CBSE Important Questions Class 10 Maths Chapter 8 – Introduction to Trigonometry

Class 10 Maths Chapter 8 Introduction to Trigonometry is part of the CBSE Class 10 Maths syllabus. It explains the six trigonometric ratios of an acute angle in a right triangle: sine, cosine, tangent, cosecant, secant and cotangent. Students learn the values of these ratios for the standard angles 0°, 30°, 45°, 60° and 90°, the relation between the ratios of complementary angles, and the three fundamental trigonometric identities.

These CBSE important questions for Class 10 Maths Chapter 8 cover finding an angle when cos(A + B) and sec B are given, evaluating expressions with standard angles, proving the identity √[(1 + sinA)/(1 − sinA)] = secA + tanA, finding the value of sinP sinR + cosP cosR, and an assertion-reason question on trigonometric identities. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. If A and B are two acute angles such that cos(A + B) = 1/2 and sec B = 2, find the value of angle A.

Answer: Given sec B = 2, so cos B = 1/2.
Since cos 60° = 1/2, we get B = 60°.
Also cos(A + B) = 1/2, so A + B = 60°.
Substituting B = 60°:
A + 60° = 60°
A = 0°

Q2. Evaluate (sin²45° + cos²45°) / tan²60°.

Answer: We know sin 45° = 1/√2, cos 45° = 1/√2 and tan 60° = √3.
Numerator = (1/√2)² + (1/√2)² = 1/2 + 1/2 = 1
(This also follows directly from the identity sin²θ + cos²θ = 1.)
Denominator = (√3)² = 3
So the value = 1/3

Q3. Prove the trigonometric identity: √[(1 + sinA) / (1 − sinA)] = secA + tanA.

Answer: Take the LHS and multiply the numerator and the denominator inside the root by (1 + sinA):
LHS = √[(1 + sinA)(1 + sinA) / (1 − sinA)(1 + sinA)]
= √[(1 + sinA)² / (1 − sin²A)]
Since 1 − sin²A = cos²A:
= √[(1 + sinA)² / cos²A]
= (1 + sinA) / cosA
= 1/cosA + sinA/cosA
= secA + tanA
= RHS. Hence proved.

Q4. In △PQR, right-angled at Q, if sin R = 1/2, then find the value of sin P sin R + cos P cos R.

Answer: Since the triangle is right-angled at Q, angles P and R are complementary: P + R = 90°.
Given sin R = 1/2, so R = 30° and therefore P = 60°.
Now substitute the values:
sin P sin R + cos P cos R
= sin 60° × sin 30° + cos 60° × cos 30°
= (√3/2)(1/2) + (1/2)(√3/2)
= √3/4 + √3/4
= √3/2

Q5. Read the statements and select the correct option.
Assertion (A): (sinθ − 2sin³θ) / (2cos³θ − cosθ) = tanθ, where θ is an acute angle.
Reason (R): For an acute angle A, tanA = sinA/cosA and sin²A + cos²A = 1.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

Answer: (a) Both A and R are true and R is the correct explanation of A
Take the LHS and factor out sinθ and cosθ:
= sinθ(1 − 2sin²θ) / cosθ(2cos²θ − 1)
Now, 1 − 2sin²θ = 1 − 2(1 − cos²θ) = 2cos²θ − 1.
So the brackets are equal and they cancel:
= sinθ/cosθ = tanθ
Both the identities in the Reason were used, so R correctly explains A.

Q6. Write the three fundamental trigonometric identities.

Answer: For any acute angle θ:
1. sin²θ + cos²θ = 1
2. 1 + tan²θ = sec²θ
3. 1 + cot²θ = cosec²θ
The second and third identities are obtained from the first by dividing throughout by cos²θ and sin²θ respectively.

Q7. In a right triangle ABC, right-angled at B, if tan A = 1/√3, find the value of sin A cos C + cos A sin C.

Answer: Since tan A = 1/√3 and tan 30° = 1/√3, we get A = 30°.
As the triangle is right-angled at B, A + C = 90°, so C = 60°.
Now:
sin A cos C + cos A sin C
= sin 30° cos 60° + cos 30° sin 60°
= (1/2)(1/2) + (√3/2)(√3/2)
= 1/4 + 3/4
= 1

Q8. Evaluate: 2 tan²45° + cos²30° − sin²60°.

Answer: Substituting the standard values: tan 45° = 1, cos 30° = √3/2 and sin 60° = √3/2.
= 2(1)² + (√3/2)² − (√3/2)²
= 2 + 3/4 − 3/4
= 2

Q9. If sin A = 3/5, find the values of cos A and tan A.

Answer: We use the identity sin²A + cos²A = 1.
cos²A = 1 − sin²A = 1 − (3/5)² = 1 − 9/25 = 16/25
cos A = 4/5 (taking the positive value, as A is acute)
Now, tan A = sin A / cos A
= (3/5) ÷ (4/5)
= 3/4
(This is the familiar 3, 4, 5 right triangle.)

Q10. Prove that (1 − cosθ) / (1 + cosθ) = (cosecθ − cotθ)².

Answer: Take the RHS and write everything in terms of sinθ and cosθ:
RHS = (1/sinθ − cosθ/sinθ)²
= [(1 − cosθ) / sinθ]²
= (1 − cosθ)² / sin²θ
Since sin²θ = 1 − cos²θ = (1 − cosθ)(1 + cosθ):
= (1 − cosθ)² / [(1 − cosθ)(1 + cosθ)]
= (1 − cosθ) / (1 + cosθ)
= LHS. Hence proved.

Q11. Write the relation between the trigonometric ratios of complementary angles. Evaluate sin 18° / cos 72°.

Answer: Two angles are complementary when they add up to 90°. For such angles:
sin(90° − A) = cos A    cos(90° − A) = sin A
tan(90° − A) = cot A    cot(90° − A) = tan A
sec(90° − A) = cosec A    cosec(90° − A) = sec A
Evaluation: Since 72° = 90° − 18°, we have cos 72° = cos(90° − 18°) = sin 18°.
So sin 18° / cos 72° = sin 18° / sin 18° = 1

Q12. Write the values of all six trigonometric ratios for the angle 30°.

Answer: For θ = 30°:
• sin 30° = 1/2
• cos 30° = √3/2
• tan 30° = 1/√3
• cosec 30° = 2
• sec 30° = 2/√3
• cot 30° = √3
Note that cosec, sec and cot are simply the reciprocals of sin, cos and tan respectively.

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