Question

Prove that : sec6 x - tan6 x - 3sec2 x. tan2 x = 1

Answer 1

Refer to the attachment

Answer 2

Step-by-step explanation:

Solution :-

On taking LHS :

Sec⁶X - Tan⁶X -3 Sec² X . Tan² X

=> (Sec² X)³ - ( Tan² X)³ - 3 Sec² X . Tan² X

We know that

a³-b³ = (a-b)(a²+ab+b²)

Where a = Sec² X and b = Tan² X

=>(Sec²X-Tan²X) (Sec⁴X+Sec²XTan²X+Tan⁴X)-3Sec²XTan²X

= (1)(Sec⁴X+Sec²XTan²X+Tan⁴X)

-3Sec²XTan²X

Since, Sec² A - Tan² A = 1

=Sec⁴X+Sec²XTan²X+Tan⁴X-3Sec²XTan²X

=> Sec⁴ X -2 Sec²X Tan²X + Tan⁴ X

=> (Sec² X)² -2 Sec²X Tan²X + (Tan² X)²

It is in the form of a²-2ab+b²

Where, a = Sec² X and b = Tan² X

We know that

(a-b)² = a²-2ab+b²

=> (Sec² X - Tan² X)²

=> (1)²

=> 1

=> RHS

=> LHS = RHS

Hence, Proved.

Answer:-

Sec⁶X - Tan⁶X -3 Sec² X . Tan² X = 1

Used formulae:-

→ a³-b³ = (a-b)(a²+ab+b²)

→ (a-b)² = a²-2ab+b²

→ , Sec² A - Tan² A = 1