CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 18

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

Access the complete set of multiple-choice questions for Chapter 08 Introduction to Trigonometry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: If \( \sin\alpha=\frac1{\sqrt2} \) and \( \tan\beta=1 \), where \( \alpha \) and \( \beta \) are acute, the value of \( \sin(\alpha+\beta) \) is:
(a) 1
(b) 0
(c) \( \frac12 \)
(d) \( \frac{1}{\sqrt2} \)
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. Acute angles with these ratios are \( \alpha=45^\circ \) and \( \beta=45^\circ \).
2. \( \sin(\alpha+\beta)=\sin90^\circ=1 \).

Teacher's Note:
1. Recall the standard angles: sin 45° = 1/√2 and tan 45° = 1.
2. Add the angles before taking the sine.

Question 2: If \( \cos\alpha=\frac12 \) and \( \tan\beta=\frac1{\sqrt3} \), with both angles acute, then \( \sin(\alpha+\beta) \) equals:
(a) \( \frac{\sqrt3}2 \)
(b) 1
(c) \( \frac12 \)
(d) 0
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. \( \cos\alpha=\frac12\Rightarrow\alpha=60^\circ \) and \( \tan\beta=\frac1{\sqrt3}\Rightarrow\beta=30^\circ \).
2. \( \sin(60^\circ+30^\circ)=\sin90^\circ=1 \).

Teacher's Note:
1. Use the table of standard angles to find α and β.
2. Do not add the sines separately; add the angles first.

Question 3: The value of \( \frac{\sin21^\circ}{\cos69^\circ}+\frac{\sec11^\circ}{\csc79^\circ} \) is:
(a) 1
(b) \( \frac12 \)
(c) 2
(d) 0
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. \( \cos69^\circ=\cos(90^\circ-21^\circ)=\sin21^\circ \), so the first term is 1.
2. \( \csc79^\circ=\sec(90^\circ-79^\circ)=\sec11^\circ \), so the second term is 1.
3. Sum = 2.

Teacher's Note:
1. Complementary angles: sin(90° − θ) = cos θ and cosec(90° − θ) = sec θ.
2. Each fraction becomes 1.

Question 4: The value of \( \left(\frac{\sin49^\circ}{\cos41^\circ}\right)^2+\left(\frac{\cos41^\circ}{\sin49^\circ}\right)^2 \) is:
(a) 1
(b) 0
(c) 4
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. \( \sin49^\circ=\cos41^\circ \), so each bracket equals 1.
2. The sum is 1 + 1 = 2.

Teacher's Note:
1. sin 49° = cos(90° − 49°) = cos 41°.
2. Each squared bracket is 1.

Question 5: The value of \( \csc^267^\circ-\tan^223^\circ \) is:
(a) 0
(b) 1
(c) 2
(d) \( \sec^223^\circ \)
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. \( \csc67^\circ=\sec(90^\circ-67^\circ)=\sec23^\circ \).
2. \( \sec^223^\circ-\tan^223^\circ=1 \).

Teacher's Note:
1. Convert cosec 67° into sec 23° using complementary angles.
2. Then use sec²θ − tan²θ = 1.

Question 6: The value of \( \sin^220^\circ+\sin^270^\circ-\tan^245^\circ \) is:
(a) 1
(b) 2
(c) 0
(d) −1
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. \( \sin70^\circ=\cos20^\circ \), so \( \sin^220^\circ+\sin^270^\circ=1 \).
2. \( \tan45^\circ=1 \), so the value is 1 − 1 = 0.

Teacher's Note:
1. Use sin(90° − θ) = cos θ and sin²θ + cos²θ = 1.
2. tan 45° = 1.

Question 7: The value of \( \sec50^\circ\sin40^\circ+\cos40^\circ\csc50^\circ \) is:
(a) 1
(b) 0
(c) \( \frac12 \)
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. \( \sec50^\circ=\csc40^\circ \) and \( \csc50^\circ=\sec40^\circ \).
2. \( \csc40^\circ\sin40^\circ+\cos40^\circ\sec40^\circ=1+1=2 \).

Teacher's Note:
1. sec(90° − θ) = cosec θ and cosec(90° − θ) = sec θ.
2. Reciprocal pairs multiply to 1.

Question 8: The value of \( \sin48^\circ\sec42^\circ+\cos48^\circ\csc42^\circ \) is:
(a) 1
(b) 2
(c) 0
(d) \( \sqrt2 \)
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \( \sec42^\circ=\csc48^\circ \) and \( \csc42^\circ=\sec48^\circ \).
2. \( \sin48^\circ\csc48^\circ+\cos48^\circ\sec48^\circ=1+1=2 \).

Teacher's Note:
1. Replace sec 42° and cosec 42° by cosec 48° and sec 48°.
2. Each product is 1.

Question 9: The value of \( \sin63^\circ\cos27^\circ+\cos63^\circ\sin27^\circ \) is:
(a) 0
(b) \( \frac12 \)
(c) 1
(d) 2
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. \( \sin63^\circ=\cos27^\circ \) and \( \cos63^\circ=\sin27^\circ \).
2. \( \cos^227^\circ+\sin^227^\circ=1 \).

Teacher's Note:
1. Complementary angles turn the expression into sin² + cos².
2. Alternatively it is sin(63° + 27°) = sin 90°.

Question 10: The value of \( \frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\csc31^\circ \) is:
(a) 1
(b) 0
(c) \( \frac12 \)
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. \( \cos80^\circ=\sin10^\circ \) and \( \cos59^\circ=\sin31^\circ \).
2. Each term is 1, so the sum is 2.

Teacher's Note:
1. cos(90° − θ) = sin θ.
2. cosec 31° is the reciprocal of sin 31°.

Question 11: For any angle \( \theta \), the value of \( \sin\theta\sin(90^\circ-\theta)-\cos\theta\cos(90^\circ-\theta) \) is:
(a) 0
(b) 1
(c) −1
(d) \( \sin2\theta \)
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. \( \sin(90^\circ-\theta)=\cos\theta \) and \( \cos(90^\circ-\theta)=\sin\theta \).
2. \( \sin\theta\cos\theta-\cos\theta\sin\theta=0 \).

Teacher's Note:
1. Replace each complementary ratio first.
2. The two products are identical and cancel.

Question 12: For \( 0^\circ\lt \theta\lt 90^\circ \), \( \frac{\sin\theta}{\sin(90^\circ-\theta)}+\frac{\cos\theta}{\cos(90^\circ-\theta)} \) equals:
(a) \( \sin\theta\cos\theta \)
(b) \( \sec\theta\csc\theta \)
(c) \( \tan\theta-\cot\theta \)
(d) 1
Show Answer & Explanation

Answer: (b) \( \sec\theta\csc\theta \)

Explanation:
1. \( \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} \).
2. \( =\frac{1}{\sin\theta\cos\theta}=\sec\theta\csc\theta \).

Teacher's Note:
1. Convert the complementary angles and combine the fractions.
2. sin²θ + cos²θ = 1 in the numerator.

Question 13: For \( 0^\circ\lt \theta\lt 90^\circ \), \( \sin^2(90^\circ-\theta)\left(1+\cot^2(90^\circ-\theta)\right) \) equals:
(a) 0
(b) \( \tan^2\theta \)
(c) 1
(d) \( \sec^2\theta \)
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. \( \sin^2(90^\circ-\theta)=\cos^2\theta \) and \( 1+\cot^2(90^\circ-\theta)=1+\tan^2\theta=\sec^2\theta \).
2. \( \cos^2\theta\sec^2\theta=1 \).

Teacher's Note:
1. Use 1 + tan²θ = sec²θ.
2. cos²θ and sec²θ are reciprocals.

Question 14: For \( 0^\circ\lt A\lt 90^\circ \), \( \frac{\tan(90^\circ-A)\cot A}{\csc^2A}-\cos^2A \) equals:
(a) 0
(b) 1
(c) \( \sin^2A \)
(d) −1
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. \( \tan(90^\circ-A)=\cot A \), so the fraction is \( \frac{\cot^2A}{\csc^2A}=\cos^2A \).
2. Hence the value is \( \cos^2A-\cos^2A=0 \).

Teacher's Note:
1. cot²A / cosec²A = cos²A / sin²A × sin²A = cos²A.
2. The two terms cancel.

Question 15: The value of \( \sec^210^\circ-\cot^280^\circ+\frac{\sin15^\circ\cos75^\circ+\cos15^\circ\sin75^\circ}{\cos\theta\sin(90^\circ-\theta)+\sin\theta\cos(90^\circ-\theta)} \) is:
(a) 1
(b) 2
(c) 3
(d) 0
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \( \cot80^\circ=\tan10^\circ \), so \( \sec^210^\circ-\tan^210^\circ=1 \).
2. Numerator: \( \sin(15^\circ+75^\circ)=1 \). Denominator: \( \cos^2\theta+\sin^2\theta=1 \), so the fraction is 1.
3. Total = 1 + 1 = 2.

Teacher's Note:
1. Use the addition formula sin(A + B) for the numerator.
2. cos θ sin(90° − θ) + sin θ cos(90° − θ) = cos²θ + sin²θ.

Question 16: The value of \( \sin(50^\circ+\theta)-\cos(40^\circ-\theta)+\tan1^\circ\tan10^\circ\tan20^\circ\tan70^\circ\tan80^\circ\tan89^\circ \) is:
(a) 0
(b) 2
(c) 1
(d) −1
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. \( \cos(40^\circ-\theta)=\sin(90^\circ-40^\circ+\theta)=\sin(50^\circ+\theta) \), so the first two terms cancel.
2. The tangents pair as \( \tan1^\circ\tan89^\circ=\tan10^\circ\tan80^\circ=\tan20^\circ\tan70^\circ=1 \), so the product is 1.

Teacher's Note:
1. Use cos x = sin(90° − x) for the first pair.
2. Pair tangents whose angles add to 90°.

Question 17: A round balloon of radius \( r \) subtends an angle \( \alpha \) at the eye of the observer, and the angle of elevation of its centre is \( \beta \). The height of the centre of the balloon above the observer's eye level is:
(a) \( r\cos\beta\csc\frac\alpha2 \)
(b) \( r\sin\beta\sec\frac\alpha2 \)
(c) \( r\sin\beta\csc\alpha \)
(d) \( r\sin\beta\csc\frac\alpha2 \)
Show Answer & Explanation

Answer: (d) \( r\sin\beta\csc\frac\alpha2 \)

Explanation:
1. Let the eye be at O and the centre at C. The tangents from O make an angle \( \alpha \), so the line OC bisects it and the radius to a point of contact is perpendicular to the tangent.
2. In the right triangle formed, \( \sin\frac\alpha2=\frac{r}{OC}\Rightarrow OC=r\csc\frac\alpha2 \).
3. The height of C is \( OC\sin\beta=r\sin\beta\csc\frac\alpha2 \).

Teacher's Note:
1. The line from the eye to the centre bisects the angle subtended by the balloon.
2. Height = distance × sin(elevation).

Question 18: If \( \tan\theta=\frac8{15} \) and \( 0^\circ\lt \theta\lt 90^\circ \), then \( \sin\theta \) equals:
(a) \( \frac8{17} \)
(b) \( \frac{15}{17} \)
(c) \( \frac8{23} \)
(d) \( \frac{17}{8} \)
Show Answer & Explanation

Answer: (a) \( \frac8{17} \)

Explanation:
1. Take a right triangle with perpendicular 8 and base 15. Hypotenuse \( =\sqrt{64+225}=17 \).
2. \( \sin\theta=\frac{8}{17} \).

Teacher's Note:
1. Draw the triangle and use Pythagoras to get the hypotenuse.
2. sin = perpendicular / hypotenuse.

Question 19: If \( \sin\theta=\frac8{17} \) and \( 0^\circ\lt \theta\lt 90^\circ \), then \( \tan\theta \) equals:
(a) \( \frac{15}{8} \)
(b) \( \frac8{15} \)
(c) \( \frac{8}{17} \)
(d) \( \frac{17}{15} \)
Show Answer & Explanation

Answer: (b) \( \frac8{15} \)

Explanation:
1. Base \( =\sqrt{17^2-8^2}=15 \).
2. \( \tan\theta=\frac{8}{15} \).

Teacher's Note:
1. Use Pythagoras to find the third side.
2. tan = perpendicular / base.

Question 20: If \( \sin A=\frac{24}{25} \) and \( 0^\circ\lt A\lt 90^\circ \), then \( \tan A+\sec A \) equals:
(a) \( \frac{7}{25} \)
(b) \( \frac{49}{24} \)
(c) 7
(d) \( \frac{24}{7} \)
Show Answer & Explanation

Answer: (c) 7

Explanation:
1. Base \( =\sqrt{25^2-24^2}=7 \).
2. \( \tan A=\frac{24}{7},\ \sec A=\frac{25}{7} \), so \( \tan A+\sec A=\frac{49}{7}=7 \).

Teacher's Note:
1. Find the third side first.
2. Add the two ratios that share the denominator 7.

Question 21: If \( \tan\theta=\frac34 \), then \( \frac{1-\cos\theta}{1+\cos\theta} \) equals:
(a) \( \frac19 \)
(b) \( \frac15 \)
(c) \( \frac{1}{3} \)
(d) \( \frac{9}{1} \)
Show Answer & Explanation

Answer: (a) \( \frac19 \)

Explanation:
1. \( \cos\theta=\frac45 \) (sides 3, 4, 5).
2. \( \frac{1-\frac45}{1+\frac45}=\frac{1/5}{9/5}=\frac19 \).

Teacher's Note:
1. Get cos θ from the 3-4-5 triangle.
2. Substitute into the fraction.

Question 22: If \( \tan\theta=\frac{12}5 \), then \( \frac{1+\sin\theta}{1-\sin\theta} \) equals:
(a) 5
(b) 25
(c) \( \frac{1}{25} \)
(d) \( \frac{25}{13} \)
Show Answer & Explanation

Answer: (b) 25

Explanation:
1. \( \sin\theta=\frac{12}{13} \) (sides 5, 12, 13).
2. \( \frac{1+\frac{12}{13}}{1-\frac{12}{13}}=\frac{25/13}{1/13}=25 \).

Teacher's Note:
1. Use the 5-12-13 triangle.
2. The denominators 13 cancel.

Question 23: If \( \tan A=\frac12 \) and \( \tan B=\frac13 \) (both acute), then \( A+B \) equals:
(a) \( 30^\circ \)
(b) \( 60^\circ \)
(c) \( 45^\circ \)
(d) \( 75^\circ \)
Show Answer & Explanation

Answer: (c) \( 45^\circ \)

Explanation:
1. \( \tan(A+B)=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\frac{5/6}{5/6}=1 \).
2. Since both angles are acute and \( \tan(A+B)=1 \), \( A+B=45^\circ \).

Teacher's Note:
1. Use tan(A + B) = (tan A + tan B)/(1 − tan A tan B).
2. tan 45° = 1.

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Practice MCQs for Class 10 Mathematics Chapter 08 Introduction to Trigonometry

About Chapter 08 Introduction to Trigonometry MCQs for Class 10 Mathematics

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FAQs

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