How to write this in perfect square form?

Question

I got $e^{-2x}\sec^2-2e^{-2x}\tan x$ I would need to put them in a form $$e^{-2x}(a+b\tan x)^2$$ how do I do it? I got to the point where I need to complete the square $\sec^2 x-2\tan x$ but how to do that?

Answer 1

Hint:

$$ \sec^2x-2\tan x= \frac{1}{\cos^2 x}-2\tan x= \frac{\sin^2 x+\cos^2 x}{\cos^2 x}-2\tan x=$$ $$=\tan^2 x+1 - 2\tan x = (1-\tan x)^2 $$

Answer 2

Use $\sec^2x = \tan^2 x + 1$.

Then we have $$ e^{-2x}\left( \tan^2 x + 1 -2\tan x \right) = e^{-2x}\left( \tan^2 x -2\tan x + 1 \right) = e^{-2x} (\tan x -1)^2 $$

Answer 3

You have $e^{-2x}\sec^2x-2e^{-2x}\tan x$

This can be factorised as $e^{-2x}\left (\sec^2x-2\tan x \right)$

Note that $\cos^2 x+\sin^2x \equiv 1 \Rightarrow 1+\tan^2x \equiv \sec^2x$

That gives you $e^{-2x}\left (1+\tan^2x-2\tan x \right)=e^{-2x}\left (\tan^2x-2\tan x +1\right)$