Kerala SSLC · Mathematics · Class 10

Trigonometry: Important Questions with Answers

These are 10 important multiple-choice questions from the Kerala SSLC Mathematics chapter “Trigonometry” (Kerala SCERT syllabus). Each one shows the correct answer and a short explanation of why it is right. Try answering before you read the answer. ExamSummary has 20 practice questions on this chapter in total.

1. Q1.If sin θ = 3/5 and θ is an acute angle, then the value of cos θ is:

   • A)3/5
   • B)4/5
   • C)5/3
   • D)5/4

   Answer: B) 4/5

   Using the identity sin²θ + cos²θ = 1, we substitute sin θ = 3/5 to get cos²θ = 1 - 9/25 = 16/25. Taking the square root gives cos θ = 4/5 since θ is acute.

2. Q2.The value of sin(60°)cos(30°) + cos(60°)sin(30°) is equal to:

   • A)1
   • B)1/2
   • C)√3/2
   • D)0

   Answer: A) 1

   This expression follows the formula sin(A+B) = sin A cos B + cos A sin B. Here A=60° and B=30°, so the sum is sin(90°), which equals 1.

3. Q3.If tan θ = 1, then the value of θ is:

   • A)30°
   • B)45°
   • C)60°
   • D)90°

   Answer: B) 45°

   The tangent of an angle is 1 when the opposite side equals the adjacent side in a right-angled isosceles triangle, which corresponds to the standard angle 45°.

4. Q4.Simplify the expression (1 - sin²θ) / cos²θ.

   • A)1
   • B)0
   • C)tan²θ
   • D)sec²θ

   Answer: A) 1

   Since 1 - sin²θ is equivalent to cos²θ based on the fundamental identity, substituting this into the numerator results in cos²θ / cos²θ, which simplifies to 1.

5. Q5.Which of the following statements is NOT true?

   • A)sin²θ + cos²θ = 1
   • B)1 + tan²θ = sec²θ
   • C)sin θ = 2
   • D)cosec θ = 1/sin θ

   Answer: C) sin θ = 2

   The sine function ranges between -1 and 1 for real angles, so sin θ cannot equal 2. The other three options are valid trigonometric identities.

6. Q6.In a right-angled triangle ABC, right-angled at B, if AB = BC, then angle C is:

   • A)30°
   • B)45°
   • C)60°
   • D)90°

   Answer: B) 45°

   Since AB = BC, the triangle is isosceles. In a right-angled triangle, the two acute angles must sum to 90°. Therefore, each acute angle is 90°/2 = 45°.

7. Q7.If sec θ = 5/4, then the value of tan θ is:

   • A)3/4
   • B)4/3
   • C)3/5
   • D)5/3

   Answer: A) 3/4

   Using the identity sec²θ - tan²θ = 1, we find tan²θ = (25/16) - 1 = 9/16. Taking the square root gives tan θ = 3/4.

8. Q8.If sin(A + B) = 1 and cos(A - B) = 1, where 0 < A + B ≤ 90°, then A and B are:

   • A)A = 30°, B = 60°
   • B)A = 45°, B = 45°
   • C)A = 60°, B = 30°
   • D)A = 90°, B = 0°

   Answer: B) A = 45°, B = 45°

   sin(A+B)=1 implies A+B=90°, and cos(A-B)=1 implies A-B=0°. Solving these simultaneous equations yields A = 45° and B = 45°.

9. Q9.The maximum value of sin θ is:

   • A)0
   • B)1
   • C)2
   • D)Infinity

   Answer: B) 1

   The sine of an angle represents the ratio of the opposite side to the hypotenuse in a right triangle. Since the hypotenuse is always the longest side, this ratio cannot exceed 1.

10. Q10.Simplify: cos θ / (1 + sin θ) + (1 + sin θ) / cos θ.

    • A)2 sec θ
    • B)2 cosec θ
    • C)1
    • D)0

    Answer: A) 2 sec θ

    Combining the fractions over a common denominator gives [cos²θ + (1+sinθ)²] / [cosθ(1+sinθ)]. Using identities, the numerator becomes 2(1+sinθ), which cancels with the denominator to leave 2/cosθ or 2 sec θ.

All Kerala SSLC Mathematics chapters →