Question

(sec A)^6 - (tan A)^6 = 1 + 3tan² A + 3tan⁴ A
Prove this equation.​

Answer 1

LHS

(sec A)^6 - (tan A)^6

(sec^2A)^3 - (tan^2A)^3

(sec^2A - Tan^2A)^3 + 3sec^2Atan^2A (sec^2A - tan^2A)

1 +3 (1+tan^2A)tan^2A

1 + 3tan^2A + 3tan^4A

Hence

LHS = RHS

proved

Answer 2

Hey mate here is your answer.

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Step-by-step explanation:

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sec^6 A-tan^6 A =1+3 tan^2 A+3 tan^4 A,

We know that a^3-b^3=(a-b)(a^2+ab+b^2),

∴sec^6 A-tan^6 A=(sec^2 A-tan^2 A)[(sec^2 A)^2+sec^2 A tan^2 A+(tan^2 A)^2],

sec^6 A-tan^6 A:

[(1+tan^2 A)-tan^2 A][(1+tan^2 A)^2+(1+tan^2 A)tan^2 A+tan^4 A]  {∵sec^2A+1+tan^2A},

sec^6 A-tan^6 A:

(1+tan^2 A-tan^2 A)(1+tan^4 A+2tan^2 A+tan^2 A+tan^4 A+tan^4 A),

sec^6 A-tan^6 A=1+3tan^2 A+3tan^4 A.

L.H.S=R.H.S,proved.

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Hope it helps you.✔✔✔

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