# How do you prove (1-tanx)^2=sec^2x-2tanx?

Use trig identities: $1 + {\tan}^{2} x = \frac{1}{\cos} ^ 2 x = {\sec}^{2} x$
$\left(1 - \tan x\right) = 1 + {\tan}^{2} x - 2 \tan x$ = ${\sec}^{2} x - 2 \tan x$