There.

Only one. And clear.

tan^2t/sect = sect - cost

tan^2(T)/secT = secT -cosT

We develop the Lefthand Side (LHS) while the RHS stays.

Convert the tan and sec into sin and/or cos,

LHS = tan^2(T)/secT

= [(sinT / cosT)^2] / [1 / cosT]

= [sin^2(T) / cos^2(T)]*[cosT / 1]

= sin^2(T) / cosT

Since sin^2(T) +cos^2(T) = 1, then,

= [1 -cos^2(T)] / cosT

= 1/cosT -cos^2(T)/cosT

= secT -cosT

And that is the same as the RHS.

Therefore, proven.