Important Questions Class 10 Maths Chapter 8 Introduction to Trigonometry

Trigonometry is a Greek terminology that is made up of three words, “Tri Gon Metron” which means Three Sides Measure. It is believed that this branch of mathematics developed near 3 BC when Greek mathematicians were learning to study astronomy by applying geometry. However, while unearthing the history of mathematics it was found that the Indian mathematician Aryabhata in his work “Aryabhatiyam” created the table for trigonometric ratios of trigonometric functions far earlier than the Greeks. It is the study of the correlation of sides and angles of a triangle. It is vastly used in the field of Astronomy, Navigation, Surveying, Optics, etc.

The important question for class 10 maths Chapter 8 Introduction to Trigonometry will focus on the entities like :

• Trigonometric ratios of specific angles
• Trigonometric identities
• The trigonometric ratio of complementary angles

Important Questions Chapter 8 – Introduction to Trigonometry

Q1. If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.

Solution:

As we know, for any given triangle, the sum of all its interior angles is equals to 180°.

Thus,

A + B + C = 180° ….(1)

Now we can write the above equation as;

⇒ B + C = 180° – A

Dividing by 2 on both the sides;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Now, put sin function on both sides.

⇒ sin (B + C)/2 = sin (90° – A/2)

Since,

sin (90° – A/2) = cos A/2

Therefore,

sin (B + C)/2 = cos A/2

Q2. If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.

Solution:

Given,

sin θ + cos θ = √3

Squaring on both sides,

(sin θ + cos θ)^2 = (√3)^2

sin^2θ + cos^2θ + 2 sin θ cos θ = 3

Using the identity sin^2A + cos^2A = 1,

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 3 – 1

2 sin θ cos θ = 2

sin θ cos θ = 1

sin θ cos θ = sin^2θ + cos^2θ

⇒ (sin^2θ + cos^2θ)/(sin θ cos θ) = 1

⇒ [sin^2θ/(sin θ cos θ)] + [cos^2θ/(sin θ cos θ)] = 1

⇒ (sin θ/cos θ) + (cos θ/sin θ) = 1

⇒ tan θ + cot θ = 1

Hence proved.

Q3. What is the value of (cos^2 67° – sin^2 23°)?

Solution:

(cos^2 67° – sin^2 23°) = cos^2(90° – 23°) – sin^2 23°

We know that cos(90° – A) = sin A

= sin^2 23° – sin^2 23°

= 0

Therefore, (cos^2 67° – sin^2 23°) = 1.

Q4. Prove that (sin A – 2 sin^3 A)/(2 cos^3 A – cos A) = tan A.

Solution:

LHS = (sin A – 2 sin^3 A)/(2 cos^3 A – cos A)

= [sin A(1 – 2 sin^2 A)]/ [cos A(2 cos^2 A – 1]

Using the identity sin^2 θ + cos^2 θ = 1,

= [sin A(sin^2 A + cos^2 A – 2 sin^2A)]/ [cos A(2 cos^2 A – sin^2 A – cos^2 A]

= [sin A(cos^2 A – sin^2 A)]/ [cos A(cos^2 A – sin^2 A)]

= sin A/cos A

= tan A

= RHS

Hence proved.

Q5. Evaluate 2 tan^2 45° + cos^2 30° – sin^2 60°.

Solution: 

Since we know,

tan 45° = 1

cos 30° = √3/2

sin 60° = √3/2

Therefore, putting these values in the given equation:

2(1)^2 + (√3/2)^2 – (√3/2)^2

= 2 + 0

= 2

Q6. If tan θ + cot θ = 5, find the value of tan2θ + cotθ

Solution:

tan θ + cot θ = 5 … [Given

tan^2θ + cot^2θ + 2 tan θ cot θ = 25 … [Squaring both sides

tan^2θ + cot^2θ + 2 = 25

∴ tan^2θ + cot^2θ = 23

Q7. If sec θ + tan θ = 7, then evaluate sec θ – tan θ

Solution:

We know that,

sec^2θ – tan^2θ = 1

(sec θ + tan θ) (sec θ – tan θ) = 1

(7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7; (Given)

∴ sec θ – tan θ = 1/7

Q8. Evaluate: sin^2 19° + sin^2 71°

Solution:

sin^2 19° + sin^2 71°

= sin^219° + sin^2 (90° – 19°)…[∵ sin(90° – θ) = cos θ

= sin^2 19° + cos^2 19° = 1 …[∵ sin^2 θ + cos^2 θ = 1

Q9. Express cot 75° + cosec 75° in terms of trigonometric ratios of angles between 0° and 30°. 

Solution:

cot 75° + cosec 75°

= cot(90° – 15°) + cosec(90° – 15°)

= tan 15° + sec 15° …[cot(90°-A) = tan A

cosec(90° – A) = sec A

Q10. If cos x = cos 40° . sin 50° + sin 40°. cos 50°, then find the value of x

Solution:

cos x = cos 40° sin 50° + sin 40° cos 50°

cos x = cos 40° sin(90° – 40°) + sin 40°.cos(90° – 40°)

cos x = cos^2 40° + sin^2 40°

cos x = 1 …[∵ cos^2 A + sin^2 A = 1

cos x = cos 0° ⇒ x = 0°