Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 08 Introduction to Trigonometry
Explore reliable objective questions for Chapter 08 Introduction to Trigonometry tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.
Practice Chapter 08 Introduction to Trigonometry MCQs for Class 10 Mathematics
View or download the dedicated Chapter 08 Introduction to Trigonometry MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.
Multiple Choice Questions
(a) \( \sin\theta \)
(b) \( \sec\theta \)
(c) 1
(d) \( \cot\theta \)
Show Answer & Explanation
Answer: (c) 1
Explanation:
1. (1 + sin θ)(1 − sin θ) = 1 − sin²θ = cos²θ.
2. So k = sec²θ × cos²θ = 1.
Teacher's Note:
1. Use (a + b)(a − b) = a² − b² first.
2. sec θ and cos θ are reciprocals, so sec²θ × cos²θ = 1.
(a) 1
(b) \( \frac{225}{64} \)
(c) \( \frac{156}{7} \)
(d) −1
Show Answer & Explanation
Answer: (b) \( \frac{225}{64} \)
Explanation:
1. Expression = \( \frac{2(1+\sin\theta)(1-\sin\theta)}{2(1+\cos\theta)(1-\cos\theta)} = \frac{1-\sin^2\theta}{1-\cos^2\theta} \).
2. This equals \( \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta \).
3. Hence the value is \( \left(\frac{15}{8}\right)^2 = \) \( \frac{225}{64} \).
Teacher's Note:
1. Take 2 common from numerator and denominator.
2. Both pairs become differences of squares, which turn into cos²θ and sin²θ.
(a) \( \frac{a^2+b^2}{a^2-b^2} \)
(b) \( a^2-b^2 \)
(c) ab
(d) \( \frac{a}{b} \)
Show Answer & Explanation
Answer: (c) ab
Explanation:
1. x/a = cosec θ + cot θ and y/b = (1 − cos θ)/sin θ = cosec θ − cot θ.
2. Multiplying, xy/(ab) = cosec²θ − cot²θ = 1.
3. So xy = ab.
Teacher's Note:
1. Split the fraction in y into two terms.
2. cosec²θ − cot²θ = 1 is the key identity.
(a) 1
(b) \( a^2+b^2 \)
(c) 0
(d) 2
Show Answer & Explanation
Answer: (c) 0
Explanation:
1. Squaring and adding: a² + b² = p² + q², so p² + q² − a² − b² = 0.
2. Then (p − a)(p + a) + (q − b)(q + b) = 0.
3. Dividing by (q + b)(p − a): \( \frac{p+a}{q+b} + \frac{q-b}{p-a} = \) 0.
Teacher's Note:
1. Square both given equations and add them.
2. Regroup the result as a sum of two differences of squares.
(a) \( r^2 = x^2+y^2+z^2 \)
(b) \( r^2 = 2xy \)
(c) \( r^2 = x+y+z \)
(d) \( r^2 = y^2+z^2+2xy \)
Show Answer & Explanation
Answer: (a) \( r^2 = x^2+y^2+z^2 \)
Explanation:
1. x² + y² = r² sin²A (cos²C + sin²C) = r² sin²A.
2. Adding z² = r² cos²A gives x² + y² + z² = r²(sin²A + cos²A) = r².
Teacher's Note:
1. Combine x² and y² first so that C disappears.
2. Then use sin²A + cos²A = 1.
(a) \( (2-a^2) \)
(b) \( (2-a^2)^{1/2} \)
(c) \( (2-a^2)^{2/3} \)
(d) \( (2-a^2)^{3/2} \)
Show Answer & Explanation
Answer: (d) \( (2-a^2)^{3/2} \)
Explanation:
1. tan³θ cosec θ = tan²θ × (sin θ/cos θ)(1/sin θ) = tan²θ sec θ.
2. So the expression = sec θ (1 + tan²θ) = sec³θ.
3. sec²θ = 1 + tan²θ = 2 − a², so sec³θ = (2 − a²)^{3/2}.
Teacher's Note:
1. Write tan θ = sin θ/cos θ and cancel sin θ.
2. 1 + tan²θ = sec²θ turns the expression into sec³θ.
(a) \( (a-b)\sin\theta\cos\theta \)
(b) \( (a-b)^2\sin^2\theta\cos^2\theta \)
(c) \( (a-b)^2\sin\theta\cos\theta \)
(d) \( (a-b)\sin^2\theta\cos^2\theta \)
Show Answer & Explanation
Answer: (b) \( (a-b)^2\sin^2\theta\cos^2\theta \)
Explanation:
1. x − a = b sin²θ − a sin²θ = (b − a) sin²θ.
2. b − x = b cos²θ − a cos²θ = (b − a) cos²θ.
3. Product = (b − a)² sin²θ cos²θ = (a − b)² sin²θ cos²θ.
Teacher's Note:
1. Replace a by a(cos²θ + sin²θ) and b by b(cos²θ + sin²θ).
2. (b − a)² = (a − b)².
(a) 1
(b) \( \frac{16}{65} \)
(c) \( \frac{65}{16} \)
(d) 0
Show Answer & Explanation
Answer: (c) \( \frac{65}{16} \)
Explanation:
1. sin A = \( \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \).
2. cot A = cos A / sin A = 3/4.
3. 9 cot²A − 1 = 9 × \( \frac{9}{16} \) − 1 = \( \frac{81-16}{16} = \) \( \frac{65}{16} \).
Teacher's Note:
1. Find sin A from the identity first.
2. Then cot A = cos A / sin A.
(a) −1
(b) 0
(c) 1
(d) \( \frac{1}{\sqrt{2}} \)
Show Answer & Explanation
Answer: (b) 0
Explanation:
1. The product contains the factor cos 90°.
2. Since cos 90° = 0, the whole product is 0.
Teacher's Note:
1. Look for a factor that is zero.
2. The angles run from 1° to 179°, so 90° is included.
(a) equal to 1
(b) less than 1
(c) greater than or equal to 2
(d) equal to 2
Show Answer & Explanation
Answer: (c) greater than or equal to 2
Explanation:
1. Let s = sin²θ, where 0 < s ≤ 1. Then the expression is s + 1/s.
2. s + 1/s − 2 = (1 − s)²/s ≥ 0, so s + 1/s ≥ 2.
3. Hence the expression is greater than or equal to 2 (equal to 2 only at θ = 90°).
Teacher's Note:
1. A number plus its reciprocal is at least 2 for positive numbers.
2. Equality holds only when sin²θ = 1.
(a) \( x^2-y^2 = p^2q^2 \)
(b) \( x^2q^2-y^2p^2 = pq \)
(c) \( x^2q^2-y^2p^2 = \frac{1}{p^2q^2} \)
(d) \( x^2q^2-y^2p^2 = p^2q^2 \)
Show Answer & Explanation
Answer: (d) \( x^2q^2-y^2p^2 = p^2q^2 \)
Explanation:
1. x/p = sec θ and y/q = tan θ, so x²/p² − y²/q² = sec²θ − tan²θ = 1.
2. Multiplying by p²q²: x²q² − y²p² = p²q².
Teacher's Note:
1. Use sec²θ − tan²θ = 1.
2. Clear the denominators by multiplying with p²q².
(a) \( \frac{a-b}{a^2+b^2} \)
(b) \( \frac{a+b}{a^2+b^2} \)
(c) \( \frac{a^2+b^2}{a^2-b^2} \)
(d) \( \frac{a^2-b^2}{a^2+b^2} \)
Show Answer & Explanation
Answer: (d) \( \frac{a^2-b^2}{a^2+b^2} \)
Explanation:
1. Divide numerator and denominator by cos θ: \( \frac{a\tan\theta - b}{a\tan\theta + b} \).
2. Put tan θ = a/b: \( \frac{a^2/b - b}{a^2/b + b} = \frac{a^2-b^2}{a^2+b^2} \).
3. The value is \( \frac{a^2-b^2}{a^2+b^2} \).
Teacher's Note:
1. Dividing by cos θ brings tan θ into the expression.
2. Multiply top and bottom by b to remove the small fractions.
(a) \( 1-2\cos^2 A \)
(b) \( 2\sin^2 A-1 \)
(c) \( \sin^2 A-\cos^2 A \)
(d) \( 2\cos^2 A-1 \)
Show Answer & Explanation
Answer: (d) \( 2\cos^2 A-1 \)
Explanation:
1. cos⁴A − sin⁴A = (cos²A − sin²A)(cos²A + sin²A) = cos²A − sin²A.
2. = cos²A − (1 − cos²A) = 2cos²A − 1.
Teacher's Note:
1. Treat it as a difference of squares.
2. cos²A + sin²A = 1 removes one bracket.
(a) \( \frac{\sin\theta}{1-\cos\theta} \)
(b) \( \frac{\sin\theta}{1-\cos\phi} \)
(c) \( \frac{\sin\phi}{\sin\theta} \)
(d) \( \frac{\sin\theta}{\sin\phi} \)
Show Answer & Explanation
Answer: (d) \( \frac{\sin\theta}{\sin\phi} \)
Explanation:
1. From the first: cot θ = (1 − a cos φ)/(a sin φ) = 1/(a sin φ) − cot φ, so cot θ + cot φ = 1/(a sin φ).
2. From the second: cot φ + cot θ = 1/(b sin θ).
3. So a sin φ = b sin θ and a/b = sin θ / sin φ.
Teacher's Note:
1. Take reciprocals of both given equations.
2. Both give the same sum cot θ + cot φ, so the two right sides are equal.
(a) \( (a^2b)^{2/3}+(ab^2)^{2/3}=1 \)
(b) \( (ab^2)^{2/3}+(a^2b^2)^{2/3}=1 \)
(c) \( a^2+b^2=1 \)
(d) \( b^2-a^2=1 \)
Show Answer & Explanation
Answer: (a) \( (a^2b)^{2/3}+(ab^2)^{2/3}=1 \)
Explanation:
1. a = cosec x − sin x = cos²x / sin x and b = sec x − cos x = sin²x / cos x.
2. a²b = (cos⁴x / sin²x)(sin²x / cos x) = cos³x, so cos²x = (a²b)^{2/3}.
3. Similarly ab² = sin³x, so sin²x = (ab²)^{2/3}.
4. Adding: (a²b)^{2/3} + (ab²)^{2/3} = 1.
Teacher's Note:
1. Simplify a and b to a single fraction each.
2. Cubes of cos x and sin x appear in a²b and ab².
(a) \( \sin 30^\circ \)
(b) \( \cos 60^\circ \)
(c) \( \frac{1}{2} \)
(d) \( \frac{\sqrt{3}}{2} \)
Show Answer & Explanation
Answer: (d) \( \frac{\sqrt{3}}{2} \)
Explanation:
1. tan 30° = 1/√3, so tan²30° = 1/3.
2. Numerator = 2/√3 and denominator = 4/3.
3. Value = (2/√3) × (3/4) = 3/(2√3) = \( \frac{\sqrt{3}}{2} \).
Teacher's Note:
1. Put tan 30° = 1/√3 and simplify the denominator first.
2. The result equals sin 60°, not sin 30°.
(a) \( \sec\theta \)
(b) \( \tan\theta \)
(c) \( \sqrt{\sec\theta-1} \)
(d) \( \cot\theta \)
Show Answer & Explanation
Answer: (b) \( \tan\theta \)
Explanation:
1. Numerator = sin θ (1 − 2sin²θ) and denominator = cos θ (2cos²θ − 1).
2. 1 − 2sin²θ = 1 − 2(1 − cos²θ) = 2cos²θ − 1, so the brackets cancel.
3. Value = sin θ / cos θ = tan θ.
Teacher's Note:
1. Take sin θ common above and cos θ common below.
2. Show that the two brackets are equal.
(a) \( \sec^2\theta+\operatorname{cosec}^2\theta \)
(b) \( \cot^2\theta-\tan^2\theta \)
(c) \( \cos^2\theta-\sin^2\theta \)
(d) \( \tan^2\theta-\cot^2\theta \)
Show Answer & Explanation
Answer: (d) \( \tan^2\theta-\cot^2\theta \)
Explanation:
1. Split the fraction: tan θ/(sin θ cos θ) − cot θ/(sin θ cos θ).
2. tan θ/(sin θ cos θ) = 1/cos²θ = sec²θ and cot θ/(sin θ cos θ) = 1/sin²θ = cosec²θ.
3. Difference = sec²θ − cosec²θ = (1 + tan²θ) − (1 + cot²θ) = tan²θ − cot²θ.
Teacher's Note:
1. Divide each term separately.
2. Use sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ.
(a) \( \tan 90^\circ \)
(b) 1
(c) \( \sin 45^\circ \)
(d) 0
Show Answer & Explanation
Answer: (d) 0
Explanation:
1. tan 45° = 1, so tan²45° = 1.
2. Value = (1 − 1)/(1 + 1) = 0/2 = 0.
Teacher's Note:
1. Substitute tan 45° = 1.
2. tan 90° is not defined, so option (a) cannot be a value.
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Show Answer & Explanation
Answer: (a) 0°
Explanation:
1. At A = 0°: sin 0° = 0 and 2 sin 0° = 0, so the equation holds.
2. At 30°: sin 60° = √3/2 but 2 sin 30° = 1. At 45°: 1 against √2. At 60°: √3/2 against √3.
3. Only A = 0° works.
Teacher's Note:
1. Test each option by substitution.
2. Both sides are 0 only when sin A = 0.
(a) \( \cos 60^\circ \)
(b) \( \sin 60^\circ \)
(c) \( \tan 60^\circ \)
(d) \( \sin 30^\circ \)
Show Answer & Explanation
Answer: (c) \( \tan 60^\circ \)
Explanation:
1. Numerator = 2/√3 and denominator = 1 − 1/3 = 2/3.
2. Value = (2/√3) × (3/2) = √3.
3. tan 60° = √3, so the answer is tan 60°.
Teacher's Note:
1. The only change from the earlier expression is the minus sign in the denominator.
2. Compare the value √3 with the standard values.
(a) 1
(b) 9
(c) 8
(d) 0
Show Answer & Explanation
Answer: (b) 9
Explanation:
1. Take 9 common: 9(sec²A − tan²A).
2. sec²A − tan²A = 1, so the value is 9.
Teacher's Note:
1. Factor out 9 first.
2. sec²A − tan²A = 1 is a basic identity.
(a) 0
(b) 1
(c) 2
(d) −1
Show Answer & Explanation
Answer: (c) 2
Explanation:
1. Write in sin and cos: \( \frac{(\cos\theta+\sin\theta+1)(\sin\theta+\cos\theta-1)}{\sin\theta\cos\theta} \).
2. Numerator = (sin θ + cos θ)² − 1 = 2 sin θ cos θ.
3. So the value is 2.
Teacher's Note:
1. Convert every ratio into sin θ and cos θ.
2. (x + 1)(x − 1) = x² − 1 with x = sin θ + cos θ.
(a) \( \sec A \)
(b) \( \sin A \)
(c) \( \operatorname{cosec} A \)
(d) \( \cos A \)
Show Answer & Explanation
Answer: (d) \( \cos A \)
Explanation:
1. sec A + tan A = (1 + sin A)/cos A.
2. (1 + sin A)(1 − sin A)/cos A = (1 − sin²A)/cos A = cos²A / cos A = cos A.
Teacher's Note:
1. Write sec A and tan A with cos A in the denominator.
2. (1 + sin A)(1 − sin A) = cos²A.
(a) \( \sec^2 A \)
(b) −1
(c) \( \cot^2 A \)
(d) \( \tan^2 A \)
Show Answer & Explanation
Answer: (d) \( \tan^2 A \)
Explanation:
1. 1 + tan²A = sec²A and 1 + cot²A = cosec²A.
2. sec²A / cosec²A = sin²A / cos²A = tan²A.
Teacher's Note:
1. Use the two identities that start with 1 +.
2. Reciprocals turn sec/cosec into sin/cos.
(a) −1
(b) 0
(c) 1
(d) 2
Show Answer & Explanation
Answer: (b) 0
Explanation:
1. sin 30° + cos 30° = 1/2 + √3/2.
2. sin 60° + cos 60° = √3/2 + 1/2.
3. Both brackets are equal, so the difference is 0.
Teacher's Note:
1. Write each value first.
2. sin 30° = cos 60° and cos 30° = sin 60°.
(a) \( \frac{1}{\sqrt{2}} \)
(b) \( \frac{1}{\sqrt{3}} \)
(c) \( \sqrt{3} \)
(d) 1
Show Answer & Explanation
Answer: (d) 1
Explanation:
1. cot 60° = 1/√3 and tan 30° = 1/√3.
2. The ratio is 1.
Teacher's Note:
1. cot 60° = tan 30° because 30° + 60° = 90°.
2. Equal numbers divide to give 1.
(a) \( \frac{1}{\sqrt{2}} \)
(b) \( \sqrt{2} \)
(c) \( \frac{\sqrt{3}}{2} \)
(d) 1
Show Answer & Explanation
Answer: (b) \( \sqrt{2} \)
Explanation:
1. sin 45° = cos 45° = 1/√2.
2. Sum = 2/√2 = √2.
Teacher's Note:
1. Both values are equal.
2. 2/√2 = √2 after rationalising.
(a) \( \frac{b}{\sqrt{b^2-a^2}} \)
(b) \( \frac{b}{a} \)
(c) \( \frac{\sqrt{b^2-a^2}}{b} \)
(d) \( \frac{a}{\sqrt{b^2-a^2}} \)
Show Answer & Explanation
Answer: (c) \( \frac{\sqrt{b^2-a^2}}{b} \)
Explanation:
1. cos θ = \( \sqrt{1-\sin^2\theta} = \sqrt{1-\frac{a^2}{b^2}} \).
2. = \( \frac{\sqrt{b^2-a^2}}{b} \), which is option (c).
Teacher's Note:
1. Use sin²θ + cos²θ = 1.
2. Take b² as the common denominator under the root.
(a) 1
(b) \( \frac{1}{2} \)
(c) 2
(d) 3
Show Answer & Explanation
Answer: (a) 1
Explanation:
1. sin A = 1 − sin²A = cos²A.
2. cos²A + cos⁴A = sin A + (sin A)² = sin A + sin²A.
3. This equals 1, the given value.
Teacher's Note:
1. Replace cos²A by sin A.
2. cos⁴A is then sin²A, which brings back the given expression.
(a) \( \tan B = 1 \)
(b) \( B = 30^\circ \)
(c) \( A = 45^\circ \)
(d) \( \cos A = \frac{1}{2} \)
Show Answer & Explanation
Answer: (c) \( A = 45^\circ \)
Explanation:
1. sin(A + B) = √3/2 gives A + B = 60°.
2. sin 2B = 1/2 gives 2B = 30°, so B = 15°.
3. Then A = 60° − 15° = 45°.
4. tan 15° ≠ 1, B ≠ 30° and cos 45° ≠ 1/2, so only option (c) holds.
Teacher's Note:
1. Find A + B and B separately from the two equations.
2. Subtract to get A.
Free study material for Mathematics
Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 08 Introduction to Trigonometry
Class 10 Mathematics Chapter 08 Introduction to Trigonometry Objective Test Questions
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FAQs
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By solving our CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 14, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.
Yes, Mathematics MCQs for Class 10 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.
Yes, you can also access online interactive tests for CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 14 on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.