Class 10 Maths Chapter 8 Question Answers - Introduction to Trigonometry

Q1: It is given that tan(θ₁ + θ₂) = tan θ₁ + tan θ₂1 - tan θ₁ tan θ₂, where θ₁ and θ₂ are acute angles. Calculate θ₁ + θ₂ when tan θ₁ = 12 and tan θ₂ = 13.

Sol:  tan θ₁ = 12 and tan θ₂ = 13

∴ tan(θ₁ + θ₂) = tan θ₁ + tan θ₂1 - tan θ₁ tan θ₂

= 12 + 131 - 12 × 13

= 3 + 261 - 16

= 56 ÷ 6 - 16

= 56 × 65 = 1

Now, tan (θ₁ + θ₂) = 1 ⇒ θ₁ + θ₂ = 45°.

Q2: Prove that:   tan²A - tan²B =  sin²A - sin²Bcos²A · cos²B
Sol:  LHS = tan²A - tan²B = sin²Acos²A - sin²Bcos²B

= cos²B · sin²A - sin²B · cos²Acos²A · cos²B

= sin²A · (1 - sin²B) - sin²B · (1 - sin²A)cos²A · cos²B

= sin²A - sin²A · sin²B - sin²B + sin²A · sin²Bcos²A · cos²B

using tanθ=sinθ/cosθ and cos²θ=1-sin²θ   as sin²θ+cos²θ=1
= sin²A- sin²Bcos²A.cos²B = R.H.S

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

= 1cos²θ + 1sin²θ

= sin²θ + cos²θcos²θ · sin²θ = 1cos²θ · sin²θ

= 1cos²θ × 1sin²θ

= sec²θ × cosec²θ = R.H.S.

Q5: Prove that: 1sec x − tan x − 1cos x = 1cos x − 1sec x + tan x
Sol:  We have:

1sec x − tan x − 1cos x = 1cos x − 1sec x + tan x

Transposing the terms, we get:

1sec x − tan x + 1sec x + tan x = 1cos x + 1cos x = 2cos x

Now, L.H.S:

1sec x − tan x + 1sec x + tan x = sec x + tan x + sec x − tan x(sec x − tan x)(sec x + tan x)

= sec x + sec xsec² x − tan² x = 2 sec x1 [∵ sec² x − tan² x = 1]

= 2cos x ∴ sec x = 1cos x = R.H.S.

Q6: Show that: cos³θ + sin³θcos θ + sin θ + cos³θ − sin³θcos θ − sin θ = 2
Sol: L.H.S. =  cos³θ + sin³θcos θ + sin θ + cos³θ − sin³θcos θ − sin θ 

= (cos θ + sin θ)(cos²θ − cos θ sin θ + sin²θ)cos θ + sin θ + (cos θ − sin θ)(cos²θ + cos θ sin θ + sin²θ)cos θ − sin θ

= (cos θ + sin θ)(1 − cos θ sin θ)cos θ + sin θ + (cos θ − sin θ)(1 + cos θ sin θ)cos θ − sin θ

= (1 − cos θ sin θ) + (1 + cos θ sin θ)

= 1 − cos θ sin θ + 1 + cos θ sin θ

= 2 + 0 = 2 = R.H.S.

Q7: Prove that:  tan A + tan Bcot A + cot B = tan A × tan B

Sol: 

tan A + tan Bcot A + cot B = sin Acos A + sin Bcos Bcos Asin A + cos Bsin B

= sin A cos B + sin B cos Acos A cos B ÷ sin B cos A + sin A cos Bsin A sin B

=sin A sin Bcos A cos B

=sin Acos A × sin Bcos B = tan A × tan B = R.H.S.

Q8:  Prove that: tan²A - tan²B = sin² A - sin² Bcos² A cos² B

Sol: L.H.S = sin² Acos² A - sin² Bcos² B

= sin² A cos² B - sin² B cos² Acos² A cos² B

= sin² A (1 - sin² B) - sin² B (1 - sin² A)cos² A cos² B

= sin² A - sin² A sin² B - sin² B + sin² A sin² Bcos² A cos² B

= sin² A - sin² Bcos² A cos² B = R.H.S.

Q9:  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

Q10:  Prove that: a² + b² = x² + y² when a cos θ − b sin θ = x and a sin θ + b cos θ =y.

Sol:  R.H.S. = x² + y²

= [a cos θ - b sin θ]² + [a sin θ + b cos θ]²

= a² cos² θ + b² sin² θ - 2ab sin θ cos θ + a² sin² θ + b² cos² θ + 2ab sin θ cos θ

= a² cos² θ + b² sin² θ + a² sin² θ + b² cos² θ

= a² [cos² θ + sin² θ] + b² [sin² θ + cos² θ]

= a² [1] + b² [1]   [∵ sin² θ + cos² θ = 1]

= a² + b²

= L.H.S.

Q11: Prove that: √sec θ - 1√sec θ + 1 + √sec θ + 1√sec θ - 1 = 2 cosec θ

Sol: 

= √sec θ - 1 + √sec θ + 1√sec θ + 1 × √sec θ - 1

= (√sec θ - 1)(√sec θ - 1) + (√sec θ + 1)(√sec θ + 1)√sec θ + 1 × √sec θ - 1

= (sec θ - 1)² + (sec θ + 1)²√sec θ + 1 × √sec θ - 1

= sec θ - 1 + sec θ + 1√sec² θ - 1²

= 2 sec θ√tan² θ = 2 sec θtan θ

= 2cos θ × cos θsin θ = 2sin θ

= 2 cosec θ = R.H.S.

Q12:  Prove that: 1 + cos Asin A + sin A1 + cos A = 2 cosec A.

Sol: L.H.S. = 1 + cos Asin A + sin A1 + cos A

= (1 + cos A)(1 + cos A) + sin A × sin Asin A(1 + cos A)

= (1 + cos A)² + (sin A)²sin A(1 + cos A)

= 1 + cos² A + 2 cos A + sin² Asin A(1 + cos A)

= 1 + (cos² A + sin² A) + 2 cos Asin A(1 + cos A)

= 2 + 2 cos Asin A(1 + cos A)

= 2(1 + cos A)sin A(1 + cos A) = 2sin A

= 2 cosec A = R.H.S.

Q13: Prove that: sin θcot θ + cosec θ = 2 + sin θcot θ - cosec θ
Sol: We have:
sin θcot θ + cosec θ = 2 + sin θcot θ - cosec θ

⇒ sin θcot θ + cosec θ - sin θcot θ - cosec θ = 2

L.H.S. = sin θcot θ + cosec θ - sin θcot θ - cosec θ

= sin θ (cot θ - cosec θ) - sin θ (cot θ + cosec θ)(cot θ + cosec θ)(cot θ - cosec θ)

= sin θ cot θ - sin θ cosec θ - sin θ cot θ - sin θ cosec θcot² θ - cosec² θ

= -2 sin θ cosec θ-1 | ∵ cot² θ - cosec² θ = -1

= 2 × sin θ1 x  1sin θ = 2

= R.H.S.

Q14: Show that: 1 + sec θ - tan θ1 + sec θ + tan θ = 1 - sin θcos θ
Sol: L.H.S. = (1) + sec θ - tan θ1 + sec θ + tan θ

= (sec² θ - tan² θ) + sec θ - tan θ1 + sec θ + tan θ  ∵ [1 - sec² θ = tan² θ]

= (sec θ - tan θ)(sec θ + tan θ) + (sec θ - tan θ)1 + sec θ + tan θ  ∵[ a² - b² = (a + b)(a - b)]

= (sec θ - tan θ)((sin θ + tan θ) + 1)sec θ + tan θ + 1

= sin θ - tan θsec θ + tan θ = 1cos θ × sin θcos θ

= 1 - sin θcos θ = R.H.S.

Q15: For an acute angle θ, show that: (sin θ − cosec θ) (cos θ − sec θ) = 1tan θ + cot θ.

Sol:  L.H.S. = (sin θ − 1sin θ) (cos θ − 1cos θ)

= (sin² θ − 1)sin θ (cos² θ − 1)cos θ

= −1(1 − sin² θ) × (−1)(1 − cos² θ)sin θ cos θ

= (−cos² θ) × (−sin² θ)sin θ × cos θ

= sin θ × cos θ = cos θ × sin θ1

= cos θ × sin θsin θ × cos θ = cos θ × sin θsin² θ + cos² θ | ∵ 1 = sin² θ + cos² θ

= sin θ × cos θsin θ × cos θ = 1tan θ + cot θ.

= R.H.S.

Q16:  Show that: tan A1 + sec A − tan A1 − sec A = 2 cosec A.
Sol:  L.H.S. = tan A1 + sec A − tan A1 − sec A

= tan A1 + sec A − tan A1 − sec A = tan A [(1 − sec A) − (1 + sec A)(1 + sec A)(1 − sec A)]

= tan A − 2 sec A(1 + sec A)(1 − sec A) =  tan A [−2 sec A1 − tan² A]

= 2 sec Atan A = 2 1cos A × cos Asin A

= 2sin A = 2 cosec A.

= R.H.S.

Q17: 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°

⇒ A = 105°2 = 52.5°

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

Q18: Simplify: sin³θ + cos³θsinθ + cosθ + sinθ cosθ

Sol:  We have:

sin³θ + cos³θsinθ + cosθ + sinθ cosθ

= (sinθ + cosθ)(sin²θ + cos²θ − sinθ cosθ)(sinθ + cosθ) + sinθ cosθ

= (1 − sinθ cosθ) + sinθ cosθ

[using a³ + b³ = (a + b)(a² + b² − ab)]

= (1 − sinθ cosθ) + sinθ cosθ

= 1 − sinθ cosθ + sinθ cosθ = 1 − 0 = 1

Q19: If cot θ = 158, then evaluate: (2 + 2 sin θ)(1 − sin θ)(1 + cos θ)(2 − 2 cos θ)

Sol: 

(2 + 2 sin θ)(1 − sin θ)(1 + cos θ)(2 − 2 cos θ) = 2(1 + sin θ)(1 − sin θ)(1 + cos θ) 2(1 − cos θ)

= 2(1 − sin² θ)2(1 − cos² θ) = cos² θsin² θ

= cot² θ

Since cot θ = cos θsin θ : cot² θ = 15²8² = 22564

Q20: 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

= 12 BC = 12 × 2a = a

Now, using Pythagoras theorem, in right ΔABD,

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

⇒ AD = √3a

∴ sin 60° = ADAB = √3a2a = √32

Thus, sin 60° = √32