Prove:sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1

1.	Prove:sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Answer» From trig. Identities we have, sec² θ − tan² θ = 1 On cubing both sides, (sec²θ − tan²θ)³ = 1 sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ(sec² θ − tan² θ)= 1 [Since, (a – b)³ = a³ – b³ – 3ab(a – b)] sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ = 1 ⇒ sec⁶ θ = tan⁶ θ + 3sec² θ tan² θ + 1 Hence, L.H.S = R.H.S Hence proved

Answer»

From trig. Identities we have,

sec² θ − tan² θ = 1

On cubing both sides,

(sec²θ − tan²θ)³ = 1

sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ(sec² θ − tan² θ)= 1

[Since, (a – b)³ = a³ – b³ – 3ab(a – b)]

sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ = 1

⇒ sec⁶ θ = tan⁶ θ + 3sec² θ tan² θ + 1

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

Hence proved

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