Question

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

Answer 1

Answer with Step-by-step explanation:

Given :  

sec⁶θ = tan⁶θ + 3tan²θsec²θ + 1

[By using an identity,  sec²θ − tan²θ = 1]

On cubing  both sides of the above identity,  

(sec²θ − tan²θ)³ = 1

(sec²θ)³ −(tan²θ)³ - 3sec²θtan²θ(sec²θ − tan²θ) = 1³                                               [By using an identity, (a - b)³ = a³ − b³ = (a − b) - 3ab(a -b)]

sec⁶θ − tan⁶θ –3sec²θtan²θ = 1

[By using an identity, sec² θ - tan² θ =1]

sec⁶θ = tan⁶θ  + 3sec²θtan²θ +1

Hence Proved..

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Answer 2

According to the Question:

Refer to the attachment for step-by-step-explanation with answer.

⇒ sec³⁺³ = tan³⁺³ ∅ + 3sec² ∅ tan² ∅ + 1

Hence Proved! __________[ANSWER]