Question

prove that sec6 theta=tan6 theta+3tan2theta.sec2 theta+1​

Answer 1

Step-by-step explanation:

Proof:-

We have to prove sec⁶θ = tan⁶θ +3tan²θ sec²θ+1

On taking LHS :

sec⁶ θ

It can be written as

=> (sec²θ)³

We know that

sec² A - tan² A = 1

=> sec² A = 1+tan² A

now,

(sec² θ)³

=> (1+tan² θ)³

It is in the form of (a+b)³

Where, a = 1 and b = tan² θ

We know that

(a+b)³ = a³+b³+3ab(a+b)

Now, (1+tan² θ)³

=>1³+(tan² θ)³+3(1)(tan² θ)(1+tan² θ)

=> 1+tan⁶ θ+3tan²θ(1+tan²θ)

=> 1+tan⁶θ +3tan²θ sec²θ

=> RHS

=> LHS = RHS

Therefore,

sec⁶θ = tan⁶θ +3tan²θ sec²θ+1

Hence, Proved.

Used Identities:-

→sec² A - tan² A = 1

→ (a+b)³ = a³+b³+3ab(a+b)

Answer 2

Question:-

Prove that: sec⁶θ = tan⁶θ + 3tan²θ sec²θ + 1.

Given:-

• sec⁶θ = tan⁶θ + 3tan²θ sec²θ + 1.

To Prove:-

• sec⁶θ = tan⁶θ + 3tan²θ sec²θ + 1.

Solution:-

From Trigonometric Identities we have, 

sec²θ − tan²θ = 1.

On cubing both sides, 

(sec²θ − tan²θ)³ = 1.

sec⁶θ − tan⁶θ − 3sec²θ tan²θ(sec²θ − tan²θ) = 1.

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

sec⁶θ − tan⁶θ − 3sec²θ tan²θ = 1.

⇒ sec⁶θ = tan⁶θ + 3sec²θ tan²θ + 1.

Answer:-

⇒ sec⁶θ = tan⁶θ + 3sec²θ tan²θ + 1.

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

Hence, Proved.

Hope you have satisfied. ⚘