CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 19

Download CBSE MCQs for Class 10 Mathematics: Chapter 08 Introduction to Trigonometry

Access targeted multiple-choice questions for Chapter 08 Introduction to Trigonometry designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Chapter 08 Introduction to Trigonometry

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

Question 1: If \( 4\tan\theta=3 \), then \( \frac{4\sin\theta-2\cos\theta}{4\sin\theta+3\cos\theta} \) equals:
(a) \( \frac{1}{3} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{3}{2} \)
(d) \( \frac16 \)
Show Answer & Explanation

Answer: (d) \( \frac16 \)

Explanation:
1. Divide by \( \cos\theta \): \( \frac{4\tan\theta-2}{4\tan\theta+3} \).
2. With \( 4\tan\theta=3 \): \( \frac{3-2}{3+3}=\frac16 \).

Teacher's Note:
1. Notice that 4 tan θ = 3 appears directly.
2. Divide numerator and denominator by cos θ.

Question 2: If \( \csc\theta=\frac{13}{12} \) (acute), then \( \frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta} \) equals:
(a) 3
(b) 2
(c) \( \frac13 \)
(d) \( \frac{9}{7} \)
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. \( \sin\theta=\frac{12}{13},\ \cos\theta=\frac5{13} \).
2. \( \frac{24-15}{48-45}=\frac93=3 \).

Teacher's Note:
1. Find sin θ from the reciprocal, then cos θ from Pythagoras.
2. Multiply numerator and denominator by 13 to clear fractions.

Question 3: The angle \( \frac{3\pi}5 \) radians, expressed in degrees, is:
(a) \( 72^\circ \)
(b) \( 108^\circ \)
(c) \( 120^\circ \)
(d) \( 144^\circ \)
Show Answer & Explanation

Answer: (b) \( 108^\circ \)

Explanation:
1. \( \pi\ \text{rad}=180^\circ \), so \( \frac{3\pi}{5}=\frac35\times180^\circ=108^\circ \).

Teacher's Note:
1. Multiply the radian measure by 180°/π.
2. π radians equal 180°.

Question 4: If \( \sin\theta=\frac5{13} \) (acute), then \( \tan\theta \) and \( \sec\theta \) are:
(a) \( \frac{12}{5} \) and \( \frac{13}{5} \)
(b) \( \frac5{12} \) and \( \frac{12}{13} \)
(c) \( \frac{5}{13} \) and \( \frac{13}{12} \)
(d) \( \frac5{12} \) and \( \frac{13}{12} \)
Show Answer & Explanation

Answer: (d) \( \frac5{12} \) and \( \frac{13}{12} \)

Explanation:
1. Base \( =\sqrt{13^2-5^2}=12 \).
2. \( \tan\theta=\frac5{12} \) and \( \sec\theta=\frac{13}{12} \).

Teacher's Note:
1. Draw a right triangle with perpendicular 5 and hypotenuse 13.
2. sec = hypotenuse / base.

Question 5: If \( \tan\theta=\frac xy \), then \( \frac{x\sin\theta+y\cos\theta}{x\sin\theta-y\cos\theta} \) equals:
(a) \( \frac{x^2+y^2}{x^2-y^2} \)
(b) \( \frac{x^2-y^2}{x^2+y^2} \)
(c) \( \frac{x+y}{x-y} \)
(d) \( \frac{x^2+y^2}{x^2+y^2} \)
Show Answer & Explanation

Answer: (a) \( \frac{x^2+y^2}{x^2-y^2} \)

Explanation:
1. Divide numerator and denominator by \( \cos\theta \): \( \frac{x\tan\theta+y}{x\tan\theta-y} \).
2. Put \( \tan\theta=\frac xy \): \( \frac{\frac{x^2}y+y}{\frac{x^2}y-y}=\frac{x^2+y^2}{x^2-y^2} \).

Teacher's Note:
1. Convert to tan θ by dividing by cos θ.
2. Multiply by y to clear the inner fractions.

Question 6: If \( 5\tan\theta=4 \), then \( \frac{5\sin\theta-3\cos\theta}{5\sin\theta+2\cos\theta} \) equals:
(a) \( \frac15 \)
(b) \( \frac16 \)
(c) \( \frac{1}{3} \)
(d) \( \frac{7}{6} \)
Show Answer & Explanation

Answer: (b) \( \frac16 \)

Explanation:
1. \( \frac{5\tan\theta-3}{5\tan\theta+2}=\frac{4-3}{4+2}=\frac16 \).

Teacher's Note:
1. Divide numerator and denominator by cos θ.
2. 5 tan θ = 4 can be substituted directly.

Question 7: If \( \tan\theta=\frac34 \) and \( 0^\circ\lt \theta\lt 90^\circ \), then \( \sin\theta\cos\theta \) equals:
(a) \( \frac{3}{5} \)
(b) \( \frac{4}{5} \)
(c) \( \frac{7}{25} \)
(d) \( \frac{12}{25} \)
Show Answer & Explanation

Answer: (d) \( \frac{12}{25} \)

Explanation:
1. \( \sin\theta=\frac35,\ \cos\theta=\frac45 \).
2. \( \sin\theta\cos\theta=\frac{12}{25} \).

Teacher's Note:
1. Use the 3-4-5 triangle.
2. Multiply the two ratios.

Question 8: If \( 8\tan x=15 \) and \( 0^\circ\lt x\lt 90^\circ \), then \( \sin x-\cos x \) equals:
(a) \( \frac7{17} \)
(b) \( \frac{23}{17} \)
(c) \( \frac{1}{17} \)
(d) \( \frac{7}{15} \)
Show Answer & Explanation

Answer: (a) \( \frac7{17} \)

Explanation:
1. Sides 15, 8, 17: \( \sin x=\frac{15}{17},\ \cos x=\frac8{17} \).
2. \( \sin x-\cos x=\frac{7}{17} \).

Teacher's Note:
1. tan x = 15/8 gives hypotenuse 17.
2. Subtract the two ratios.

Question 9: If \( \tan\theta=\frac43 \) (acute), then \( \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} \) equals:
(a) \( \frac{1}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{1}{9} \)
(d) \( \frac13 \)
Show Answer & Explanation

Answer: (d) \( \frac13 \)

Explanation:
1. \( \sin\theta=\frac45 \).
2. \( \sqrt{\frac{1-\frac45}{1+\frac45}}=\sqrt{\frac{1/5}{9/5}}=\frac13 \).

Teacher's Note:
1. Get sin θ from the 3-4-5 triangle.
2. Take the square root at the end.

Question 10: The value of \( \tan75^\circ \) is:
(a) \( 2-\sqrt3 \)
(b) \( 2+\sqrt3 \)
(c) \( \sqrt3-1 \)
(d) \( \frac{1}{2+\sqrt3} \)
Show Answer & Explanation

Answer: (b) \( 2+\sqrt3 \)

Explanation:
1. \( \tan75^\circ=\tan(45^\circ+30^\circ)=\frac{1+\frac{1}{\sqrt3}}{1-\frac{1}{\sqrt3}}=\frac{\sqrt3+1}{\sqrt3-1} \).
2. Rationalising: \( \frac{(\sqrt3+1)^2}{2}=2+\sqrt3 \).

Teacher's Note:
1. Write 75° as 45° + 30° and use the addition formula.
2. tan 75° is greater than tan 60° = √3, which rules out 2 − √3.

Question 11: If \( \tan\theta=\frac ax \), then \( \frac{x}{\sqrt{a^2+x^2}} \) equals:
(a) \( \sin\theta \)
(b) \( \tan\theta \)
(c) \( \cos\theta \)
(d) \( \sec\theta \)
Show Answer & Explanation

Answer: (c) \( \cos\theta \)

Explanation:
1. In a right triangle with perpendicular a and base x, the hypotenuse is \( \sqrt{a^2+x^2} \).
2. \( \frac{x}{\sqrt{a^2+x^2}}=\frac{\text{base}}{\text{hypotenuse}}=\cos\theta \).

Teacher's Note:
1. Draw the triangle and label the sides.
2. cos = base / hypotenuse.

Question 12: If \( 3\sin x+5\cos x=5 \), then \( (3\cos x-5\sin x)^2 \) equals:
(a) 25
(b) 34
(c) 16
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. \( (3\sin x+5\cos x)^2+(3\cos x-5\sin x)^2=9+25=34 \).
2. So \( (3\cos x-5\sin x)^2=34-25=9 \).

Teacher's Note:
1. Expand: the cross terms cancel and sin² + cos² = 1.
2. The sum of the two squares is always 9 + 25.

Question 13: The value of \( (\sin A+\cos A)^2+(\sin A-\cos A)^2 \) is:
(a) 2
(b) 1
(c) 0
(d) \( 2\sin A\cos A \)
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. Expanding, the cross terms cancel: \( 2(\sin^2A+\cos^2A)=2 \).

Teacher's Note:
1. Use (a + b)² + (a − b)² = 2a² + 2b².
2. sin²A + cos²A = 1.

Question 14: For \( 0^\circ\lt A\lt 90^\circ \), \( \sqrt{\frac{1+\sin A}{1-\sin A}} \) equals:
(a) \( \sec A-\tan A \)
(b) \( \sec A+\tan A \)
(c) \( \csc A+\cot A \)
(d) \( \sin A+\cos A \)
Show Answer & Explanation

Answer: (b) \( \sec A+\tan A \)

Explanation:
1. Multiply numerator and denominator by \( 1+\sin A \): \( \sqrt{\frac{(1+\sin A)^2}{1-\sin^2A}}=\frac{1+\sin A}{\cos A} \).
2. \( =\sec A+\tan A \).

Teacher's Note:
1. Rationalise by multiplying by (1 + sin A).
2. 1 − sin²A = cos²A.

Question 15: For \( 0^\circ\lt x\lt 90^\circ \), \( \sqrt{\frac{1+\cos x}{1-\cos x}} \) equals:
(a) \( \csc x-\cot x \)
(b) \( \sec x+\tan x \)
(c) \( \csc x+\cot x \)
(d) \( \cos x+\sin x \)
Show Answer & Explanation

Answer: (c) \( \csc x+\cot x \)

Explanation:
1. Multiply numerator and denominator by \( 1+\cos x \): \( \frac{1+\cos x}{\sin x} \).
2. \( =\csc x+\cot x \).

Teacher's Note:
1. Rationalise using (1 + cos x).
2. 1 − cos²x = sin²x.

Question 16: The value of \( \frac{\cot\theta}{\cot\theta-\cot3\theta}+\frac{\tan\theta}{\tan\theta-\tan3\theta} \) is:
(a) 0
(b) 2
(c) \( \frac12 \)
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. Rewrite the second term using \( \tan\theta=\frac{1}{\cot\theta} \): \( \frac{\tan\theta}{\tan\theta-\tan3\theta}=\frac{\cot3\theta}{\cot3\theta-\cot\theta} \).
2. \( \frac{\cot\theta}{\cot\theta-\cot3\theta}-\frac{\cot3\theta}{\cot\theta-\cot3\theta}=1 \).

Teacher's Note:
1. Convert tan to cot in the second fraction.
2. The two fractions share the denominator cot θ − cot 3θ.

Question 17: The value of \( \frac{\sin A+\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A-\sin B} \) is:
(a) 0
(b) 1
(c) 2
(d) \( \tan\frac{A+B}{2} \)
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. \( \frac{\sin A+\sin B}{\cos A+\cos B}=\tan\frac{A+B}2 \) and \( \frac{\cos A-\cos B}{\sin A-\sin B}=-\tan\frac{A+B}2 \).
2. The sum is 0.

Teacher's Note:
1. Use the sum-to-product formulas for numerator and denominator.
2. The two terms are equal and opposite.

Question 18: The value of \( \sin15^\circ \) is:
(a) \( \frac{\sqrt3+1}{2\sqrt2} \)
(b) \( \frac{\sqrt3-1}{2\sqrt2} \)
(c) \( \frac{\sqrt3-1}{2} \)
(d) \( \frac{1}{4} \)
Show Answer & Explanation

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

Explanation:
1. \( \sin15^\circ=\sin(45^\circ-30^\circ)=\frac{1}{\sqrt2}\cdot\frac{\sqrt3}2-\frac{1}{\sqrt2}\cdot\frac12=\frac{\sqrt3-1}{2\sqrt2} \).

Teacher's Note:
1. Write 15° as 45° − 30°.
2. sin 15° ≈ 0.259, which is less than 1/2.

Question 19: The value of \( \sin40^\circ-\cos50^\circ \) is:
(a) 1
(b) \( \frac12 \)
(c) 0
(d) −1
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. \( \cos50^\circ=\sin(90^\circ-50^\circ)=\sin40^\circ \), so the difference is 0.

Teacher's Note:
1. cos(90° − θ) = sin θ.
2. The two terms are equal.

Question 20: The value of \( \cot15^\circ\cot16^\circ\cot17^\circ\cdots\cot73^\circ\cot74^\circ\cot75^\circ \) is:
(a) 1
(b) 0
(c) 2
(d) \( \frac{1}{2} \)
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. The angles run from 15° to 75°; those that add to 90° pair up, and \( \cot\theta\cot(90^\circ-\theta)=1 \).
2. The middle term is \( \cot45^\circ=1 \), so the product is 1.

Teacher's Note:
1. Pair 15° with 75°, 16° with 74° and so on.
2. The unpaired middle angle is 45°.

Free study material for Mathematics

Practice MCQs for Class 10 Mathematics Chapter 08 Introduction to Trigonometry

Download Multiple Choice Questions: Chapter 08 Introduction to Trigonometry (Class 10 Mathematics)

Test your conceptual understanding of Chapter 08 Introduction to Trigonometry with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Mathematics, these problem sets build accuracy and prepare students for objective exams.

Concept Clarification for Chapter 08 Introduction to Trigonometry

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FAQs

Where can I access latest CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 19?

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