1. ## prove the identity

i have to prove tan 2x - sec 2x = tan (x - pi/4)

well i started with the right side and used identities to simplify it down to tan x - 1/(1 + tan x)

im not sure where to go from here? any suggestions?

2. Hi

$\displaystyle \tan 2x - \sec 2x = \frac{\sin 2x}{\cos 2x} - \frac{1}{\cos 2x} = \frac{2 \sin x \cos x - 1}{2 \cos^2x - 1}$

$\displaystyle \tan \left(x+\frac{\pi}{4}\right) = \frac{\tan x - 1}{1 + \tan x} = \frac{\sin x - \cos x}{\cos x + \sin x}$

$\displaystyle (2 \cos^2 x - 1)(\sin x - \cos x) = 2 \sin x \cos^2x - 2 \cos^3x - \sin x + \cos x$

$\displaystyle (2 \cos^2 x - 1)(\sin x - \cos x) = 2 \sin x \cos^2x - 2 \cos x (1 - \sin^2x) - \sin x + \cos x$

$\displaystyle (2 \cos^2 x - 1)(\sin x - \cos x) = 2 \sin x \cos^2x + 2 \cos x \sin^2x - \sin x - \cos x$

and

$\displaystyle (2 \sin x \cos x - 1)(\cos x + \sin x) = 2 \sin x \cos^2x + 2 \cos x \sin^2x - \sin x - \cos x$

Therefore
$\displaystyle (2 \cos^2 x - 1)(\sin x - \cos x) = (2 \sin x \cos x - 1)(\cos x + \sin x)$

$\displaystyle \frac{2 \sin x \cos x - 1}{2 \cos^2x - 1} = \frac{\sin x - \cos x}{\cos x + \sin x}$

$\displaystyle \tan 2x - \sec 2x = \tan \left(x+\frac{\pi}{4}\right)$

3. Originally Posted by slapmaxwell1
i have to prove tan 2x - sec 2x = tan (x - pi/4)

well i started with the right side and used identities to simplify it down to tan x - 1/(1 + tan x)

im not sure where to go from here? any suggestions?
Hi all

4. Hello, slapmaxwell1!

This is a messy one . . . You're off to a good start!

Prove: .$\displaystyle \tan2x - \sec2x \:=\:\tan\left(x - \tfrac{\pi}{4}\right)$

I started with the right side and used identities
. . to simplify it down to: .$\displaystyle \frac{\tan x - 1}{1 + \tan x}$

I'm not sure where to go from here? Any suggestions?

You had: .$\displaystyle \boxed{\frac{\tan x - 1}{1 + \tan x}}\;=\;\frac{\frac{\sin x}{\cos x} - 1}{1 + \frac{\sin x}{\cos x}}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad \frac{\sin x - \cos x}{\cos x + \sin x} \;=\;-\frac{\cos x - \sin x}{\cos x + \sin x}$

Multiply by $\displaystyle \frac{\cos x - \sin x}{\cos x - \sin x}\!:$

. . $\displaystyle -\frac{\cos x - \sin x}{\cos x + \sin x}\cdot{\color{blue}\frac{\cos x - \sin x}{\cos x - \sin x}} \;=\;-\frac{(\cos x-\sin x)^2}{(\cos x +\sin x)(\cos x - \sin x)}$

. . . . $\displaystyle =\;-\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}} - \overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{\ \underbrace{\cos^2\!x - \sin^2\!x}_{\text{This is }\cos2x}}$

We have: . $\displaystyle -\frac{1-\sin2x}{\cos2x} \;\;=\;\;-\frac{1}{\cos2x} + \frac{\sin2x}{\cos2x}$

. . . . . . $\displaystyle =\;\;-\sec2x + \tan2x \;\;=\;\; \boxed{\tan2x - \sec2x}$

5. Originally Posted by slapmaxwell1
i have to prove tan 2x - sec 2x = tan (x - pi/4)

well i started with the right side and used identities to simplify it down to tan x - 1/(1 + tan x)

im not sure where to go from here? any suggestions?
If you use the identities

$\displaystyle cos2x=\frac{1-tan^2x}{1+tan^2x}$

$\displaystyle sin2x=\frac{2tanx}{1+tan^2x}$

we obtain

$\displaystyle tan2x-sec2x=\frac{sin2x}{cos2x}-\frac{1}{cos2x}=\frac{sin2x-1}{cos2x}$

$\displaystyle =\frac{\frac{2tanx}{1+tan^2x}-1}{\left(\frac{1-tan^2x}{1+tan^2x}\right)}=\frac{\left(\frac{2tanx-1-tan^2x}{1+tan^2x}\right)}{\left(\frac{1-tan^2x}{1+tan^2x}\right)}$

$\displaystyle =\frac{tanx-1+tanx-tan^2x}{1+tan^2}\ \left(\frac{1+tan^2x}{1-tan^2x}\right)=\frac{(tanx-1)-tanx(tanx-1)}{(1-tanx)(1+tanx)}$

$\displaystyle =\frac{(tanx-1)(1-tanx)}{(1+tanx)(1-tanx)}=\frac{tanx-1}{1+tanx}$

$\displaystyle =\frac{tanx-tan\left(\frac{{\pi}}{4}\right)}{1+tan\left(\frac{ {\pi}}{4}\right)tanx}=tan\left(x-\frac{{\pi}}{4}\right)$

6. wow i didnt think of that approach...nice! thanks alot..so doing math i have remain open minded? LOL :O)

7. soroban thank you for keeping it where i could follow! whatever program your using, i want it. LOL thanks everybody for all your help..i stopped doing the problem too soon and got frustrated..and wow this was one messy problem!!!! geez

Originally Posted by Soroban
Hello, slapmaxwell1!

This is a messy one . . . You're off to a good start!

You had: .$\displaystyle \boxed{\frac{\tan x - 1}{1 + \tan x}}\;=\;\frac{\frac{\sin x}{\cos x} - 1}{1 + \frac{\sin x}{\cos x}}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad \frac{\sin x - \cos x}{\cos x + \sin x} \;=\;-\frac{\cos x - \sin x}{\cos x + \sin x}$

Multiply by $\displaystyle \frac{\cos x - \sin x}{\cos x - \sin x}\!:$

. . $\displaystyle -\frac{\cos x - \sin x}{\cos x + \sin x}\cdot{\color{blue}\frac{\cos x - \sin x}{\cos x - \sin x}} \;=\;-\frac{(\cos x-\sin x)^2}{(\cos x +\sin x)(\cos x - \sin x)}$

. . . . $\displaystyle =\;-\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}} - \overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{\ \underbrace{\cos^2\!x - \sin^2\!x}_{\text{This is }\cos2x}}$

We have: . $\displaystyle -\frac{1-\sin2x}{\cos2x} \;\;=\;\;-\frac{1}{\cos2x} + \frac{\sin2x}{\cos2x}$

. . . . . . $\displaystyle =\;\;-\sec2x + \tan2x \;\;=\;\; \boxed{\tan2x - \sec2x}$