Trigonometric Identities

**casey_k** · November 5th 2008, 05:54 PM

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

**skeeter** · November 5th 2008, 06:19 PM

use this Pythagorean identity ...

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

**casey_k** · November 5th 2008, 07:11 PM

(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.

**tdschool** · November 5th 2008, 07:39 PM

**skeeter** · November 6th 2008, 04:00 PM

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.

**tdschool** · November 6th 2008, 04:31 PM

My method is more fun, is not?

Thread: Trigonometric Identities

LinkBack

Thread Tools

Display