1. Trigonometric Identities

I'm having trouble proving this equation.
Prove:
(1+ tan x)^2 - sec^2 x = 2 tan x

2. use this Pythagorean identity ...

$1+\tan^2{x} = \sec^2{x}$

3. Reply: I'm still a little stuck

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

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

= 1 + 2 tan x + tan^2 x + 1 - sec^2 x

= 1 + (2 sin x/cos x) + (sin^2 x/cos^2 x) - (1/cos^2 x)

= 1 + (2 sin x/cos x) + (sin^2 x - 1/cos^2 x)

= 1 + (2 sin x/cos x) + (sin^2 x - 1/sin^2 x - 1)

i cancel out the sin^2 x - 1 and i`m left with

1 + (2 sin x/cos x) which is close to 2 tan x

but how do i get rid of the one? any help will be appreciated.

4. Originally Posted by casey_k
I'm having trouble proving this equation.
Prove:
(1+ tan x)^2 - sec^2 x = 2 tan x
this may be a less arduous method ...

$(1 + 2\tan{x} + \tan^2{x}) - (1 + \tan^2{x}) =$

$1 + 2\tan{x} + \tan^2{x} - 1 - \tan^2{x}) = 2\tan{x}$

you're done.

5. My method is more fun, is not?