Question

Prove the following trigonometric identities. $cosec^{6}\Theta =cot^{6}\Theta +3cot^{2}\Theta cosec^{2}\Theta+1$

Answer 1

Answer with Step-by-step explanation:

Given :  

cosec⁶θ = cot⁶θ + 3cot²θ cosec²θ + 1

[By using an identity,  cosec²θ − cot²θ = 1]

On cubing  both sides of the above identity,  

(cosec²θ − cot²θ)³ = 1³

(cosec²θ)³ −(cot²θ)³ - 3cosec²θcot²θ(cosec²θ − cot²θ) = 1              

[By using an identity, (a - b)³ = a³ − b³ = (a − b) - 3ab(a -b)]

cosec⁶θ − cot⁶θ –3cosec²θcot²θ = 1

[By using an identity, cosec² θ - cot² θ = 1]

cosec⁶θ = cot⁶θ + 3cosec²θcot²θ + 1

Hence Proved..

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

According to the Question:

» cosec⁶θ = cot⁶θ + 3cot²θ cosec²θ + 1

Note: Using identity - cosec²θ − cot²θ = 1

Now on cubing the above extracted identity, we have :-

» cosec²θ − cot²θ)³ = 1³

» (cosec²θ)³ −(cot²θ)³ - 3cosec²θcot²θ(cosec²θ − cot²θ) = 1

Note: Using identity - (a - b)³ = a³ − b³ = (a − b) - 3ab(a -b)

» cosec⁶θ − cot⁶θ –3cosec²θcot²θ = 1

Note: Using identity - cosec² θ - cot² θ = 1

We have,

» cosec⁶θ = cot⁶θ + 3cosec²θcot²θ + 1

_______________[VERIFIED]