# Prove this identity: $\tan(2x)-\sec(2x) =\tan(x-\pi/4)$

I've been having a time with this problem. I tried to start with the left side but I hit a dead end quick... I then tried the right side and had a little more luck but I've hit a block. I first used the tan difference identity then converted everything into sins and cosines. Then I used a difference identity on sin and cosine and then converted into exact values. $\dfrac{\pi}{4}$ for sine and cosine is $\dfrac{\sqrt{2}}{2}$. All the radicals cancelled out leaving me with $\dfrac{\sin x - \cos x }{\cos x + \sin x}$... I can't think anything to do with this... where did I go wrong?

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You're almost there. Dividing the numerator and denominator by $\cos x$ gives $$\frac{\sin x - \cos x}{\cos x + \sin x} = \frac{\tan x - 1}{1 + \tan x},$$ and then observe that $\tan \frac{\pi}{4} = 1$. Then recall the tangent addition identity $$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}.$$ For what suitable choices of $\alpha$, $\beta$, does the identity match the expression you have?

Okay, so since there's some confusion about the direction in which you're going, and since you already basically have done the proof and just need to see it put all together, here it is: \begin{align*} \tan(x - \tfrac{\pi}{4}) &= \frac{\tan x - \tan \frac{\pi}{4}}{1 + \tan x \tan \frac{\pi}{4}} \\ &= \frac{\tan x - 1}{1 + \tan x} \\ &= \frac{\sin x - \cos x}{\cos x + \sin x} \\ &= \frac{-(\sin x - \cos x)^2}{\cos^2 x - \sin^2 x} \\ &= \frac{- \sin^2 x + 2 \sin x \cos x - \cos^2 x}{\cos 2x} \\ &= \frac{-1 + \sin 2x}{\cos 2x} \\ &= \tan 2x - \sec 2x. \end{align*} Going in the other direction, we might write \begin{align*} \tan 2x - \sec 2x &= \frac{\sin 2x}{\cos 2x} - \frac{1}{\cos 2x} \\ &= \frac{2 \sin x \cos x - 1}{\cos^2 x - \sin^2 x} \\ &= \frac{2 \sin x \cos x - \sin^2 x - \cos^2 x}{(\cos x + \sin x)(\cos x - \sin x)} \\ &= \frac{-(\cos x - \sin x)^2}{(\cos x + \sin x)(\cos x - \sin x)} \\ &= \frac{\sin x - \cos x}{\cos x + \sin x} \\ &= \frac{\tan x - 1}{1 + \tan x} \\ &= \frac{\tan x - \tan \frac{\pi}{4}}{1 + \tan x \tan \frac{\pi}{4}} \\ &= \tan(x - \tfrac{\pi}{4}).\end{align*}

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Wait... I'm a little confused now... I understand dividing by cos x... but how do I get tan pi/4 out of that? – user450 Apr 21 '14 at 3:20
$$\frac{\tan x - 1}{1 + \tan x} = \frac{\tan x - \tan \frac{\pi}{4}}{1 + \tan x \tan \frac{\pi}{4}}.$$ Can you do the rest? – heropup Apr 21 '14 at 3:26
Ok! I follow you... But I can't see how I get the sec 2x... I know that sec x equals 1/cos x and tan equals sin x over cos x... Do I convert all the tans to sins and cosines? – user450 Apr 21 '14 at 3:27
Oh...I thought you started from the LHS and got to $(\sin x - \cos x)/(\cos x + \sin x)$... Well, if you started from the RHS and you want to get to the left, then all you do is multiply the numerator and denominator by $(\cos x - \sin x)$. This gives you $$\frac{-\sin^2 x + 2 \sin x \cos x - \cos^2 x}{\cos^2 x - \sin^2 x} = \frac{\sin 2x - 1}{\cos 2x}.$$ – heropup Apr 21 '14 at 3:33
ugh... I'm so sorry but now I'm still not there yet.... Do I use double angle identities again now? – user450 Apr 21 '14 at 3:36

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

Using Weierstrass substitution, this becomes $$\frac{\dfrac{2t}{1+t^2}-1}{\dfrac{1-t^2}{1+t^2}}=-\frac{(1-t)^2}{1-t^2}$$ where $t=\tan x$

If $1-t\ne0\iff t\ne1$ i.e.,$\tan x\ne1$ this can be reduces to $$-\frac{1-t}{1+t}=\frac{t-1}{t+1}=\frac{\tan x-\tan\dfrac\pi4}{1+\tan x\tan\dfrac\pi4}=\cdots$$

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\begin{align} \tan(2x)-\sec(2x) &=\frac{\sin(2x)-1}{\cos(2x)}\tag{1}\\[6pt] &=\frac{\cos\left(2\left(\frac\pi4-x\right)\right)-1}{\sin\left(2\left(\frac\pi4-x\right)\right)}\tag{2}\\ &=\frac{-2\sin^2\left(\frac\pi4-x\right)}{2\sin\left(\frac\pi4-x\right)\cos\left(\frac\pi4-x\right)}\tag{3}\\[9pt] &=-\tan\left(\frac\pi4-x\right)\tag{4}\\[18pt] &=\tan\left(x-\frac\pi4\right)\tag{5} \end{align} Explanation:
$(1)$: $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ and $\sec(\theta)=\frac1{\cos(\theta)}$
$(2)$: $\sin\left(\frac\pi2-\theta\right)=\cos(\theta)$ and $\cos\left(\frac\pi2-\theta\right)=\sin(\theta)$
$(3)$: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and $\cos(2\theta)=1-2\sin^2(\theta)$
$(4)$: $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$
$(5)$: $\tan(-\theta)=-\tan(\theta)$
Starting where you left off, we can use the formulas for the sine and cosine of a difference to get \begin{align} \frac{\sin(x)-\cos(x)}{\cos(x)+\sin(x)} &=\frac{\sqrt2\sin\left(x-\frac\pi4\right)}{\sqrt2\cos\left(x-\frac\pi4\right)}\\ &=\tan\left(x-\frac\pi4\right)\tag{6} \end{align}