proving Trig Identitity: (1- tanx)^2 = sec^2x - 2tanx

Hockeyman

Junior Member
Joined
Dec 8, 2005
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79
not sure how to start this one, i have tried it a few different ways and i still can't get it.

(1- tanx)^2 = sec^2x - 2tanx
 
from the right side ...

sec<sup>2</sup>x - 2tanx =

use a Pythagorean identity ...

1 + tan<sup>2</sup>x - 2tanx =

1 - 2tanx + tan<sup>2</sup>x =

factor the quadratic ...

(1 - tanx)(1 - tanx) =

(1 - tanx)<sup>2</sup>
 
Hello, Hockeyman!

Going the other way . . .


\(\displaystyle (1\,-\,\tan x)^2 \:= \:\sec^2x\,-\,2\cdot\tan x\)

We have: \(\displaystyle \:(1\,-\,\tan x)^2\;=\;1\,-\,2\cdot\tan x\,+\,\tan^2x \;=\;\underbrace{1\,+\,\tan^2x}\,-\,2\cdot\tan x\)
. . . . . . . . . Since \(\displaystyle 1\,+\,\tan^2x\:=\:\sec^2x\), we have:. . . \(\displaystyle \sec^2x\,-\,2\cdot\tan x\)

 
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