Question

sec^6x-tan^6x=1+3tan^2x+3tan^4x

Answer 1

LHS= (sec^2)^3-(tan^2)^3
= (sec^2-tan^2)^3+3sec^2*tan^2 (sec^2-tan^2)
= (1)^3+3 (1+tan^2x)tan^2x (1)
(sec^2x-tan^2x)=1
=1+3 (tan^2x+tan^4x)
=1+3tan^2x+3tan^4x
=RHS

Answer 2

Answer:

LHS= (sec^2)^3-(tan^2)^3

= (sec^2-tan^2)^3+3sec^2*tan^2 (sec^2-tan^2)

= (1)^3+3 (1+tan^2x)tan^2x (1)

(sec^2x-tan^2x)=1

=1+3 (tan^2x+tan^4x)

=1+3tan^2x+3tan^4x

=RHStep-by-step explanation: