Question

Prove that. Sec^6A=tan^6A + 3tan²A sec²A+1

Answer 1

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

Answer 2

Your answer is in the attachment.

HOPE IT HELPS..... :-)