Question

sec^6A= tan^6 A + 3 tan^2A sec^2A + 1​

Answer 1

To prove :-

sec⁶A = tan⁶ A + 3 tan²A.sec²A + 1​

Proof :-

RHS :

tan⁶A + 3 tan²A.sec²A + 1​

= tan⁶A + 3 tan²A.(tan²A + 1) + 1    [sec²A = tan²A + 1]

= (tan²A)³ + 3 tan²A.(tan²A + 1) + (1)³

[This is of the form x³ + 3xy(x + y) + y³]

= (tan²A + 1)³

[As x³ + 3xy(x + y) + y³ = (x + y)³]

= (sec²A)³

= sec⁶A = LHS

Proved

Answer 2

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$\huge\rm\underline\purple{Question :-}$

sec^6A= tan^6 A + 3 tan^2A sec^2A + 1

$\huge\rm\underline\purple{Solution :-}$

Here using formula,

a^6 - b^6 = ( a^2 - b^2 )^3 + 3a^2b^2(a^2+b^2)

let a=secA & tanA=b,

by substituting value of a & b , you will get the equation given in the question .

sec^6A - tan^6A = (sec^2A - tan^2A)^3 +3tan^2Asec^2A(sec^2A - tan^2A)

here we know that sec^2A-tan^2A=1

than we will get,

sec^6A=tan^6A +3tan^2Asec^2A + 1

PROVED

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