Chapter 6: Trigonometric Identities Exercise – 6.1

Question: 1

(1 - cos² A) cosec² A = 1 

Solution: 

(1 - cos² A)cosec² A

 = Sin² A cosec² A

= (Sin A cosec A)²

 = (Sin A × (1/Sin A))²

 = (1)² 

= 1 

 

Question: 2

(1 + Cot² A) Sin² A = 1 

Solution: 

We know,

cosec²A - Cot ² A = 1

So, (1 + Cot² A) Sin² A

= Cosec² A Sin² A

= (Cosec A Sin A)² 

= ((1/Sin A) × Sin A)² 

= (1)² 

= 1 

 

Question: 3

tan²θ cos²θ = 1 − cos²θ 

Solution: 

We know,

 sin² θ + cos² θ = 1 

So, tan² θ cos² θ

= (tan θ × cos θ)²

= (sin θ)² sin²θ 1 - cos²θ 

 

Question: 4

Solution: 

We know,

 sin² θ + cos² θ = 1

= 1

 

Question: 5

(sec²θ − 1)(cosec²θ − 1) = 1 

Solution:

We know that, (sec²θ − tan²θ) = 1 (cosec²θ − cot²θ) = 1

So, (sec²θ - 1)(cosec²θ - 1)

= tan²θ × cot²θ

= (tan θ × cot θ)²

= 1²

= 1

 

Question: 6

Solution: 

We know that, (sec²θ − tan²θ) = 1

So,

 

Question: 7

Solution:

We know, sin²θ + cos²θ = 1

So, Multiplying both numerator and denominator by (1+ sin θ), we have

 

Question: 8

Solution: 

We know, sin²θ + cos²θ = 1 Multiplying both numerator and denominator by (1− sin θ), we have 

 

Question: 9

Solution: 

We know that, sin²A + cos²A = 1 cosec²A - cot²A = 1

= cosec θ + cot θ  

Therefore, LHS = RHS

Hence proved

 

Question: 49

(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ

Solution:

To prove, (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ

Considering LHS, we get,

= sec θ + cosec θ

Therefore, LHS = RHS Hence proved

 

Question: 50

Solution:

To prove,

Considering LHS, we get,

= 2 cosec A

Therefore, LHS = RHS Hence proved 

 

Question: 51

Solution:

Considering LHS, we get,

= 1 + cosec θ - 1                [∴   (a + b)(a - b) = a²- b²]

= cosec θ                           [ a = cosec θ, b = 1]

Therefore, LHS = RHS Hence, proved

 

Question: 52

Solution:

Therefore, LHS = RHS Hence, proved

 

Question: 53

Solution:

Therefore, LHS = RHS. Hence Proved.

 

Question: 54

sin²A cos²B - cos²A sin²B = sin²A - sin²B 

Solution:

LHS = sin²Acos²B - cos²A sin²B

= sin²A− sin²A sin² B − sin² B + sin²A sin² B 

= sin²A − sin² B

RHS Hence Proved. 

 

Question: 55

Solution:

= cot A tan B

= RHS Hence Proved.

= tan A tan B

= RHS Hence Proved.

 

Question: 56

cot²A cosec²B − cot²B cosec²A = cot²A − cot²B

Solution:

LHS = cot²A cosec²B − cot² B cosec²A

= cot²A + cot²A cot²B − cot²B − cot²B cot²A 

= cot²A - cot²B

RHS Hence Proved.

 

Question: 57

tan²A sec²B − sec²A tan²B = tan²A − tan²B

Solution:

LHS = tan²A sec²B − sec²A tan²B 

= tan²A(1 + tan²B) − sec²A(tan²A)

= tan²A + tan²A tan²B − tan²B − tan²A tan²B 

= tan²A − tan²B 

= RHS Hence Proved.

 

Question: 58

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, Prove that x² - y² = a² - b². 

Solution:

LHS = x² - y² = (a sec θ + b tan θ)² − (a tan θ + b sec θ)² 

= a²sec²θ + b²tan²θ + 2 a b sec θ tan θ − a² tan²θ − b²sec²θ – 2 ab sec θ tan θ 

= a²sec²θ + b²tan²θ − a²tan²θ − b²sec²θ 

= a²sec²θ − b²sec²θ + b²tan²θ − a²tan²θ 

= sec²θ (a² − b²) + tan²θ (b² − a²) 

= sec²θ (a² − b²) − tan²θ (a² − b²) 

= (sec²θ − tan²θ) (a² − b²) = a² - b² 

RHS Hence Proved.

 

Question: 59

If 3 sin θ + 5 cos θ = 5,

Prove that 5 sin θ – 3 cos θ = ± 3.

Solution:

Given 3 sin θ + 5 cos θ = 5

3 sin θ = 5 – 5 cos θ 

3 sin θ = 5(1 – cos θ)

3+ 3 cos θ = 5sin θ

5 sin θ - 3 cos θ = ±3

= RHS Hence Proved. 

5 sin θ – 3 cos θ = ± 3. 

 

Question: 60

If cosec θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1. 

Solution:

LHS = m n = (cosec θ + cot θ) (cosec θ – cot θ) 

= cosec²θ − cot²θ 

= 1

= RHS Hence Proved. 

 

Question: 61

If T_n = sinⁿ θ + cos_n θ, Prove that 

Solution:

= sin²θ cos²θ 

LHS = RHS

Hence proved.

 

Question: 62

Solution:

(tan θ + sec θ)² + (tan θ – sec θ)² = tan²θ + sec²θ + 2 tan θ sec θ + tan² θ + sec²θ – 2 tan θ sec θ

= 2 tan 2θ + 2 sec 2θ 

= 2[tan²θ + sec²θ]

LHS = RHS   Hence proved.

 

Question: 63

Solution:

 LHS = RHS   Hence proved.

 

Question: 64

Solution:

LHS = RHS Hence proved.

LHS = RHS  

Hence proved. 

 

Question: 65

(sec A + tan A - 1) (sec A - tan A + 1) = 2 tan A 

Solution:

= (sec A + tan A − {sec²A - tan²A}) [sec A – tan A + (sec²A − tan²A)]

= (sec A + tan A − (sec A + tan A) (sec A – tan A)) [sec A – tan A + (sec A – tan A)(sec A + tan A)]

 = (sec A + tan A) (1 − (sec A – tan A)) (sec A – tan A) (1 + (sec A + tan A)) 

= (sec²A − tan²A) (1 – sec A + tan A) (1 + sec A + tan A) 

=2 tan A

LHS = RHS  Hence proved.

 

Question: 66

(1 + cot A - cosec A)(1 + tan A + sec A) = 2 

Solution:

LHS = (1 + cot A - cosec A) (1 + tan A + sec A)

LHS = RHS Hence proved.

 

Question: 67

(cosec θ – sec θ) (cot θ – tan θ) = (cosec θ + sec θ) (sec θ cosec θ − 2) 

Solution:

LHS = (cosec θ – sec θ) (cot θ – tan θ)

RHS = (cosec θ + sec θ) (sec θ cosec θ - 2)

LHS = RHS   Hence proved. 

 

Question: 68

Solution:

= cosec A - sec A 

= RHS = LHS Hence proved.

 

Question: 69

Solution:

= LHS = RHS Hence proved.

 

Question: 70

Solution:

= sin A × cos³ A +cos A × sin³A 

= sin A cos A (cos²A + sin²A) 

= sin A cos A

LHS = RHS Hence proved.

 

Question: 71

sec⁴A (1 − sin⁴A) – 2 tan²A = 1

Solution:

Given, L.H.S = sec⁴A (1 - sin⁴A) - 2tan²A

= sec⁴A − sec⁴A × sin⁴A − 2tan⁴A

 

= sec⁴A - tan⁴A – 2 tan⁴A

= (sec²A)² - tan⁴A – 2 tan⁴A 

= (1+ tan²A)² − tan⁴A − 2tan⁴A

= 1 + tan⁴A + 2tan²A − tan⁴A − 2tan⁴A 

= 1

Hence, L.H.S = R.H.S     

 

Question: 72

Solution:

Here, sin²A + cos²A = 1

Solving, RHS ⟹

Multiplying Nr. And Dr. with (1 + Sin A)

Hence, LHS = RHS

 

Question: 73

Solution:

Given, L.H.S

= (1 + cot A + tan A) (sin A – cos A) 

⟹ sin A – cos A + cot A sin A – cot A cos A + sin A tan A – tan A cos A

⟹ sin A tan A – cos A cot A

Hence, L.H.S = R.H.S 

 

Question: 74

Solution:

 

Question: 75

If cosec θ – sin θ = a³, sec θ – cos θ = b³, Prove that a²b² (a²+ b²) = 1 

Solution:

Given, cosec θ – sin θ = a³

Here cos²θ A + sin² A = 1

Squaring on both sides

Squaring on both sides

= 1

Hence, L.H.S = R.H.S

 

Question: 76

If a cos³ θ + 3 a cos θ sin²θ = m, a sin³θ + 3 a cos²θ sin θ = n,

Solution:

Substitute the values of m and n in the above equation

⟹ a cos³θ + 3 a cos θ sin² θ + a sin³ θ + 3 a cos²θ sin^2/3 θ + a cos³ θ + 3 a cos θ sin² θ – a sin³ θ – 3 a cos² θ sin^2/3 θ

⟹ (a)^2/3 cos³θ + 3 cos θ sin² θ + sin³θ + 3 cos² θ sin^2/3θ + (a)^2/3 cos³ θ + 3 cos θ sin 2θ - sin³ θ – 3 cos² θ sin^2/3θ

⟹ (a)^2/3((cos θ + sin θ)³)^2/3 + (a)^2/3 ((cos θ – sin θ)³)^2/3

⟹ (a)^2/3[(cos θ + sin θ)²] + (a)^2/3[(cos θ – sin θ)²]

⟹ (a)^2/3((cos²θ + sin²θ + 2 sin θ cos θ)) + (a)^2/3((cos²θ + sin²θ – 2 sin θ cos θ))

⟹ (a)^2/3[ 1 + 2 sin θ cos θ] + (a)^2/3[1 – 2 sin θ cos θ]

 ⟹ (a)^2/3[1 + 2 sin θ cos θ] + 1 – 2 sin θ cos θ]

⟹ (a)^2/3(1 + 1)

⟹ 2(a)^2/3 

Hence, L.H.S = R.H.S 

 

Question: 77

If x = a cos³θ, y = b sin³θ,

Solution:

x = a cos³ θ: y = b sin³ θ 

= 1

Hence proved.

 

Question: 78

If a cos θ + b sin θ = m and a sin θ – b cos θ = n, Prove that a² + b² = m² + n² 

Solution:

R.H.S = m² + n² = m² + n²

= (a cos θ + b sin θ)² + (a sin θ – b cos θ)²

= a² cos² θ + b² sin²θ + 2ab sin θ cos θ + a² + sin²θ + b² cos²θ – 2ab sin θ cos θ

= a² cos²θ + b² cos²θ + b² sin²θ + a² sin²θ

= a²(sin² θ + cos² θ) + b²(sin² θ + cos² θ)

= a² + b² (∵ sin² θ + cos²θ = 1)

 

Question: 79

If cos A + cos²A = 1, Prove that sin²A + sin⁴A = 1 

Solution:

Given - cos A + cos² A = 1

We have to prove sin² A + sin⁴ A = 1

Now, cos A + cos² A = 1

cos A = 1 - cos² A

cos A = sin² A

sin² A = cos A

Therefore,

We have sin² A + sin⁴ A = cos A + (cos A)²

= cos A + cos² A = 1

Hence proved.

 

Question: 80

If cos θ + cos² θ = 1, Prove that sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2 = 1

Solution:

cos θ + cos² θ = 1 

cos θ = 1 − cos² θ 

cos θ = sin² θ (i)

Now, sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2

= (sin⁴ θ)³ + 3 sin⁴ θ. sin² θ (sin⁴ θ + sin² θ) + (sin² θ)³ + 2(sin² θ)² + 2 sin² θ − 2

Using (a + b)³ = a³ + b³ + 3 ab(a + b) and also from(i) cos θ = sin² θ

(sin⁴ θ + sin² θ)³ + 2cos² θ + 2 cos θ − 2 

((sin² θ)² + sin² θ)³ + 2 cos² θ + 2 cos θ – 2

(cos² θ + sin² θ)³ + 2 cos² θ + 2 cos θ − 2 

1 + 2(cos² θ + sin² θ) − 2 

= 1 + 2(1) −2

= 1 

L.H.S = R.H.S Hence proved.

 

Question: 81

Given that: (1 + cos α)(1 + cos β)(1 + cos γ) = (1 – cos α)(1 – cos β)(1 – cos γ). Show that one of the values of each member of this equality is sin α sin β sin γ. 

Solution:

We know that 

Multiply (i) with sin α sin β sin γ and divide it with same we get

Hence sin α sin β sin γ is the member of equality.

 

Question: 82

If sin θ + cos θ = x,

Solution:

sin θ + cos θ = x

Squaring on both sides 

(sin θ + cos θ)² = x² 

⇒ sin²θ + cos²θ + 2 sin θ cos θ = x²

We know that sin² θ + cos² θ = 1 

Cubing on both sides (sin² θ + cos² θ)³ = 1³

sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ (sin² θ + cos² θ) = 1

⇒ sin⁶ θ + cos⁶ θ = 1 – 3 sin² θ cos²θ

 

Question: 83

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan ϕ, 

Solution:

Given, x = a sec θ cos ϕ, y = b sec θ sin ϕ, z = c tan ϕ squaring x, y, z on the sides

x² = a²sec²θ cos² ϕ x²/a² = sec²θ cos² ϕ    --- 1 

y² = b² sec²θ sin² ϕ y²/b² = sec²θ sin² ϕ    --- 2 

z² = c² tan² ϕ z²/c² = tan² ϕ             --- 3

Substitute equation 1, 2, 3 in x²/a² + y²/b² − z²/c² 

⟹ x²/a² + y²/b² − z²/c²

 ⟹ Sec² θ cos² ϕ + sec² θ sin² ϕ - tan² ϕ 

⟹ sec²θ(cos² ϕ + sin² ϕ) - tan² ϕ 

We know that, 

cos² ϕ + sin² ϕ = 1

⟹ sec²θ (1) - tan² ϕ and sec²θ − tan²θ = 1

⟹ 1

Hence, L.H.S= R.H.S

 

Question: 84

If sin θ + 2 cos θ. Prove that 2 sin θ – cos θ = 2 

Solution:

Given, sin θ + 2 cos θ = 1

Squaring on both sides

⟹ (sin θ + 2 cos θ)² = 1² 

⟹ sin²θ + 4 cos²θ + 4 sin θ cos θ = 1

⟹ 4 cos² θ + 4 sin θ cos θ = 1 - sin²θ 

Here, 1 - sin²θ = cos²θ 

⟹ 4 cos²θ + 4 sin θ cos θ - cos²θ = 0

⟹ 3 cos²θ + 4 sin θ cos θ = 0     ---- 1

We have, 

2 sin θ – cos θ = 2

Squaring L.H.S

 (2 sin θ – cos θ)² 

= 4 sin²θ + cos²θ – 4 sin θ cos θ 

Here, 4 sin θ cos θ = 3 cos²θ

= 4 sin²θ + cos²θ + 3 cos²θ

= 4 sin² θ + 4 cos² θ 

= 4(sin²θ + cos²θ)

= 4(1)

= 4 (2 sin θ – cos θ)² = 4

⟹ 2sin θ – cos θ 

= 2

Hence proved