Short Answer Questions: Introduction to Trigonometry

# Class 10 Maths Chapter 8 Question Answers - Introduction to Trigonometry

Q1: It is given that tan (θ1 + θ2) =   where θ1 and θ2 are acute angles.
Calculate θ1 + θ2 when tan θ1

Sol:

Now, tan (θ1 + θ2) = 1 ⇒ θ1 + θ2 = 45°.
Q2: Prove that:
Sol:

using tanθ=sinθ/cosθ  and cos2θ=1-sin2θ   as sin2θ+cos2θ=1

Q3: Prove that:  (sin4 θ – cos4 θ +1) cosec2 θ = 2
Sol:  L.H.S.
= (sin4 θ – cos4 θ + 1) cosec2 θ
= [(sin2 θ)2 – (cos2 θ)2 + 1] cosec2 θ             as    [a2-b2=(a-b)(a+b)]
= [(sin2 θ – cos2 θ) (sin2 θ + cos2 θ) + 1] cosec2 θ        as   [ sin2 θ + cos2 θ = 1]
= [(sin2 θ – cos2 θ) *1 + 1] cosec2 θ
= [sin2 θ – cos2 θ+1]  cosec2 θ
= [(sin2 θ + (1 –cos2 θ)] cosec2 θ   [ 1 – cos2 θ = sin2 θ]
= [sin2 θ + sin2 θ] cosec2 θ
= 2 sin2 θ . cosec2 θ
= 2 = RHS [∵ sin θ . cosec θ = 1]
Q4: Prove that: sec2 θ + cosec2 θ = sec2 θ · cosec2 θ

Sol: L.H.S. = sec2 θ + cosec2 θ

Q5:
Sol:

Q6:  Given that α + β = 90°, show that:
Sol: ∵ α + β =90°
∵ β = (90 – a)

Q7:

Sol:

using a3+b3=(a+b)(a2+b2-ab)   and a3-b3=(a-b)(a2+b2+ab) in numerator of these terms

and also sin2θ +cos2θ =1

Q8:

Sol:

using tanθ=sinθ/cosθ  and then taking LCM

Q9:

Sol:

Q10:  Prove that: sin6θ + cos6θ + 3sin2θ cos2θ = 1.

Sol: ∵ sinθ + cos2θ = 1
∵ (sin2 θ + cos2 θ)3 = (1)3 = 1
⇒ (sin2 θ) 3 + (cos2 θ)3 + 3 sin2 θ . cos2 θ (sin2 θ + cos2 θ) = 1
⇒ sin6 θ + cos6 θ + 3 sin2 θ . cos2θ (1) = 1
⇒ sin6 θ + cos6 θ + 3 sin2 θ . cos2 θ = 1

Q11:  Prove that: a2 + b2 = x2 + y2 when a cos θ − b sin θ = x and a sin θ + b cos θ =y.

Sol:

Q12:

Sol:

Q13:

Sol:

Q14:

Sol:

Q15:

Sol:

Q16: For an acute angle θ, show that: (sin θ − cosec θ) (cos θ − sec θ)
Sol:

Q17:

Sol:

Q18:

Sol:

Q19:

Sol:

Q20: Without using trigonometric tables evaluate:

Sol:

Q21: If tan (A + B) = √3 and tan (A − B) = 1, 0° < A + B < 90°; A > B, then find A and B.

Sol: We have: tan (A + B) = 3 (Given)
tan 60° = 3 (From the table)
⇒ A + B = 60° ...(1)
Also,tan (A − B)= 1 [Given]
and cosec 60° = 2 and cos 90° = 0
⇒ A − B = 45 ...(2)
2A = 60° + 45° = 105°

From (2), 52.5° − B = 45°
⇒ B = 52.5° − 45° = 7.5°
Thus, A = 52.5° and B = 7.5°.

Q22: If tan (2A) = cot (A − 21°), where 2A is an acute angle, then find the value of A.

Sol: We have: tan (2A) = cot (A − 21°)
∵ cot (90°− θ) = tan θ
∴ cot (90°− 2A) = tan 2A
⇒ cot (90°− 2A) = cot (A − 21)°
⇒ 90 − 2A = A − 21°
⇒ − 2A − A = − 21°− 90°
⇒ − 3A = − 111°

⇒

Q23: If sin 3A = cos (A − 10°), then find the value of A, where 3A is an acute angle.

Sol: We have:
sin 3A = cos (A − 10°)
∵ cos (90° − θ) = sin θ
∴ cos (90° − 3A) = sin 3A
⇒ 90° − 3A = A − 10°
⇒ −3A − A = −10° − 90°
⇒ −4A = −100°

Q24: If sec 2A = cosec (A − 27°), then find the value of A, where 2A is an acute angle.

Sol: We have:

sec 2A = cosec (A − 27°) ...(1)
∵ sec θ = cosec (90° − θ)
∴ sec 2A = cosec (90° − 2A) ...(2)

From (1) and (2), we get
A − 27° = 90° − 2A
⇒ A + 2A = 90 + 27° = 117°
⇒ 3A = 117°
⇒

Q25: Simplify:

+ sin θ cos θ

Sol: We have:

Q26:

Sol:

Q27:

Sol:

= 1 = R.H.S.

Q28: Without using trigonometrical tables, evaluate:

Sol:

[∵ sin (90° − θ) = cos θ, cos (90° − θ) = sin θ, cosec (90° − θ) = sec θ, and tan (90° − θ) = cot θ]

Q29: Using Geometry, find the value of sin 60°.

Sol: Let us consider an equilateral ΔABC and draw AD ⊥ BC.
Since, each angle of an equilateral triangle = 60°
∴∠A = ∠B = ∠C = 60°

Let AB = BC = AC = 2a

In ΔABD and ΔACD, we have:
AB = AC   [Given]
⇒ ΔABD ≅ ΔACD
⇒ BD = CD

Now, using Pythagoras theorem, in right ΔABD,

= (2a)2 − a2
= 4a2 − a2
= 3a2

∴

Thus,

The document Class 10 Maths Chapter 8 Question Answers - Introduction to Trigonometry is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 8 Question Answers - Introduction to Trigonometry

 1. What is trigonometry?
Ans. Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It involves the study of trigonometric functions such as sine, cosine, and tangent, and their applications in solving various problems related to angles and distances.
 2. How is trigonometry useful in real life?
Ans. Trigonometry has various real-life applications, such as in architecture, engineering, navigation, and physics. It is used to calculate distances, heights, angles, and trajectories in these fields. For example, trigonometry helps engineers design and construct buildings, bridges, and roads.
 3. What are the basic trigonometric ratios?
Ans. The basic trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). These ratios represent the relationships between the sides of a right triangle. Sine is the ratio of the length of the side opposite the angle to the hypotenuse, cosine is the ratio of the length of the adjacent side to the hypotenuse, and tangent is the ratio of the length of the opposite side to the adjacent side.
 4. How do you find the value of trigonometric functions?
Ans. The value of trigonometric functions can be found using a calculator or trigonometric tables. For example, to find the sine of an angle, divide the length of the side opposite the angle by the length of the hypotenuse. Similarly, to find the cosine, divide the length of the adjacent side by the length of the hypotenuse. The tangent can be found by dividing the length of the opposite side by the length of the adjacent side.
 5. What are the applications of trigonometry in solving triangles?
Ans. Trigonometry is used to solve triangles by finding the unknown sides or angles. It is particularly useful in solving right triangles, where one angle is 90 degrees. By using the trigonometric ratios and the known values of sides or angles, we can determine the missing values. This is helpful in various fields such as surveying, navigation, and physics.

## Mathematics (Maths) Class 10

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