CBSE X Maths — Introduction to Trigonometry: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit V: Trigonometry

Introduction to Trigonometry

Trigonometric ratios connect the angles of a right triangle to the ratios of its sides. Once you know a handful of exact values at standard angles, you can evaluate many expressions without a calculator.

1. The Six Trigonometric Ratios

For an acute angle θ in a right triangle, with the side opposite θ, the side adjacent to θ, and the hypotenuse:

sinθ = opposite/hypotenuse

cosθ = adjacent/hypotenuse

tanθ = opposite/adjacent = sinθ/cosθ

cosecθ, secθ, and cotθ are the reciprocals of sinθ, cosθ, and tanθ respectively.

2. Standard Angle Values

θ	0°	30°	45°	60°	90°
sinθ	0	1/2	1/√2	√3/2	1
cosθ	1	√3/2	1/√2	1/2	0
tanθ	0	1/√3	1	√3	undefined

Worked example — finding ratios from one given ratio If sinθ = 3/5, find cosθ and tanθ.
Using a right triangle with opposite=3, hypotenuse=5, the adjacent side = √(5²−3²) = √16 = 4 (Pythagoras).
cosθ = 4/5, tanθ = 3/4

Worked example — evaluating an expression Evaluate sin30°cos60° + cos30°sin60°.
= (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1

Worked example — a mixed evaluation Evaluate 2tan²45° + cos²30° − sin²60°.
tan45° = 1, so 2(1)² = 2. cos²30° = 3/4, sin²60° = 3/4, so these cancel.
Result = 2 + 3/4 − 3/4 = 2

Common mistakes to watch for

Mixing up which side is “opposite” and which is “adjacent” — this depends on which angle you’re measuring from.
Forgetting that tan90° is undefined (cos90°=0, and you can’t divide by zero).
Writing sin²θ as sinθ² — the square applies to the ratio’s value, so sin²θ means (sinθ)².

Key Vocabulary Recap

Trigonometric ratio — a ratio of two sides of a right triangle, linked to an angle

Reciprocal ratios — cosec, sec, cot are 1/sin, 1/cos, 1/tan

Syllabus Reference

This note covers CBSE Class X Mathematics Unit V, Introduction to Trigonometry: trigonometric ratios of an acute angle, proof of their existence, values of the ratios at 0°, 30°, 45°, 60°, and 90°, and relationships between the ratios.

Practice questions

CBSE Class X Mathematics · Unit V: Introduction to Trigonometry — Practice Question Bank (no answers)

Introduction to Trigonometry — Practice

Covers: trigonometric ratios and evaluating expressions at standard angles. 10 questions.

If sinθ = 3/5, find cosθ and tanθ.

Evaluate sin30°cos60° + cos30°sin60°.

Evaluate 2tan²45° + cos²30° − sin²60°.

If tanA = 1, find the value of sinA + cosA.

Evaluate (sin30° + cos60°)/(sin60° + cos30°).

If cosθ = 12/13, find sinθ and tanθ.

Evaluate sin60°cos30° + cos60°sin30°.

Find the value of tan45°/cosec30° + sec60°/cot45°.

If A = 30°, verify that sin2A = 2sinAcosA.

Q10

Find the value of θ (0° ≤ θ ≤ 90°) if 2sinθ = √3.

More Practice

Q11

If A and B are acute angles such that sinA = cosB, prove that A + B = 90°.

Q12

Evaluate: (sin35°/cos55°) + (cos60°×cosec30°) − (√3×tan30°).

Q13

If tanθ = a/b, find the value of (sinθ+cosθ)/(sinθ−cosθ) in terms of a and b.

Q14

If √3 tanθ = 1, find the value of sin²θ − cos²θ.

Q15

Prove that (sinA+cosecA)² + (cosA+secA)² = 7 + tan²A + cot²A.

Q16

If cosecθ − cotθ = 1/3, find cosecθ + cotθ.

Q17

Find the value of sinθ if 7sin²θ + 3cos²θ = 4.

Q18

Evaluate: [5cos²60° + 4sec²30° − tan²45°] / [sin²30° + cos²30°].

Q19

If sinA = 1/2, verify that 3cosA − 4cos³A = cos3A.

Q20

If tanA + cotA = 2, find the value of tan²A + cot²A.

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit V: Introduction to Trigonometry — Solutions

Introduction to Trigonometry — Practice: Solutions

sinθ=3/5. Find cosθ, tanθ.

Use Pythagoras: adjacent=√(hyp²−opp²)

√(25−9)=√16=4

✓cosθ = 4/5, tanθ = 3/4

sin30cos60+cos30sin60.

Substitute standard values

(1/2)(1/2)+(√3/2)(√3/2)=1/4+3/4

✓1

2tan²45+cos²30−sin²60.

Substitute standard values

2(1)+3/4−3/4

✓2

tanA=1. Find sinA+cosA.

tanA=1 means A=45°

sin45+cos45=1/√2+1/√2

✓√2

(sin30+cos60)/(sin60+cos30).

Substitute standard values

(1/2+1/2)/(√3/2+√3/2)=1/√3

✓1/√3

cosθ=12/13. Find sinθ, tanθ.

Use Pythagoras: opposite=√(hyp²−adj²)

√(169−144)=√25=5

✓sinθ = 5/13, tanθ = 5/12

sin60cos30+cos60sin30.

Substitute standard values

(√3/2)(√3/2)+(1/2)(1/2)=3/4+1/4

✓1 (note: this is sin(30°+60°)=sin90°=1)

tan45/cosec30 + sec60/cot45.

Substitute standard values (tan45=1,cosec30=2,sec60=2,cot45=1)

1/2 + 2/1

✓5/2

Verify sin2A=2sinAcosA for A=30°.

LHS: sin60°

√3/2

RHS: 2sin30°cos30°

2(1/2)(√3/2)=√3/2

✓LHS = RHS = √3/2, so the identity is verified

Q10

2sinθ=√3. Find θ.

sinθ=√3/2

θ = sin⁻¹(√3/2)

✓θ = 60°

More Practice — Solutions

Q11

Prove A+B=90° given sinA=cosB.

✓Since cosB=sin(90°−B) (co-function identity), the condition sinA=cosB becomes sinA=sin(90°−B). Since A and B are both acute, A=90°−B, i.e. A+B=90°.

Q12

Evaluate sin35/cos55 + cos60×cosec30 − √3tan30.

sin35°=cos55° (co-function), so the first term = 1

cos60×cosec30=(1/2)(2)=1; √3×tan30=√3×(1/√3)=1

✓1+1−1 = 1

Q13

tanθ=a/b. Find (sinθ+cosθ)/(sinθ−cosθ).

Divide numerator and denominator by cosθ

(tanθ+1)/(tanθ−1)

✓(a+b)/(a−b)

Q14

√3tanθ=1. Find sin²θ−cos²θ.

tanθ=1/√3 → θ=30°

sin²30−cos²30=1/4−3/4

✓−1/2

Q15

Prove (sinA+cosecA)²+(cosA+secA)²=7+tan²A+cot²A.

✓Expand: sin²A+2+cosec²A+cos²A+2+sec²A = (sin²A+cos²A)+4+cosec²A+sec²A = 1+4+(1+cot²A)+(1+tan²A) = 7+tan²A+cot²A. Proved.

Q16

cosecθ−cotθ=1/3. Find cosecθ+cotθ.

(cosecθ−cotθ)(cosecθ+cotθ)=cosec²θ−cot²θ=1

cosecθ+cotθ=1÷(1/3)

✓3

Q17

7sin²θ+3cos²θ=4. Find sinθ.

Rewrite using cos²θ=1−sin²θ

7sin²θ+3(1−sin²θ)=4 → 4sin²θ=1

✓sinθ = 1/2

Q18

Evaluate the mixed expression.

Numerator: 5(1/4)+4(4/3)−1=5/4+16/3−1

common denom 12: 15/12+64/12−12/12=67/12

Denominator: sin²30+cos²30=1

67/12 ÷ 1

✓67/12

Q19

sinA=1/2, verify 3cosA−4cos³A=cos3A.

A=30°, cosA=√3/2. LHS=3(√3/2)−4(√3/2)³

=3√3/2 − 3√3/2 = 0

✓RHS=cos90°=0. LHS=RHS — verified

Q20

tanA+cotA=2. Find tan²A+cot²A.

Square both sides: tan²A+2(tanA·cotA)+cot²A=4

tan²A+2(1)+cot²A=4

✓tan²A+cot²A = 2

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