Question

Prove that
Cosec^6 A =Cot^6A+3Cot^2ACosec^2A+1​

Answer 1

Step-by-step explanation:

cosec^6 x - cot^6x

= (cosec^2 x)^3- (cot ^2 x)^3

= (cosec^2 x - cot^2 x)(cosec^4 x + cosec^2 x cot ^2 x + cot^4 x)

= (cosec^4 x + cosec^2 x cot ^2 x + cot^4 x) as (cosec^2 x - cot^2 x) = 1

= (cot^2 x +1 )^2 + (cot^2 x+1) cot^2 x + cot^ 4x

= cot ^4 x + 2 cot^2 x + 1 + cot^4 x + cot^2 x + cot^4 x

= 1+ 3 cot^2 x + cot^4 x

Answer 2

QUESTION:

Prove that Cosec⁶ θ = Cot⁶ θ + 3 Cot² θ Cosec² θ + 1.

GIVEN:

• Cosec⁶ θ = Cot⁶ θ + 3 Cot² θ Cosec² θ + 1.

TO PROVE:

• Cosec⁶ θ = Cot⁶ θ + 3 Cot² θ Cosec² θ + 1.

PROOF:

Take Cosec⁶ θ as L.H.S.

Take Cot⁶ θ + 3 Cot² θ Cosec² θ + 1 as R.H.S.

L.H.S:

Cosec⁶ θ = (Cosec² θ)³

$\boxed{\bold{ \large{\gray{cosec^2 \ \theta - cot^2 \ \theta= 1}}}}$

$\boxed{\bold{ \large{\gray{cosec^2 \ \theta= 1 + cot^2 \ \theta}}}}$

Cosec⁶ θ = (1 + Cot² θ)³

$\boxed{\bold{ \large{\gray{(A+B)^3=A^3+B^3+3AB(A+B)}}}}$

Cosec⁶ θ = 1³ + (Cot² θ)³ + 3(1)(Cot² θ)(1 + Cot² θ)

Cosec⁶ θ = 1 + Cot⁶ θ + 3Cot² θ(1 + Cot² θ)

$\boxed{\bold{ \large{\gray{cosec^2 \ \theta= 1 + cot^2 \ \theta}}}}$

Cosec⁶ θ = 1 + Cot⁶ θ + 3Cot² θ(Cosec² θ)

Cosec⁶ θ = 1 + Cot⁶ θ + 3 Cot² θ Cosec² θ

Cosec⁶ θ = Cot⁶ θ + 3 Cot² θ Cosec² θ + 1

R.H.S:

Cot⁶ θ + 3 Cot² θ Cosec² θ + 1

L.H.S = R.H.S

Cosec⁶ θ = Cot⁶ θ + 3 Cot² θ Cosec² θ + 1

HENCE PROVED.

VERIFICATION:

Cosec⁶ θ = Cot⁶ θ + 3 Cot² θ Cosec² θ + 1

Substitute θ = 45°

Cosec⁶ 45° = Cot⁶ 45° + 3 Cot² 45° Cosec² 45° + 1

(√2)⁶ = 1 + 3(1)²(√2)² + 1

(√2)⁶ = (√2 ×√2 × √2 × √2 × √2 × √2)

(√2)⁶ = (2 × 2 × 2)

(√2)⁶ = 8

8 = 1 + 3(2) + 1

8 = 1 + 6 + 1

8 = 8

HENCE VERIFIED.