how it differs from a trigonometric equation.

A trigonometric identity is an equation involving trigonometric ratios of an angle that is true for every value of the angle for which each ratio is defined. A trigonometric equation, such as cos A = (1)/(2), is true only for particular angles.

Q: If cos A = (5)/(13) and A is acute, find sin A and tan A.

sin2 A = 1 - (25)/(169) = (144)/(169), so sin A = (12)/(13) and tan A = (12)/(5).

Q: If cosec A = (17)/(8) and A is acute, find cot A.

cot2 A = (289)/(64) - 1 = (225)/(64), so cot A = (15)/(8).

Q: Prove that (cos A)/(1 - sin A) = (1 + sin A)/(cos A).

Multiply the numerator and denominator of the LHS by 1 + sin A: (cos A(1 + sin A))/(1 - sin2 A) = (cos A(1 + sin A))/(cos2 A) = (1 + sin A)/(cos A) = RHS.

ICSE Class 10 · Mathematics · Chapter 21

What this chapter covers

The six ratios and how they are linked
What a trigonometric identity is
The first identity: sin²A + cos²A = 1
The second identity: 1 + tan²A = sec²A
The third identity: 1 + cot²A = cosec²A
Where the identities hold: 0° ≤ A ≤ 90°
Writing every ratio in terms of one ratio
How to prove an identity
Proving identities: worked examples
Using identities to find values and solve
Common mistakes and how to avoid them

Key points

The reciprocal relations are cosec A = (1)/(sin A), sec A = (1)/(cos A), cot A = (1)/(tan A); the quotient relations are tan A = (sin A)/(cos A) and cot A = (cos A)/(sin A).
An identity is true for every value of the angle at which its ratios are defined; an equation is true only for particular angles.
sin² A + cos² A = 1 follows from Pythagoras' theorem and holds for 0^° ≤ A ≤ 90^°.
1 + tan² A = sec² A comes from dividing the first identity by cos² A and holds for 0^° ≤ A < 90^°.
1 + cot² A = cosec² A comes from dividing the first identity by sin² A and holds for 0^° < A ≤ 90^°.
(sec A - tan A)(sec A + tan A) = 1 and (cosec A - cot A)(cosec A + cot A) = 1.
For an acute angle every ratio is positive, so square roots are taken as positive.
To prove an identity, start from the more complicated side, change to sines and cosines, combine fractions, factorise, and use the identities; never assume the result.
Multiplying by the partner 1 mp sin A, 1 mp cos A, sec A mp tan A or cosec A mp cot A turns the denominator into a single squared ratio.
The identities also give values without tables, eliminate an angle between two equations, and turn an equation in two ratios into one in a single ratio that can be solved.

Questions answered

Define a trigonometric identity and explain how it differs from a trigonometric equation.

A trigonometric identity is an equation involving trigonometric ratios of an angle that is true for every value of the angle for which each ratio is defined. A trigonometric equation, such as cos A = (1)/(2), is true only for particular angles.

Starting from sin² A + cos² A = 1, derive 1 + cot² A = cosec² A and state the condition you need.

Divide every term by sin² A to get 1 + (cos² A)/(sin² A) = (1)/(sin² A), that is 1 + cot² A = cosec² A. The division needs sin A ≠ 0, so A ≠ 0^°.

If cos A = (5)/(13) and A is acute, find sin A and tan A.

sin² A = 1 - (25)/(169) = (144)/(169), so sin A = (12)/(13) and tan A = (12)/(5).

If cosec A = (17)/(8) and A is acute, find cot A.

cot² A = (289)/(64) - 1 = (225)/(64), so cot A = (15)/(8).

Without tables, find the value of 5sec² 38^° - 5tan² 38^°.

5(sec² 38^° - tan² 38^°) = 5 × 1 = 5.

Simplify (1 + cot² A)(1 - cos² A).

cosec² A · sin² A = 1.

Prove that (1)/(1 + sin A) + (1)/(1 - sin A) = 2sec² A.

LHS = ((1 - sin A) + (1 + sin A))/(1 - sin² A) = (2)/(cos² A) = 2sec² A = RHS.

Prove that (cos A)/(1 - sin A) = (1 + sin A)/(cos A).

Multiply the numerator and denominator of the LHS by 1 + sin A: (cos A(1 + sin A))/(1 - sin² A) = (cos A(1 + sin A))/(cos² A) = (1 + sin A)/(cos A) = RHS.

Past board exam questions

ICSE 2027 sample paper · 4 marks

Prove the following trigonometric identity: (cos²θ)/(1 - tanθ) + (sin²θ)/(1 - cotθ) = 1 + cosθsinθ

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LHS = (cos²θ)/(1 - (sinθ)/(cosθ)) + (sin²θ)/(1 - (cosθ)/(sinθ)) = (cos³θ)/(cosθ - sinθ) + (sin³θ)/(sinθ - cosθ) = (cos³θ - sin³θ)/(cosθ - sinθ) = ((cosθ - sinθ)(cos²θ + sin²θ + cosθsinθ))/(cosθ - sinθ) = 1 + cosθsinθ = RHS. Hence proved.

ICSE 2026 sample paper · 1 mark

Assertion (A): If secθ + tanθ = a and secθ - tanθ = b then ab = 1 Reason (R): sec²θ - tan²θ = 1

(A) is true and (R) is false.
(A) is false and (R) is true.
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Show answer

Both (A) and (R) are true and (R) is the correct explanation of (A).

ICSE 2026 board paper · 3 marks

Prove that: (secθ - cosθ)(cscθ - sinθ) = sinθcosθ

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LHS = ((1)/(cosθ) - cosθ)((1)/(sinθ) - sinθ) = (1 - cos²θ)/(cosθ) · (1 - sin²θ)/(sinθ) = (sin²θ)/(cosθ) · (cos²θ)/(sinθ) = sinθcosθ = RHS. Hence proved.

ICSE 2025 board paper · 1 mark

Assertion(A): If sin² A + sin A = 1 then cos⁴ A + cos² A = 1 Reason(R): 1 - sin² A = cos² A

(A) is true, (R) is false.
(A) is false, (R) is true.
Both (A) and (R) are true, and (R) is the correct reason for (A).
Both (A) and (R) are true, and (R) is the incorrect reason for (A).

Show answer

Both (A) and (R) are true, and (R) is the correct reason for (A).

ICSE 2025 compartment paper · 4 marks

Prove that: ((sin A - sin³ A))/((cos³ A - cos A)) × (sec A - cosec A) = cosec A(cot A - 1)

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(sin A - sin³ A)/(cos³ A - cos A) = (sin A(1 - sin² A))/(cos A(cos² A - 1)) = (sin A cos² A)/(-cos A sin² A) = -(cos A)/(sin A) = -cot A. LHS = -cot A(sec A - cosec A) = -(cos A)/(sin A) · (1)/(cos A) + cot A cosec A = -cosec A + cot A cosec A = cosec A(cot A - 1) = RHS. Hence proved.

ICSE 2025 sample paper · 4 marks

Prove the following trigonometry identity: (sinθ + cosθ)(cosec θ - secθ) = cosec θ · secθ - 2tanθ

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LHS = (sinθ + cosθ)((1)/(sinθ) - (1)/(cosθ)) = (sinθ + cosθ)(cosθ - sinθ)/(sinθcosθ) = (cos²θ - sin²θ)/(sinθcosθ) = (1 - 2sin²θ)/(sinθcosθ) = (1)/(sinθcosθ) - (2sin²θ)/(sinθcosθ) = cosec θ · secθ - 2tanθ = RHS.

ICSE 2025 board paper · 4 marks

Prove that: ((cot A + tan A - 1)(sin A + cos A))/(sin³ A + cos³ A) = sec A · cosec A

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cot A + tan A - 1 = (cos A)/(sin A) + (sin A)/(cos A) - 1 = (cos² A + sin² A - sin A cos A)/(sin A cos A) = (1 - sin A cos A)/(sin A cos A) sin³ A + cos³ A = (sin A + cos A)(sin² A - sin A cos A + cos² A) = (sin A + cos A)(1 - sin A cos A) LHS = ((1 - sin A cos A)(sin A + cos A))/(sin A cos A (sin A + cos A)(1 - sin A cos A)) = (1)/(sin A cos A) = sec A · cosec A = RHS.

ICSE 2024 board paper · 1 mark

Statement 1: sin²θ + cos²θ = 1 Statement 2: cosec²θ + cot²θ = 1 Which of the following is valid?

only 1
only 2
both 1 and 2
neither 1 nor 2

Show answer

only 1