prove the identity

**slapmaxwell1** · April 25th 2010, 07:05 AM

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?

**running-gag** · April 25th 2010, 08:59 AM

Hi

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

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

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

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

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

and

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

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

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

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

**mrmohamed** · April 25th 2010, 09:42 AM

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

**Soroban** · April 25th 2010, 10:31 AM

Hello, slapmaxwell1!

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

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

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

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

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

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

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

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

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

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

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

**Archie Meade** · April 25th 2010, 10:43 AM

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

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

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

we obtain

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

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

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

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

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

**slapmaxwell1** · April 25th 2010, 04:18 PM

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

**slapmaxwell1** · April 25th 2010, 04:22 PM

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

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

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

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

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

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

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

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