PROVE: sec^6A = tan^6A + 3tan^2A.sec^A + 1

From RHS: tan^6A + 1 + 3tan^2A sec^2A

[tan^2 A]^3 + 1^3 + 3*tan^2 A*sec^2 A

BY TRIGONOMETRIC IDENTITIES WE KNOW THAT,

sec^2 A = 1 + tan^2 A

[tan^2 A]^3 + 1^3 + 3*TAN^2 A * 1[ 1+ tan^2 A]

BY THE IDENTITIES OF [a+b]^3 = a^3 +b^3 + 3ab[a+b]

So, it is [tan^2A + 1]^3.

[sec^2A]^3 [ BY TRIGONOMETRIC IDENTITIES, i.e., sec^2A = 1 + tan^2A] ===> sec^6A [ (a^p)^q = a^p*q ]

HENCE, PROVED

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