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: ∵ sin² θ + 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)
Adding (1) and (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]
∠ADB = ∠ADC = 90°    [Construction]
AD = AD [Construction]
⇒ ΔABD ≅ ΔACD
⇒ BD = CD

Now, using Pythagoras theorem, in right ΔABD,

AD² = AB² − BD²
= (2a)² − a²
= 4a² − a²
= 3a²

⇒

∴  

Thus,