Question

sech A - tan6 A = 1 + 3 tan? A + 3 tan4​

Answer 1

sec6A−tan6A=1+3tan2A+3tan4A

We know that a3−b3=(a−b)(a2+ab+b2)

∴sec6A−tan6A=(sec2A−tan2A)((sec2A)2+sec2Atan2A+(tan2A)2)

sec6A−tan6A=((1+tan2A)−tan2A)((1+tan2A)2+(1+tan2A)tan2A+tan4A)[∵sec2A=1+tan2A]

sec6A−tan6