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CBSE › Class 10 Maths › Introduction to Trigonometry › MCQs

CBSE CLASS 10 · CHAPTER 8

Introduction to Trigonometry: MCQs with Answers

Practise 56 selected MCQs from official CBSE Basic and Standard board papers and sample papers. Open each explanation for the formula, symbol meanings, substitution and reasoning.

38 board-paper questions · 18 official sample-paper questions

Checked against NCERT Chapter 8 and the 2026–27 CBSE syllabus. Ordinary MCQs are presented here; Assertion–Reason and written case-study questions are kept separate.

Introduction to Trigonometry Assertion–Reason previous year questions

Right-triangle ratios Standard-angle values Relationships between ratios Trigonometric identities Finding angles

Right-triangle ratios

Question 1 · Standard

In the diagram, RS : SQ = 1 : 2. Find $\frac{tan α}{tan β}$ .

CBSE 2025 supplementary examination · Standard · 30/S/1 · Q14

Diagram redrawn from the official question · not to scale.

A. 1B. 2C. 3D. 4

Show Answer & Explanation

Formula guide · Team TutorVertex

Tangent and the right triangle

tan θ = $\frac{opposite}{adjacent}$ ; hypotenuse² = opposite² + adjacent²

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.
Here α=∠PSR and β=∠PQR; set RS=k and SQ=2k, so RQ=3k.

Given here: ∠PRQ=90°; S lies between R and Q; α=∠PSR and β=∠PQR; RS:SQ=1:2.

Official CBSE answer · Solution by Team TutorVertex

In triangle PRS, tan α = $\frac{PR}{RS}$ .

In triangle PRQ, tan β = $\frac{PR}{RQ}$ .

Therefore $\frac{tan α}{tan β}$ = $\frac{RQ}{RS}$ .

Let RS = k and SQ = 2k, where k > 0.

RQ = RS + SQ = 3k.

Hence $\frac{tan α}{tan β}$ = $\frac{3k}{k}$ = 3.

Correct answer: C — 3

Question 2 · Standard

Triangle ABC is right-angled at A. If cos B = $\frac{1}{4}$ , the value of cot B is:

CBSE 2026 second board examination · Standard · 30/7/1 · Q11

\frac{1}{4}

\frac{1}{√15}

\frac{√15}{1}

\frac{4}{√15}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

sin θ = $\frac{opposite}{hypotenuse}$ ; cos θ = $\frac{adjacent}{hypotenuse}$ ; tan θ = $\frac{opposite}{adjacent}$ ; cot θ = $\frac{adjacent}{opposite}$ ; sec θ = $\frac{hypotenuse}{adjacent}$ ; cosec θ = $\frac{hypotenuse}{opposite}$

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: ∠A=90°; cos B= $\frac{1}{4}$ .

Official CBSE answer · Solution by Team TutorVertex

For angle B, cos B = $\frac{AB}{BC}$ = $\frac{1}{4}$ .

Take AB = k and BC = 4k, where k > 0.

By Pythagoras, AC = √(BC² − AB²).

AC = √(16k² − k²) = √15 k.

cot B = $\frac{AB}{AC}$ = $\frac{k}{√15 k}$ = $\frac{1}{√15}$ .

Correct answer: B — $\frac{1}{√15}$

TutorVertex illustration · added to explain the solution; not to scale.

Question 3 · Standard

In the figure, AD = DB, ∠B = 90°, ∠BCD = θ, AC = b and AB = a. Then tan θ equals:

CBSE 2026 second board examination · Standard · 30/8/1 · Q11

Diagram redrawn from the official question · not to scale.

\frac{2a}{√(b² − a²)}

\frac{a}{2√(b² − a²)}

\frac{a}{2√(b² + a²)}

\frac{a}{2(b² − a²)}

Show Answer & Explanation

Formula guide · Team TutorVertex

Tangent and the right triangle

tan θ = $\frac{opposite}{adjacent}$ ; hypotenuse² = opposite² + adjacent²

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.
Here θ=∠BCD, a=AB and b=AC; D is the midpoint of AB, so BD=a/2. The lengths satisfy b>a>0.

Given here: Right triangle ABC; D is the midpoint of AB; AC=b and AB=a, with b>a>0.

Official CBSE answer · Solution by Team TutorVertex

In right triangle ABC, BC² = AC² − AB².

BC = √(b² − a²).

Since AD = DB and AB = a, DB = $\frac{a}{2}$ .

For θ = ∠BCD in triangle CBD, tan θ = $\frac{DB}{BC}$ .

Substituting, tan θ = $\frac{a}{2√(b² − a²)}$ .

Correct answer: B — $\frac{a}{2√(b² − a²)}$

Question 4 · Basic

If 5 cos A − 4 = 0, then tan A equals:

CBSE 2024 board examination · Basic · 430/2/3 · Q10

\frac{3}{4}

\frac{4}{3}

\frac{3}{5}

\frac{4}{5}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: cos A= $\frac{4}{5}$ .

Official CBSE answer · Solution by Team TutorVertex

5 cos A − 4 = 0 gives cos A = $\frac{4}{5}$ .

Take adjacent = 4k and hypotenuse = 5k, where k > 0.

Opposite = √(25k² − 16k²) = 3k.

tan A = $\frac{opposite}{adjacent}$ = $\frac{3k}{4k}$ = $\frac{3}{4}$ .

Correct answer: A — $\frac{3}{4}$

TutorVertex illustration · added to explain the solution; not to scale.

Question 5 · Basic

In right triangle ABC, ∠A = 90° and AB = AC. The value of sin C is:

CBSE 2024 board examination · Basic · 430/3/1 · Q7

A. 0B.

\frac{√3}{2}

\frac{1}{2}

\frac{1}{√2}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: ∠A=90° and the two legs AB and AC are equal.

Official CBSE answer · Solution by Team TutorVertex

Let AB = AC = k, where k > 0.

By Pythagoras, BC = √(k² + k²) = √2 k.

For angle C, sin C = $\frac{AB}{BC}$ .

sin C = $\frac{k}{√2 k}$ = $\frac{1}{√2}$ .

Correct answer: D — $\frac{1}{√2}$

Question 6 · Standard

If 4 sec θ − 5 = 0, then cot θ equals:

CBSE 2024 board examination · Standard · 30/5/1 · Q3

\frac{3}{4}

\frac{4}{5}

\frac{5}{3}

\frac{4}{3}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: sec θ= $\frac{5}{4}$ , so cos θ= $\frac{4}{5}$ .

Official CBSE answer · Solution by Team TutorVertex

4 sec θ − 5 = 0 gives sec θ = $\frac{5}{4}$ .

Take adjacent = 4k and hypotenuse = 5k, where k > 0.

Opposite = √(25k² − 16k²) = 3k.

cot θ = $\frac{adjacent}{opposite}$ = $\frac{4k}{3k}$ = $\frac{4}{3}$ .

Correct answer: D — $\frac{4}{3}$

Question 7 · Standard

If 5 tan θ − 12 = 0, then sin θ equals:

CBSE 2024 board examination · Standard · 30/5/3 · Q3

\frac{5}{12}

\frac{12}{13}

\frac{5}{13}

\frac{12}{5}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: tan θ= $\frac{12}{5}$ .

Official CBSE answer · Solution by Team TutorVertex

5 tan θ − 12 = 0 gives tan θ = $\frac{12}{5}$ .

Take opposite = 12k and adjacent = 5k, where k > 0.

Hypotenuse = √(144k² + 25k²) = 13k.

sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{12k}{13k}$ = $\frac{12}{13}$ .

Correct answer: B — $\frac{12}{13}$

TutorVertex illustration · added to explain the solution; not to scale.

Question 8 · Basic

If sec θ = √3, then tan θ equals:

CBSE 2026 board examination · Basic · 430/4/2 · Q17

A. √3 − 1B. 60°C. 2D. √2

Show Answer & Explanation

Formula guide · Team TutorVertex

Secant–tangent identity

sec²θ − tan²θ = 1; (sec θ + tan θ)(sec θ − tan θ) = 1

What the symbols mean

For acute θ, tan θ is the positive square root of sec²θ−1.

Given here: sec θ=√3 and θ is acute.

Official CBSE answer · Solution by Team TutorVertex

sec²θ − tan²θ = 1.

tan²θ = sec²θ − 1 = 3 − 1 = 2.

Since θ is acute, tan θ = √2.

Correct answer: D — √2

Question 9 · Basic

In triangle ABC, ∠B = 90°. If $\frac{AB}{AC}$ = $\frac{1}{2}$ , then cos C equals:

CBSE 2025 board examination · Basic · 430/6/1 · Q17

\frac{3}{2}

\frac{1}{2}

\frac{√3}{2}

\frac{1}{√3}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: ∠B=90°, AB/AC= $\frac{1}{2}$ .

Official CBSE answer · Solution by Team TutorVertex

Let AB = k and AC = 2k, where k > 0.

AC is the hypotenuse, so BC = √(AC² − AB²).

BC = √(4k² − k²) = √3 k.

For angle C, cos C = $\frac{BC}{AC}$ .

cos C = $\frac{√3 k}{2k}$ = $\frac{√3}{2}$ .

Correct answer: C — $\frac{√3}{2}$

TutorVertex illustration · added to explain the solution; not to scale.

Question 10 · Basic

If sin A = $\frac{1}{2}$ , then sec A equals:

CBSE 2022–23 official sample · Basic · Q10

\frac{2}{√3}

\frac{1}{√3}

C. √3D. 1

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: A is acute and sin A= $\frac{1}{2}$ .

Official CBSE answer · Solution by Team TutorVertex

sin A = $\frac{1}{2}$ gives A = 30° for an acute angle.

cos A = $\frac{√3}{2}$ .

sec A = $\frac{1}{cos A}$ = $\frac{2}{√3}$ .

Correct answer: A — $\frac{2}{√3}$

Question 11 · Basic

If cos A = $\frac{4}{5}$ , then tan A equals:

CBSE 2023–24 official sample · Basic · Q10

\frac{3}{5}

\frac{3}{4}

\frac{4}{3}

\frac{1}{8}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: A is acute and cos A= $\frac{4}{5}$ .

Official CBSE answer · Solution by Team TutorVertex

Take adjacent = 4k and hypotenuse = 5k, where k > 0.

By Pythagoras, opposite = √(25k² − 16k²) = 3k.

tan A = $\frac{opposite}{adjacent}$ .

= $\frac{3k}{4k}$ = $\frac{3}{4}$ .

Correct answer: B — $\frac{3}{4}$

Question 12 · Standard

Given sin θ = $\frac{a}{b}$ , then cos θ equals:

CBSE 2023–24 official sample · Standard · Q11

\frac{b}{√(b² − a²)}

\frac{b}{a}

\frac{√(b² − a²)}{b}

\frac{a}{√(b² − a²)}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.
a is the opposite side length and b is the hypotenuse length, with 0<a<b.

Given here: In the right-triangle interpretation, opposite=a, hypotenuse=b, with 0<a<b.

Official CBSE answer · Solution by Team TutorVertex

Take opposite = a and hypotenuse = b, with 0 < a < b.

By Pythagoras, adjacent = √(b² − a²).

cos θ = $\frac{adjacent}{hypotenuse}$ .

= $\frac{√(b² − a²)}{b}$ .

Correct answer: C — $\frac{√(b² − a²)}{b}$

Question 13 · Basic

If 3 cot A = 4, where 0° < A < 90°, then sec A equals:

CBSE 2024–25 official sample · Basic · Q7

\frac{5}{4}

\frac{4}{3}

\frac{5}{3}

\frac{3}{4}

Show Answer & Explanation

Formula guide · Team TutorVertex

Right-triangle ratios

What the symbols mean

For positive side lengths, hypotenuse² = opposite² + adjacent².
k, when used, is a positive common scale factor for side lengths.

Given here: cot A= $\frac{4}{3}$ .

Official CBSE answer · Solution by Team TutorVertex

3 cot A = 4 gives cot A = $\frac{4}{3}$ .

Take adjacent = 4k and opposite = 3k, where k > 0.

Hypotenuse = √(16k² + 9k²) = 5k.

sec A = $\frac{hypotenuse}{adjacent}$ .

= $\frac{5k}{4k}$ = $\frac{5}{4}$ .

Correct answer: A — $\frac{5}{4}$

TutorVertex illustration · added to explain the solution; not to scale.

Standard-angle values

Question 14 · Standard

Triangle ABC is right-angled at C. The value of cos(A + B) is:

CBSE 2025 supplementary examination · Standard · 30(B)/S · Q17

A. 1B.

\frac{1}{2}

\frac{√3}{2}

D. 0

Show Answer & Explanation

Formula guide · Team TutorVertex

Triangle angle sum

A + B + C = 180°; cos 90° = 0

What the symbols mean

A, B and C are the three interior angles of triangle ABC, measured in degrees.
In a right triangle, the two remaining angles add to 90°.

Given here: ∠C=90°; A and B are the other two angles.

Official CBSE answer · Solution by Team TutorVertex

A + B = 180° − 90° = 90°.

cos(A + B) = cos 90° = 0.

Correct answer: D — 0

Question 15 · Basic

The value of cos 60° − cosec 30° + tan 45° is:

CBSE 2024 board examination · Basic · 430(B) · Q13

A. 1B. −

\frac{1}{2}

\frac{1}{2}

D. −1

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

sin 0°=0; cos 0°=1; sin 30°= $\frac{1}{2}$ ; cos 30°= $\frac{√3}{2}$ ; sin 45°=cos 45°= $\frac{1}{√2}$ ; sin 60°= $\frac{√3}{2}$ ; cos 60°= $\frac{1}{2}$ ; sin 90°=1; cos 90°=0. tan θ= $\frac{sin θ}{cos θ}$ ; cot θ= $\frac{cos θ}{sin θ}$ .

Given here: cos 60°= $\frac{1}{2}$ ; cosec 30°=2; tan 45°=1.

Official CBSE answer · Solution by Team TutorVertex

cos 60° − cosec 30° + tan 45° = $\frac{1}{2}$ − 2 + 1.

= $\frac{1}{2}$ − 1.

= − $\frac{1}{2}$ .

Correct answer: B — − $\frac{1}{2}$

Question 16 · Basic

The value of tan 30° tan 60° − sin 90° cos 90° is:

CBSE 2026 board examination · Basic · 430/4/2 · Q8

A. 1B. 0C. 2D. Not defined

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: tan30°=1/√3, tan60°=√3, sin90°=1 and cos90°=0.

Official CBSE answer · Solution by Team TutorVertex

tan 30° tan 60° − sin 90° cos 90° = $\frac{1}{√3}$ × √3 − 1 × 0.

= 1 − 0.

= 1.

Correct answer: A — 1

Question 17 · Standard

If x tan 45° sin 30° = cos 30° cot 60°, then x equals:

CBSE 2026 board examination · Standard · 30(B) · Q7

A. √3B.

\frac{1}{√3}

C. 1D.

\frac{1}{2}

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

What the symbols mean

x is the unknown numerical factor in the given equation.

Given here: tan45°=1, sin30°= $\frac{1}{2}$ , cos30°=√ $\frac{3}{2}$ , cot60°=1/√3.

Official CBSE answer · Solution by Team TutorVertex

Substituting standard values gives x × 1 × $\frac{1}{2}$ = $\frac{√3}{2}$ × $\frac{1}{√3}$ .

$\frac{x}{2}$ = $\frac{1}{2}$ .

Hence x = 1.

Correct answer: C — 1

Question 18 · Basic

Which statement is false?

CBSE 2025 board examination · Basic · 430/1/1 · Q10

A. tan 45° = cot 45°B. sin 90° = tan 45°C. sin 30° = cos 30°D. sin 45° = cos 45°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: Use the standard-angle values on both sides of each equality.

Official CBSE answer · Solution by Team TutorVertex

tan 45° = cot 45° = 1, so the first statement is true.

sin 90° = tan 45° = 1, so the second statement is true.

sin 30° = $\frac{1}{2}$ , whereas cos 30° = $\frac{√3}{2}$ .

Therefore the third statement is false.

sin 45° = cos 45° = $\frac{1}{√2}$ , so the fourth statement is true.

Correct answer: C — sin 30° = cos 30°

Question 19 · Basic

If tan A = 1, then 3 sin A + cos A equals:

CBSE 2025 board examination · Basic · 430/5/1 · Q12

A. 4√2B. 4C. 2√2D. 4 × 45°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: A is acute and tan A=1.

Official CBSE answer · Solution by Team TutorVertex

tan A = 1 gives A = 45° for an acute angle.

sin A = cos A = $\frac{1}{√2}$ .

3 sin A + cos A = $\frac{3}{√2}$ + $\frac{1}{√2}$ .

= $\frac{4}{√2}$ = 2√2.

Correct answer: C — 2√2

Question 20 · Standard

The value of 1 − 2 sin²60° is the same as:

CBSE 2025 board examination · Standard · 30/5/3 · Q3

A. sin 30°B. −sin 30°C. cos 60°D. −cos 30°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: sin60°=√ $\frac{3}{2}$ and sin30°= $\frac{1}{2}$ .

Official CBSE answer · Solution by Team TutorVertex

1 − 2 sin²60° = 1 − 2 × $\frac{3}{4}$ .

= 1 − $\frac{3}{2}$ = − $\frac{1}{2}$ .

Since sin 30° = $\frac{1}{2}$ , the result is −sin 30°.

Correct answer: B — −sin 30°

Question 21 · Basic

The value of cos 1° · cos 2° · cos 3° · … · cos 90° is:

CBSE 2022–23 official sample · Basic · Q12

A. 1B. 0C. −1D. 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: The product contains cos90°.

Official CBSE answer · Solution by Team TutorVertex

cos 90° = 0.

The product contains cos 90° as a factor.

Therefore the whole product is 0.

Correct answer: B — 0

Question 22 · Standard

If x tan 60° cos 60° = sin 60° cot 60°, then x equals:

CBSE 2022–23 official sample · Standard · Q7

A. cos 30°B. tan 30°C. sin 30°D. cot 30°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

What the symbols mean

x is the unknown numerical factor in the given equation.

Given here: tan60°=√3, cos60°= $\frac{1}{2}$ , sin60°=√ $\frac{3}{2}$ and cot60°=1/√3.

Official CBSE answer · Solution by Team TutorVertex

Substituting standard values gives x × √3 × $\frac{1}{2}$ = $\frac{√3}{2}$ × $\frac{1}{√3}$ .

$\frac{x√3}{2}$ = $\frac{1}{2}$ .

x = $\frac{1}{√3}$ .

Hence x = tan 30°.

Correct answer: B — tan 30°

Question 23 · Basic

$\frac{1 − tan²30°}{1 + tan²30°}$ equals:

CBSE 2024–25 official sample · Basic · Q13

A. cos 60°B. sin 60°C. 1D. tan²60°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: tan30°=1/√3, so tan²30°= $\frac{1}{3}$ .

Official CBSE answer · Solution by Team TutorVertex

tan²30° = $\frac{1}{3}$ .

The expression = $\frac{1 − 1/3}{1 + 1/3}$ .

= $\frac{2/3}{4/3}$ .

= $\frac{1}{2}$ = cos 60°.

Correct answer: A — cos 60°

Question 24 · Standard

If θ = 30°, then 3 tan θ equals:

CBSE 2024–25 official sample · Standard · Q12

A. 1B.

\frac{1}{√3}

\frac{3}{√3}

D. Not defined

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

What the symbols mean

θ is the given angle, 30°.

Given here: θ=30° and tan30°=1/√3.

Official CBSE answer · Solution by Team TutorVertex

3 tan θ = 3 tan 30°.

= 3 × $\frac{1}{√3}$ .

= $\frac{3}{√3}$ .

Correct answer: C — $\frac{3}{√3}$

Question 25 · Basic

The value of sin²60° − 2 tan²45° − cos²30° is:

CBSE 2026–27 official sample · Basic · Q14

A. −2B. −1C. 1D. 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: sin60°=√ $\frac{3}{2}$ , tan45°=1 and cos30°=√ $\frac{3}{2}$ .

Official CBSE solution · typesetting normalised

sin²60° − 2 tan²45° − cos²30° = (√ $\frac{3}{2}$ )² − 2(1)² − (√ $\frac{3}{2}$ )².

= $\frac{3}{4}$ − 2 − $\frac{3}{4}$ .

= −2.

Correct answer: A — −2

Relationships between ratios

Question 26 · Basic

If sin θ = (√2 − 1) cos θ, then:

CBSE 2024 supplementary examination · Basic · 430/S/3 · Q1

A. tan θ = √2 + 1B. cot θ = √2 + 1C. cot θ = √2 − 1D. tan θ = √2

Show Answer & Explanation

Formula guide · Team TutorVertex

Quotient and reciprocal relationships

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

Given here: sin θ=(√2−1)cos θ.

Official CBSE answer · Solution by Team TutorVertex

Dividing by cos θ gives tan θ = √2 − 1.

cot θ = $\frac{1}{tan θ}$ = $\frac{1}{√2 − 1}$ .

Rationalising, cot θ = $\frac{√2 + 1}{(√2 − 1)(√2 + 1)}$ .

Hence cot θ = √2 + 1.

Correct answer: B — cot θ = √2 + 1

Question 27 · Basic

If 5 tan A = 3, then cot A equals:

CBSE 2024 board examination · Basic · 430/4/3 · Q16

\frac{3}{5}

\frac{5}{3}

\frac{3}{4}

\frac{4}{5}

Show Answer & Explanation

Formula guide · Team TutorVertex

Quotient and reciprocal relationships

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

Given here: tan A= $\frac{3}{5}$ .

Official CBSE answer · Solution by Team TutorVertex

5 tan A = 3 gives tan A = $\frac{3}{5}$ .

cot A = $\frac{1}{tan A}$ .

Hence cot A = $\frac{5}{3}$ .

Correct answer: B — $\frac{5}{3}$

Question 28 · Standard

If 5 tan θ = 2, then $\frac{10 sin θ − 2 cos θ}{5 sin θ + 3 cos θ}$ equals:

CBSE 2024 supplementary examination · Standard · 30/S/3 · Q14

\frac{2}{5}

\frac{5}{2}

C. 1D.

\frac{46}{31}

Show Answer & Explanation

Formula guide · Team TutorVertex

Expressing ratios using sine and cosine

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

For a quotient of linear expressions in sin θ and cos θ, divide numerator and denominator by cos θ when it is nonzero.

Given here: tan θ= $\frac{2}{5}$ .

Official CBSE answer · Solution by Team TutorVertex

5 tan θ = 2 gives tan θ = $\frac{2}{5}$ .

Divide the numerator and denominator by cos θ.

The expression becomes $\frac{10 tan θ − 2}{5 tan θ + 3}$ .

Substituting tan θ = $\frac{2}{5}$ gives $\frac{10(2/5) − 2}{5(2/5) + 3}$ .

= $\frac{4 − 2}{2 + 3}$ = $\frac{2}{5}$ .

Correct answer: A — $\frac{2}{5}$

Question 29 · Standard

If sin θ = cos θ, with 0° < θ < 90°, then sec θ · sin θ equals:

CBSE 2024 board examination · Standard · 30/2/1 · Q7

\frac{1}{√2}

B. √2C. 1D. 0

Show Answer & Explanation

Formula guide · Team TutorVertex

Quotient and reciprocal relationships

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

Given here: sin θ=cos θ and θ is acute.

Official CBSE answer · Solution by Team TutorVertex

sec θ · sin θ = $\frac{sin θ}{cos θ}$ .

Since sin θ = cos θ, $\frac{sin θ}{cos θ}$ = 1.

Hence sec θ · sin θ = 1.

Correct answer: C — 1

Question 30 · Basic

If 4 tan A = 3, then $\frac{4 sin A − cos A}{4 sin A + cos A}$ equals:

CBSE 2025 supplementary examination · Basic · 430(B)/S · Q11

\frac{2}{3}

\frac{1}{3}

\frac{1}{2}

\frac{3}{4}

Show Answer & Explanation

Formula guide · Team TutorVertex

Expressing ratios using sine and cosine

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

For a quotient of linear expressions in sin θ and cos θ, divide numerator and denominator by cos θ when it is nonzero.

Given here: tan A= $\frac{3}{4}$ .

Official CBSE answer · Solution by Team TutorVertex

4 tan A = 3 gives tan A = $\frac{3}{4}$ .

Divide the numerator and denominator by cos A.

The expression becomes $\frac{4 tan A − 1}{4 tan A + 1}$ .

Substituting gives $\frac{4(3/4) − 1}{4(3/4) + 1}$ .

= $\frac{3 − 1}{3 + 1}$ = $\frac{1}{2}$ .

Correct answer: C — $\frac{1}{2}$

Question 31 · Standard

cot θ + tan θ equals:

CBSE 2025 board examination · Standard · 30/3/1 · Q9

A. cosec θ sec θB. sin θ sec θC. cos θ tan θD. sin θ cos θ

Show Answer & Explanation

Formula guide · Team TutorVertex

Expressing ratios using sine and cosine

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

For a quotient of linear expressions in sin θ and cos θ, divide numerator and denominator by cos θ when it is nonzero.

Given here: θ is an angle for which cot θ and tan θ are defined.

Official CBSE answer · Solution by Team TutorVertex

cot θ + tan θ = $\frac{cos θ}{sin θ}$ + $\frac{sin θ}{cos θ}$ .

= $\frac{cos²θ + sin²θ}{sin θ cos θ}$ .

= $\frac{1}{sin θ cos θ}$ .

= cosec θ sec θ.

Correct answer: A — cosec θ sec θ

Question 32 · Standard

For an acute angle θ, sec θ expressed in terms of cot θ is:

CBSE 2023 board examination · Standard · 30/4/1 · Q10

\frac{1 + cot²θ}{cot θ}

B. √(1 + cot²θ)C.

\frac{√(1 + cot²θ)}{cot θ}

\frac{√(1 − cot²θ)}{cot θ}

Show Answer & Explanation

Formula guide · Team TutorVertex

Trigonometric identities

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1; tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

Positive square roots apply to ratios of acute angles.

Given here: θ is acute, so cot θ>0 and cosec θ>0.

Official CBSE answer · Solution by Team TutorVertex

1 + cot²θ = cosec²θ.

For acute θ, √(1 + cot²θ) = cosec θ.

$\frac{√(1 + cot²θ)}{cot θ}$ = $\frac{cosec θ}{cot θ}$ .

= $\frac{1/sin θ}{cos θ/sin θ}$ .

= $\frac{1}{cos θ}$ = sec θ.

Correct answer: C — $\frac{√(1 + cot²θ)}{cot θ}$

Question 33 · Standard

If 5 tan β = 4, then $\frac{5 sin β − 2 cos β}{5 sin β + 2 cos β}$ equals:

CBSE 2022–23 official sample · Standard · Q18

\frac{1}{3}

\frac{2}{5}

\frac{3}{5}

D. 6

Show Answer & Explanation

Formula guide · Team TutorVertex

Expressing ratios using sine and cosine

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

For a quotient of linear expressions in sin θ and cos θ, divide numerator and denominator by cos θ when it is nonzero.

Given here: tan β= $\frac{4}{5}$ .

Official CBSE answer · Solution by Team TutorVertex

5 tan β = 4 gives tan β = $\frac{4}{5}$ .

Divide the numerator and denominator by cos β.

The expression becomes $\frac{5 tan β − 2}{5 tan β + 2}$ .

Substituting gives $\frac{4 − 2}{4 + 2}$ .

= $\frac{2}{6}$ = $\frac{1}{3}$ .

Correct answer: A — $\frac{1}{3}$

Question 34 · Standard

If tan θ = $\frac{5}{2}$ , then $\frac{4 sin θ + cos θ}{4 sin θ − cos θ}$ equals:

CBSE 2024–25 official sample · Standard · Q6

\frac{11}{9}

\frac{3}{2}

\frac{9}{11}

D. 4

Show Answer & Explanation

Formula guide · Team TutorVertex

Expressing ratios using sine and cosine

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

For a quotient of linear expressions in sin θ and cos θ, divide numerator and denominator by cos θ when it is nonzero.

Given here: tan θ= $\frac{5}{2}$ .

Official CBSE answer · Solution by Team TutorVertex

Divide the numerator and denominator by cos θ.

The expression becomes $\frac{4 tan θ + 1}{4 tan θ − 1}$ .

Substituting tan θ = $\frac{5}{2}$ gives $\frac{4(5/2) + 1}{4(5/2) − 1}$ .

= $\frac{10 + 1}{10 − 1}$ = $\frac{11}{9}$ .

Correct answer: A — $\frac{11}{9}$

Trigonometric identities

Question 35 · Basic

If x = a sin θ and y = a cos θ, then x² + y² equals:

CBSE 2026 second board examination · Basic · 430/8/1 · Q5

A. 1B. a²C. aD. 0

Show Answer & Explanation

Formula guide · Team TutorVertex

Fundamental identity

sin²θ + cos²θ = 1

What the symbols mean

a is the given constant; x and y are the quantities defined by x=a sin θ and y=a cos θ.

Given here: x=a sin θ, y=a cos θ.

Official CBSE answer · Solution by Team TutorVertex

x² + y² = a² sin²θ + a² cos²θ.

= a²(sin²θ + cos²θ).

= a² × 1 = a².

Correct answer: B — a²

Question 36 · Basic

If x = a sin θ and y = b cos θ, then b²x² + a²y² equals:

CBSE 2023 supplementary examination · Basic · 430/C/1 · Q6

A. 1B. a²b²C.

\frac{a² + b²}{a²b²}

D. a² + b²

Show Answer & Explanation

Formula guide · Team TutorVertex

Fundamental identity

sin²θ + cos²θ = 1

What the symbols mean

a and b are the given constants; x=a sin θ and y=b cos θ define x and y.

Given here: x=a sin θ and y=b cos θ.

Official CBSE answer · Solution by Team TutorVertex

b²x² + a²y² = a²b² sin²θ + a²b² cos²θ.

= a²b²(sin²θ + cos²θ).

= a²b² × 1 = a²b².

Correct answer: B — a²b²

Question 37 · Standard

$\frac{1 + tan²A}{1 + cot²A}$ equals:

CBSE 2026 board examination · Standard · 30/4/3 · Q15

A. tan²AB. −1C. −tan²AD. cot²A

Show Answer & Explanation

Formula guide · Team TutorVertex

Trigonometric identities

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1; tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

Positive square roots apply to ratios of acute angles.

Given here: A is acute, so all ratios and the denominator are defined.

Official CBSE answer · Solution by Team TutorVertex

$\frac{1 + tan²A}{1 + cot²A}$ = $\frac{sec²A}{cosec²A}$ .

= $\frac{1/cos²A}{1/sin²A}$ .

= $\frac{sin²A}{cos²A}$ .

= tan²A.

Correct answer: A — tan²A

Question 38 · Basic

The value of (sin²A + cos²A) + (sec²A − tan²A) − (cot²A − cosec²A) is:

CBSE 2025 board examination · Basic · 430/2/1 · Q11

A. 1B. −1C. 3D. −2

Show Answer & Explanation

Formula guide · Team TutorVertex

Trigonometric identities

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1; tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

Positive square roots apply to ratios of acute angles.

Given here: A is an angle for which the displayed ratios are defined.

Official CBSE answer · Solution by Team TutorVertex

sin²A + cos²A = 1.

sec²A − tan²A = 1.

cot²A − cosec²A = −1.

The expression = 1 + 1 − (−1).

= 3.

Correct answer: C — 3

Question 39 · Basic

(sin θ + cos θ)² + (sin θ − cos θ)² equals:

CBSE 2025 board examination · Basic · 430/2/2 · Q5

A. 1B. 2C. 2 + 2 sin θ cos θD. 2 + 4 sin θ cos θ

Show Answer & Explanation

Formula guide · Team TutorVertex

Fundamental identity

sin²θ + cos²θ = 1

What the symbols mean

u and v are algebraic placeholders: u=sin θ and v=cos θ.

Given here: Use (u+v)²+(u−v)²=2(u²+v²), with u=sin θ and v=cos θ.

Official CBSE answer · Solution by Team TutorVertex

Expanding both squares cancels the cross terms.

(sin θ + cos θ)² + (sin θ − cos θ)² = 2 sin²θ + 2 cos²θ.

= 2(sin²θ + cos²θ).

= 2 × 1 = 2.

Correct answer: B — 2

Question 40 · Standard

The value of (tan A cosec A)² − (sin A sec A)² is:

CBSE 2025 board examination · Standard · 30/1/3 · Q2

A. 0B. 1C. −1D. 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Trigonometric identities

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1; tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

Positive square roots apply to ratios of acute angles.

Given here: A is an angle for which the displayed ratios are defined.

Official CBSE answer · Solution by Team TutorVertex

tan A · cosec A = $\frac{sin A}{cos A}$ × $\frac{1}{sin A}$ = sec A.

sin A · sec A = $\frac{sin A}{cos A}$ = tan A.

The expression = sec²A − tan²A.

= 1.

Correct answer: B — 1

Question 41 · Standard

Which is a trigonometric identity?

CBSE 2025 board examination · Standard · 30/4/2 · Q10

A. sin²θ = 1 + cos²θB. cosec²θ + cot²θ = 1C. sec²θ = 1 + tan²θD. sin 2θ = 2 sin θ

Show Answer & Explanation

Formula guide · Team TutorVertex

Secant–tangent identity

sec²θ − tan²θ = 1; (sec θ + tan θ)(sec θ − tan θ) = 1

What the symbols mean

For acute θ, tan θ is the positive square root of sec²θ−1.

Given here: An identity must hold for every angle for which both sides are defined.

Official CBSE answer · Solution by Team TutorVertex

sin²θ + cos²θ = 1.

Dividing by cos²θ gives tan²θ + 1 = sec²θ.

Hence sec²θ = 1 + tan²θ is an identity.

The other statements fail, for example, at θ = 30°.

Correct answer: C — sec²θ = 1 + tan²θ

Question 42 · Standard

For an acute angle θ in a right triangle, which equation is not true?

CBSE 2023 board examination · Standard · 30/6/1 · Q16

A. sin θ cot θ = cos θB. cos θ tan θ = sin θC. cosec²θ − cot²θ = 1D. tan²θ − sec²θ = 1

Show Answer & Explanation

Formula guide · Team TutorVertex

Trigonometric identities

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1; tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

Positive square roots apply to ratios of acute angles.

Given here: All ratios are defined because θ is acute.

Official CBSE answer · Solution by Team TutorVertex

cot θ = $\frac{cos θ}{sin θ}$ and tan θ = $\frac{sin θ}{cos θ}$ give the first two equations.

The third equation, cosec²θ − cot²θ = 1, is an identity.

sec²θ − tan²θ = 1.

Therefore tan²θ − sec²θ = −1.

The fourth equation, which gives 1 instead of −1, is false.

Correct answer: D — tan²θ − sec²θ = 1

Question 43 · Basic

√3 cos²A + √3 sin²A equals:

CBSE 2022–23 official sample · Basic · Q11

A. 1B.

\frac{1}{√3}

C. √3D. 0

Show Answer & Explanation

Formula guide · Team TutorVertex

Fundamental identity

sin²θ + cos²θ = 1

Given here: Factor out the common coefficient √3.

Official CBSE answer · Solution by Team TutorVertex

√3 cos²A + √3 sin²A = √3(cos²A + sin²A).

= √3 × 1.

= √3.

Correct answer: C — √3

Question 44 · Standard

If sin θ + cos θ = √2, then tan θ + cot θ equals:

CBSE 2022–23 official sample · Standard · Q8

A. 1B. 2C. 3D. 4

Show Answer & Explanation

Formula guide · Team TutorVertex

Fundamental identity

sin²θ + cos²θ = 1

Given here: sin θ+cos θ=√2.

Official CBSE answer · Solution by Team TutorVertex

Squaring sin θ + cos θ = √2 gives sin²θ + cos²θ + 2 sin θ cos θ = 2.

Using sin²θ + cos²θ = 1 gives 2 sin θ cos θ = 1.

sin θ cos θ = $\frac{1}{2}$ .

tan θ + cot θ = $\frac{sin²θ + cos²θ}{sin θ cos θ}$ .

= $\frac{1}{1/2}$ = 2.

Correct answer: B — 2

Question 45 · Basic

1 − cos²A equals:

CBSE 2023–24 official sample · Basic · Q12

A. sin²AB. tan²AC. 1 − sin²AD. sec²A

Show Answer & Explanation

Formula guide · Team TutorVertex

Fundamental identity

sin²θ + cos²θ = 1

Given here: Use sin²A+cos²A=1.

Official CBSE answer · Solution by Team TutorVertex

sin²A + cos²A = 1.

Rearranging gives 1 − cos²A = sin²A.

Correct answer: A — sin²A

Question 46 · Standard

(sec A + tan A)(1 − sin A) equals:

CBSE 2023–24 official sample · Standard · Q12

A. sec AB. sin AC. cosec AD. cos A

Show Answer & Explanation

Formula guide · Team TutorVertex

Trigonometric identities

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1; tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

Positive square roots apply to ratios of acute angles.

Given here: A is acute, so cos A≠0.

Official CBSE answer · Solution by Team TutorVertex

sec A + tan A = $\frac{1 + sin A}{cos A}$ .

(sec A + tan A)(1 − sin A) = $\frac{(1 + sin A)(1 − sin A)}{cos A}$ .

= $\frac{1 − sin²A}{cos A}$ .

= $\frac{cos²A}{cos A}$ = cos A.

Correct answer: D — cos A

Question 47 · Basic

Which expression is not equal to unity, wherever all its ratios are defined?

CBSE 2025–26 official sample · Basic · Q12

A. sin²x + cos²xB. cot²x − cosec²xC. sec²x − tan²xD. tan x · cot x

Show Answer & Explanation

Formula guide · Team TutorVertex

Trigonometric identities

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1; tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

What the symbols mean

Positive square roots apply to ratios of acute angles.
x denotes the angle in the options, measured in degrees.

Given here: Unity means 1.

Official CBSE solution · typesetting normalised

cot²x − cosec²x = −1.

Therefore it is not equal to unity.

Correct answer: B — cot²x − cosec²x

Question 48 · Standard

If sec θ + tan θ = x, then sec θ − tan θ equals:

CBSE 2025–26 official sample · Standard · Q5

A. xB. x²C.

\frac{2}{x}

\frac{1}{x}

Show Answer & Explanation

Formula guide · Team TutorVertex

Secant–tangent identity

sec²θ − tan²θ = 1; (sec θ + tan θ)(sec θ − tan θ) = 1

What the symbols mean

For acute θ, tan θ is the positive square root of sec²θ−1.
x denotes sec θ+tan θ. Its product with sec θ−tan θ is 1, so x≠0.

Given here: x=sec θ+tan θ; the ratios are defined.

Official CBSE solution · typesetting normalised

(sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ.

= 1.

Hence sec θ − tan θ = $\frac{1}{x}$ .

Correct answer: D — $\frac{1}{x}$

Finding angles

Question 49 · Standard

If tan A = 3 cot A, then A equals:

CBSE 2024 board examination · Standard · 30(B) · Q11

A. 15°B. 30°C. 45°D. 60°

Show Answer & Explanation

Formula guide · Team TutorVertex

Quotient and reciprocal relationships

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

Given here: A is acute and tan A=3cot A.

Official CBSE answer · Solution by Team TutorVertex

Using cot A = $\frac{1}{tan A}$ , tan A = $\frac{3}{tan A}$ .

Multiplying by tan A gives tan²A = 3.

Since A is acute, tan A = √3.

Therefore A = 60°.

Correct answer: D — 60°

Question 50 · Standard

If cos θ = $\frac{√3}{2}$ and sin φ = $\frac{1}{2}$ , then tan(θ + φ) equals:

CBSE 2024 board examination · Standard · 30/3/1 · Q8

A. √3B.

\frac{1}{√3}

C. 1D. Not defined

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: θ and φ are the acute angles; cos θ=√ $\frac{3}{2}$ and sin φ= $\frac{1}{2}$ .

Official CBSE answer · Solution by Team TutorVertex

cos θ = $\frac{√3}{2}$ gives θ = 30°.

sin φ = $\frac{1}{2}$ gives φ = 30°.

θ + φ = 60°.

tan(θ + φ) = tan 60° = √3.

Correct answer: A — √3

Question 51 · Standard

If sin α = $\frac{√3}{2}$ and cos β = $\frac{√3}{2}$ , then tan α · tan β equals:

CBSE 2024 board examination · Standard · 30/3/3 · Q13

A. √3B.

\frac{1}{√3}

C. 1D. 0

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: α and β are acute; sin α=√ $\frac{3}{2}$ and cos β=√ $\frac{3}{2}$ .

Official CBSE answer · Solution by Team TutorVertex

sin α = $\frac{√3}{2}$ gives α = 60°.

cos β = $\frac{√3}{2}$ gives β = 30°.

tan α · tan β = √3 × $\frac{1}{√3}$ .

= 1.

Correct answer: C — 1

Question 52 · Standard

If sin θ = 1, then $\frac{1}{2}$ sin( $\frac{θ}{2}$ ) equals:

CBSE 2024 board examination · Standard · 30/4/1 · Q10

\frac{1}{2√2}

\frac{1}{√2}

\frac{1}{2}

D. 0

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: For 0°≤θ≤90°, sin θ=1.

Official CBSE answer · Solution by Team TutorVertex

sin θ = 1 gives θ = 90° in the chapter's 0°–90° range.

$\frac{θ}{2}$ = 45°.

$\frac{1}{2}$ sin( $\frac{θ}{2}$ ) = $\frac{1}{2}$ × sin 45°.

= $\frac{1}{2}$ × $\frac{1}{√2}$ = $\frac{1}{2√2}$ .

Correct answer: A — $\frac{1}{2√2}$

Question 53 · Basic

The value of θ for which sin 2θ = tan 45° is:

CBSE 2025 board examination · Basic · 430/3/3 · Q13

A. 22.5°B. 30°C. 45°D. 90°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: tan45°=1. Use the standard value sin90°=1.

Official CBSE answer · Solution by Team TutorVertex

tan 45° = 1, so sin 2θ = 1.

Taking 2θ = 90° gives θ = 45°.

Among the four listed values, only 45° satisfies the equation.

Correct answer: C — 45°

Question 54 · Standard

If θ is acute and 7 + 4 sin θ = 9, then θ equals:

CBSE 2025 board examination · Standard · 30/1/1 · Q4

A. 90°B. 30°C. 45°D. 60°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: 7+4sin θ=9.

Official CBSE answer · Solution by Team TutorVertex

7 + 4 sin θ = 9.

4 sin θ = 2.

sin θ = $\frac{1}{2}$ .

Since θ is acute, θ = 30°.

Correct answer: B — 30°

Question 55 · Basic

If 2 sin A = cosec A, then A equals:

CBSE 2023 board examination · Basic · 430(B) · Q13

A. 30°B. 60°C. 45°D. 90°

Show Answer & Explanation

Formula guide · Team TutorVertex

Quotient and reciprocal relationships

tan θ = $\frac{sin θ}{cos θ}$ ; cot θ = $\frac{1}{tan θ}$ ; sec θ = $\frac{1}{cos θ}$ ; cosec θ = $\frac{1}{sin θ}$

Given here: A is acute and 2sin A=cosec A.

Official CBSE answer · Solution by Team TutorVertex

Using cosec A = $\frac{1}{sin A}$ , 2 sin A = $\frac{1}{sin A}$ .

Multiplying by sin A gives 2 sin²A = 1.

sin²A = $\frac{1}{2}$ .

Since A is acute, sin A = $\frac{1}{√2}$ .

Hence A = 45°.

Correct answer: C — 45°

Question 56 · Standard

If tan(A + B) = √3 and tan(A − B) = $\frac{1}{√3}$ , with 0° < A + B < 90° and A > B, then A and B respectively are:

CBSE 2026–27 official sample · Standard · Q8

A. 60°, 30°B. 60°, 45°C. 45°, 15°D. 60°, 15°

Show Answer & Explanation

Formula guide · Team TutorVertex

Standard-angle values

Given here: A and B are positive angles; A+B<90° and A>B.

Official CBSE solution · typesetting normalised

A + B = 60°.

A − B = 30°.

Adding gives 2A = 90°.

A = 45°.

Subtracting gives 2B = 30°.

B = 15°.

Correct answer: C — 45°, 15°

Common questions

Which concepts are covered?

Six right-triangle ratios, standard-angle values, relationships between ratios and simple trigonometric identities. Heights and distances belong to the next chapter.

Whose solutions are shown?

Where the official marking scheme supplies working, it is labelled as an official CBSE solution. When it supplies only an answer, Team TutorVertex provides the working. Formula guides and symbol definitions are by Team TutorVertex.

Which diagrams come from the original questions?

Two questions include redrawn original diagrams. Five explanations include additional TutorVertex illustrations, labelled as teaching aids. All diagrams are marked as not to scale.

Why are 2022 written questions not included as MCQs?

The checked 2022 written papers are not converted into invented MCQs. This bank includes official sample-paper MCQs from the 2022–23 session.

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Official sources and marking schemes

Source links are collected here. Each question above identifies its year/session, level, paper code and question number.

How this question bank is prepared

This is a selected collection of 38 board-examination MCQs from 2023–2026 and 18 official sample-paper MCQs from sessions 2022–23 to 2026–27. Answers are checked against corresponding official marking schemes and independently checked mathematically. Mathematical data and choice order are retained; wording is lightly edited for clarity. Repeated questions across sets are represented selectively.

The available CBSE archive collections are screened with text extraction and OCR where needed. This is not an exhaustive transcription of every question or a claim of complete translation of every language version. Unclear or ambiguous questions are excluded. NCERT is used to check chapter relevance, notation and methods.

Introduction to Trigonometry: MCQs with Answers

Right-triangle ratios

In the diagram, RS : SQ = 1 : 2. Find tan αtan β.

Tangent and the right triangle

Triangle ABC is right-angled at A. If cos B = 14, the value of cot B is:

Right-triangle ratios

In the figure, AD = DB, ∠B = 90°, ∠BCD = θ, AC = b and AB = a. Then tan θ equals:

Tangent and the right triangle

If 5 cos A − 4 = 0, then tan A equals:

Right-triangle ratios

In right triangle ABC, ∠A = 90° and AB = AC. The value of sin C is:

Right-triangle ratios

If 4 sec θ − 5 = 0, then cot θ equals:

Right-triangle ratios

If 5 tan θ − 12 = 0, then sin θ equals:

Right-triangle ratios

If sec θ = √3, then tan θ equals:

Secant–tangent identity

In triangle ABC, ∠B = 90°. If ABAC = 12, then cos C equals:

Right-triangle ratios

If sin A = 12, then sec A equals:

Right-triangle ratios

If cos A = 45, then tan A equals:

Right-triangle ratios

Given sin θ = ab, then cos θ equals:

Right-triangle ratios

If 3 cot A = 4, where 0° < A < 90°, then sec A equals:

Right-triangle ratios

Standard-angle values

Triangle ABC is right-angled at C. The value of cos(A + B) is:

Triangle angle sum

The value of cos 60° − cosec 30° + tan 45° is:

Standard-angle values

The value of tan 30° tan 60° − sin 90° cos 90° is:

Standard-angle values

If x tan 45° sin 30° = cos 30° cot 60°, then x equals:

Standard-angle values

Which statement is false?

Standard-angle values

If tan A = 1, then 3 sin A + cos A equals:

Standard-angle values

The value of 1 − 2 sin²60° is the same as:

Standard-angle values

The value of cos 1° · cos 2° · cos 3° · … · cos 90° is:

Standard-angle values

If x tan 60° cos 60° = sin 60° cot 60°, then x equals:

Standard-angle values

1 − tan²30°1 + tan²30° equals:

Standard-angle values

If θ = 30°, then 3 tan θ equals:

Standard-angle values

The value of sin²60° − 2 tan²45° − cos²30° is:

Standard-angle values

Relationships between ratios

If sin θ = (√2 − 1) cos θ, then:

Quotient and reciprocal relationships

If 5 tan A = 3, then cot A equals:

Quotient and reciprocal relationships

If 5 tan θ = 2, then 10 sin θ − 2 cos θ5 sin θ + 3 cos θ equals:

Expressing ratios using sine and cosine

If sin θ = cos θ, with 0° < θ < 90°, then sec θ · sin θ equals:

Quotient and reciprocal relationships

If 4 tan A = 3, then 4 sin A − cos A4 sin A + cos A equals:

Expressing ratios using sine and cosine

cot θ + tan θ equals:

Expressing ratios using sine and cosine

For an acute angle θ, sec θ expressed in terms of cot θ is:

Trigonometric identities

If 5 tan β = 4, then 5 sin β − 2 cos β5 sin β + 2 cos β equals:

Expressing ratios using sine and cosine

If tan θ = 52, then 4 sin θ + cos θ4 sin θ − cos θ equals:

Expressing ratios using sine and cosine

Trigonometric identities

If x = a sin θ and y = a cos θ, then x² + y² equals:

Fundamental identity

If x = a sin θ and y = b cos θ, then b²x² + a²y² equals:

Fundamental identity

1 + tan²A1 + cot²A equals:

Trigonometric identities

The value of (sin²A + cos²A) + (sec²A − tan²A) − (cot²A − cosec²A) is:

In the diagram, RS : SQ = 1 : 2. Find $\frac{tan α}{tan β}$ .

Triangle ABC is right-angled at A. If cos B = $\frac{1}{4}$ , the value of cot B is:

In triangle ABC, ∠B = 90°. If $\frac{AB}{AC}$ = $\frac{1}{2}$ , then cos C equals:

If sin A = $\frac{1}{2}$ , then sec A equals:

If cos A = $\frac{4}{5}$ , then tan A equals:

Given sin θ = $\frac{a}{b}$ , then cos θ equals:

$\frac{1 − tan²30°}{1 + tan²30°}$ equals:

If 5 tan θ = 2, then $\frac{10 sin θ − 2 cos θ}{5 sin θ + 3 cos θ}$ equals:

If 4 tan A = 3, then $\frac{4 sin A − cos A}{4 sin A + cos A}$ equals:

If 5 tan β = 4, then $\frac{5 sin β − 2 cos β}{5 sin β + 2 cos β}$ equals:

If tan θ = $\frac{5}{2}$ , then $\frac{4 sin θ + cos θ}{4 sin θ − cos θ}$ equals:

$\frac{1 + tan²A}{1 + cot²A}$ equals:

If cos θ = $\frac{√3}{2}$ and sin φ = $\frac{1}{2}$ , then tan(θ + φ) equals:

If sin α = $\frac{√3}{2}$ and cos β = $\frac{√3}{2}$ , then tan α · tan β equals:

If sin θ = 1, then $\frac{1}{2}$ sin( $\frac{θ}{2}$ ) equals:

If tan(A + B) = √3 and tan(A − B) = $\frac{1}{√3}$ , with 0° < A + B < 90° and A > B, then A and B respectively are: