CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 21

Mathematics Objective Questions and Answers: Chapter 08 Introduction to Trigonometry

Review structured MCQ sets for Class 10 Mathematics Chapter 08 Introduction to Trigonometry. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Chapter 08 Introduction to Trigonometry MCQs with Answers

Navigate directly to the 50 objective questions for Chapter 08 Introduction to Trigonometry using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: If \( \tan\theta+\sin\theta=m \) and \( \tan\theta-\sin\theta=n \), then \( \sin\theta\cos\theta \) equals:
(a) \( \frac{(m-n)^2}{2(m+n)} \)
(b) \( \frac{(m+n)^2}{2(m-n)} \)
(c) \( \frac{m-n}{m+n} \)
(d) \( \frac{(m-n)^2}{m+n} \)
Show Answer & Explanation

Answer: (a) \( \frac{(m-n)^2}{2(m+n)} \)

Explanation:
1. sin θ = (m − n)/2 and tan θ = (m + n)/2, so cos θ = sin θ/tan θ = (m − n)/(m + n). The product is (m − n)²/(2(m + n)).

Teacher's Note:
1. Add and subtract the two equations.
2. cos θ = sin θ/tan θ.

Question 2: The value of \( \frac{\sin30^\circ}{\cos45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ} \) is:
(a) \( \frac{\sqrt2+1+2\sqrt3}{2} \)
(b) \( \frac{\sqrt2-1-2\sqrt3}{2} \)
(c) \( \frac{2\sqrt2+1-\sqrt3}{2} \)
(d) \( \frac{\sqrt2+1-2\sqrt3}{2} \)
Show Answer & Explanation

Answer: (d) \( \frac{\sqrt2+1-2\sqrt3}{2} \)

Explanation:
1. The terms are 1/√2 = √2/2, 1/2, −√3/2 and −√3/2. Their sum is (√2 + 1 − 2√3)/2.

Teacher's Note:
1. Substitute the standard values.

Question 3: The value of \( \frac{\tan45^\circ}{\operatorname{cosec}30^\circ}+\frac{\sec45^\circ}{\cot45^\circ}-\frac{2\sin90^\circ}{\cos0^\circ} \) is:
(a) \( \frac{2\sqrt2+3}{2} \)
(b) \( \frac{\sqrt2-3}{2} \)
(c) \( \frac{2\sqrt2-3}{2} \)
(d) \( \frac{2\sqrt2-1}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{2\sqrt2-3}{2} \)

Explanation:
1. The terms are 1/2, √2/1 and −2/1. Sum = 1/2 + √2 − 2 = (2√2 − 3)/2.

Teacher's Note:
1. Use cosec 30° = 2 and sec 45° = √2.

Question 4: If \( x=a\cos\theta-b\sin\theta \) and \( y=a\sin\theta+b\cos\theta \), then \( x^2+y^2 \) equals:
(a) \( a^2-b^2 \)
(b) \( a^2+b^2 \)
(c) \( (a+b)^2 \)
(d) \( 2ab \)
Show Answer & Explanation

Answer: (b) \( a^2+b^2 \)

Explanation:
1. x² + y² = a²cos² + b²sin² − 2ab sin cos + a²sin² + b²cos² + 2ab sin cos = a² + b².

Teacher's Note:
1. Square and add; the cross terms cancel.

Question 5: If \( a\sin^3\theta+b\cos^3\theta=\sin\theta\cos\theta \) and \( a\sin\theta-b\cos\theta=0 \), then \( a^2+b^2 \) equals:
(a) 1
(b) 0
(c) 2
(d) ab
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. Let a sin θ = b cos θ = k. Then the first equation gives k sin²θ + k cos²θ = k = sin θ cos θ. So a = cos θ and b = sin θ, giving a² + b² = 1.

Teacher's Note:
1. Put a sin θ = b cos θ = k.
2. Find a and b separately.

Question 6: If \( x=r\sin\theta\cos\varphi \), \( y=r\sin\theta\sin\varphi \) and \( z=r\cos\theta \), then \( x^2+y^2+z^2 \) equals:
(a) \( r \)
(b) \( 2r^2 \)
(c) \( r^2\sin^2\theta \)
(d) \( r^2 \)
Show Answer & Explanation

Answer: (d) \( r^2 \)

Explanation:
1. x² + y² = r²sin²θ (cos²φ + sin²φ) = r²sin²θ. Adding z² = r²cos²θ gives r².

Teacher's Note:
1. Add x² and y² first.

Question 7: If \( \tan\theta+\sin\theta=m \) and \( \tan\theta-\sin\theta=n \), then \( m^2-n^2 \) equals:
(a) \( 2\sqrt{mn} \)
(b) \( 4mn \)
(c) \( 4\sqrt{mn} \)
(d) \( \sqrt{mn} \)
Show Answer & Explanation

Answer: (c) \( 4\sqrt{mn} \)

Explanation:
1. m² − n² = 4 tan θ sin θ. And mn = tan²θ − sin²θ = tan²θ sin²θ (since 1/cos² − 1 = tan²), so √(mn) = tan θ sin θ.

Teacher's Note:
1. m² − n² = (m + n)(m − n).
2. mn = tan²θ − sin²θ = tan²θ sin²θ.

Question 8: If \( \cos\theta+\sin\theta=m \) and \( \sec\theta+\operatorname{cosec}\theta=n \), then \( n(m^2-1) \) equals:
(a) m
(b) \( 2m \)
(c) \( 2n \)
(d) \( m^2 \)
Show Answer & Explanation

Answer: (b) \( 2m \)

Explanation:
1. m² − 1 = 2 sin θ cos θ and n = (sin θ + cos θ)/(sin θ cos θ) = m/(sin θ cos θ). So n(m² − 1) = 2m.

Teacher's Note:
1. m² = 1 + 2 sin θ cos θ.
2. Combine sec and cosec over a common denominator.

Question 9: If \( \sec\theta=x+\frac{1}{4x} \), then \( \sec\theta+\tan\theta \) equals:
(a) \( 2x \) or \( \frac1{2x} \)
(b) \( x \) or \( \frac1{x} \)
(c) \( 4x \) or \( \frac1{4x} \)
(d) \( 2x \) only
Show Answer & Explanation

Answer: (a) \( 2x \) or \( \frac1{2x} \)

Explanation:
1. Let T = sec θ + tan θ, so sec θ − tan θ = 1/T and sec θ = (T + 1/T)/2. Comparing with x + 1/(4x) = (2x + 1/(2x))/2 gives T = 2x or 1/(2x).

Teacher's Note:
1. Use (sec + tan)(sec − tan) = 1.
2. sec θ is the average of T and 1/T.

Question 10: If \( x=a\sec\theta\cos\varphi \), \( y=b\sec\theta\sin\varphi \) and \( z=c\tan\theta \), then \( \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} \) equals:
(a) 0
(b) sec²θ
(c) −1
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. The expression is sec²θ(cos²φ + sin²φ) − tan²θ = sec²θ − tan²θ = 1.

Teacher's Note:
1. Substitute and simplify.

Question 11: The value of \( \tan1^\circ\tan2^\circ\tan3^\circ\cdots\tan89^\circ \) is:
(a) 0
(b) 89
(c) 1
(d) Not defined
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. Pair tan x with tan(90° − x), whose product is 1. The middle term tan 45° = 1. So the product is 1.

Teacher's Note:
1. tan(90° − x) = cot x.

Question 12: If \( (a^2-b^2)\sin\theta+2ab\cos\theta=a^2+b^2 \) (θ acute, a, b > 0), then \( \tan\theta \) equals:
(a) \( \frac{2ab}{a^2-b^2} \)
(b) \( \frac{a^2-b^2}{2ab} \)
(c) \( \frac{a^2+b^2}{2ab} \)
(d) \( \frac{a}{b} \)
Show Answer & Explanation

Answer: (b) \( \frac{a^2-b^2}{2ab} \)

Explanation:
1. Divide by a² + b² and put cos α = (a² − b²)/(a² + b²), sin α = 2ab/(a² + b²). The equation becomes sin(θ + α) = 1, so θ = 90° − α and tan θ = cot α = (a² − b²)/(2ab).

Teacher's Note:
1. Divide by a² + b² to get coefficients that are cos α and sin α.
2. Then use sin(θ + α) = 1.

Question 13: If \( \sin\theta+\cos\theta=\sqrt2\cos(90^\circ-\theta) \), then \( \cot\theta \) equals:
(a) \( \sqrt2-1 \)
(b) \( \frac1{\sqrt2} \)
(c) \( \frac{\sqrt3}{2} \)
(d) \( \sqrt2+1 \)
Show Answer & Explanation

Answer: (a) \( \sqrt2-1 \)

Explanation:
1. cos(90° − θ) = sin θ, so sin θ + cos θ = √2 sin θ. Hence cos θ = (√2 − 1) sin θ and cot θ = √2 − 1.

Teacher's Note:
1. cos(90° − θ) = sin θ.

Question 14: If \( \sec\theta+\tan\theta=x \), then \( \sec\theta-\tan\theta \) equals:
(a) −x
(b) \( -\frac1x \)
(c) \( \sqrt x \)
(d) \( \frac1x \)
Show Answer & Explanation

Answer: (d) \( \frac1x \)

Explanation:
1. (sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1, so the second factor is 1/x.

Teacher's Note:
1. Use the difference of squares.

Question 15: If \( x=a\sin\theta \) and \( y=b\cos\theta \), then \( b^2x^2+a^2y^2 \) equals:
(a) ab
(b) \( \frac1{a^2b^2} \)
(c) \( a^2b^2 \)
(d) \( \frac1{ab} \)
Show Answer & Explanation

Answer: (c) \( a^2b^2 \)

Explanation:
1. b²x² + a²y² = a²b²sin²θ + a²b²cos²θ = a²b².

Teacher's Note:
1. Substitute x and y.

Question 16: Let \( x=(\sec A+\tan A)(\sec B+\tan B)(\sec C+\tan C) \) and \( y=(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C) \). If x = y, the common value is:
(a) 0
(b) \( \pm1 \)
(c) \( \pm2 \)
(d) None of these
Show Answer & Explanation

Answer: (b) \( \pm1 \)

Explanation:
1. Each pair (sec + tan)(sec − tan) = 1, so xy = 1. With x = y, x² = 1 and x = ±1.

Teacher's Note:
1. Multiply x and y.

Question 17: The value of \( 2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta) \) is:
(a) 0
(b) 1
(c) −2
(d) −1
Show Answer & Explanation

Answer: (d) −1

Explanation:
1. sin⁶ + cos⁶ = 1 − 3s²c² and sin⁴ + cos⁴ = 1 − 2s²c². So 2 − 6s²c² − 3 + 6s²c² = −1.

Teacher's Note:
1. Write both sums in terms of sin²θ cos²θ.

Question 18: For acute θ, \( \sqrt{\frac{1+\sin\theta}{1-\sin\theta}} \) equals:
(a) \( \sec\theta-\tan\theta \)
(b) \( \sec^2\theta+\tan^2\theta \)
(c) \( \sec\theta+\tan\theta \)
(d) \( \sec^2\theta-\tan^2\theta \)
Show Answer & Explanation

Answer: (c) \( \sec\theta+\tan\theta \)

Explanation:
1. Multiply inside by (1 + sin θ): (1 + sin θ)²/(1 − sin²θ) = (1 + sin θ)²/cos²θ, so the root is (1 + sin θ)/cos θ = sec θ + tan θ.

Teacher's Note:
1. Rationalise by multiplying by 1 + sin θ.

Question 19: \( \sec^4A-\sec^2A \) is equal to:
(a) \( \tan^2A-\tan^4A \)
(b) \( \tan^4A+\tan^2A \)
(c) \( \tan^4A-\tan^2A \)
(d) \( \tan^2A \)
Show Answer & Explanation

Answer: (b) \( \tan^4A+\tan^2A \)

Explanation:
1. sec²A (sec²A − 1) = (1 + tan²A) tan²A = tan⁴A + tan²A.

Teacher's Note:
1. Factor sec²A.

Question 20: \( \cos^4A-\sin^4A \) is equal to:
(a) \( 2\cos^2A-1 \)
(b) \( 2\cos^2A+1 \)
(c) \( 2\sin^2A-1 \)
(d) \( 2\sin^2A+1 \)
Show Answer & Explanation

Answer: (a) \( 2\cos^2A-1 \)

Explanation:
1. cos⁴A − sin⁴A = (cos²A − sin²A)(1) = 2cos²A − 1.

Teacher's Note:
1. Factor as a difference of squares.

Question 21: The value of \( P=(1+\cot\theta-\operatorname{cosec}\theta)(1+\tan\theta+\sec\theta) \) is:
(a) 1
(b) 4
(c) 0
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. Write in sin and cos: ((s + c − 1)/s)((c + s + 1)/c) = ((s + c)² − 1)/(sc) = 2sc/(sc) = 2.

Teacher's Note:
1. Combine each bracket over sin θ and cos θ.
2. Use (a − 1)(a + 1) = a² − 1.

Question 22: The value of \( (\operatorname{cosec}\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta) \) is:
(a) 0
(b) −1
(c) 1
(d) 2
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. The brackets are cos²θ/sin θ, sin²θ/cos θ and 1/(sin θ cos θ). Their product is 1.

Teacher's Note:
1. Write everything in sin and cos.

Question 23: The expression \( \sqrt{\sin^4\alpha+4\cos^2\alpha}+\sqrt{\cos^4\alpha+4\sin^2\alpha} \) simplifies to:
(a) 2
(b) 3
(c) 4
(d) 1
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. sin⁴α + 4cos²α = (1 − cos²α)² + 4cos²α = (1 + cos²α)². Similarly the other root gives 1 + sin²α. The sum is 2 + 1 = 3.

Teacher's Note:
1. Write sin⁴α as (1 − cos²α)².

Question 24: If \( 15\sin^4\alpha+10\cos^4\alpha=6 \), then \( 8\operatorname{cosec}^4\alpha+27\sec^6\alpha \) equals (α acute):
(a) 175
(b) 250
(c) 200
(d) 220
Show Answer & Explanation

Answer: (a) 175

Explanation:
1. Divide by cos⁴α with t = tan²α: 15t² + 10 = 6(1 + t)², so 9t² − 12t + 4 = 0 and t = 2/3. cosec²α = 1 + 1/t = 5/2, so 8 cosec⁴α = 8 × 25/4 = 50. sec²α = 5/3, so 27 sec⁶α = 27 × 125/27 = 125. The total is 175.

Teacher's Note:
1. Divide the condition by cos⁴α.
2. Solve for tan²α, then evaluate.

Question 25: If \( \sin\theta \) and \( \cos\theta \) are the roots of \( ax^2-bx+c=0 \), then:
(a) \( a^2-b^2=2ac \)
(b) \( a^2+b^2=2ac \)
(c) \( a^2+b^2+2ac=0 \)
(d) \( b^2-a^2=2ac \)
Show Answer & Explanation

Answer: (d) \( b^2-a^2=2ac \)

Explanation:
1. Sum = b/a and product = c/a. Since (sin θ + cos θ)² = 1 + 2 sin θ cos θ, b²/a² = 1 + 2c/a, so b² − a² = 2ac.

Teacher's Note:
1. Square the sum of roots.

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