# How to solve this identity?

• Nov 21st 2010, 08:53 AM
doodofnood
How to solve this identity?
tan(pi/4 + x) + tan(pi/4 - x) = 2sec2x

I know that I must use compound angle formula for tan, but I'm still having heaps of trouble. Can someone show me the steps to solve this please?

• Nov 21st 2010, 09:18 AM
Also sprach Zarathustra
tan(pi/4 + x) =cotx
+
tan(pi/4 - x)=-cotx

=0

hmm.. or maybe I am fool!
• Nov 21st 2010, 09:42 AM
Unknown008
$\tan(\frac{\pi}{4} + x) + \tan(\frac{\pi}{4} - x) = \dfrac{\tan(\frac{\pi}{4}) + \tan(x)}{1-\tan(\frac{\pi}{4})\tan(x)} + \dfrac{\tan(\frac{\pi}{4}) - \tan(x)}{1+\tan(\frac{\pi}{4})\tan(x)}$

Recall that $\tan(\frac{\pi}{4}) = 1$

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

Combine the fractions.

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

Simplify;

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

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

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

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

Hm.. time to convert into the basic ratios.

$= \dfrac{\frac{2}{\cos^2(x)}}{1-\frac{\sin^2(x)}{\cos^2(x)}}$

Multiply by cos/cos

$= \dfrac{\frac{2}{\cos^2(x)}}{1-\frac{\sin^2(x)}{\cos^2(x)}} \times \dfrac{\cos^2(x)}{\cos^2(x)}$

$= \dfrac{2}{\cos^2(x)-\sin^2(x)}$

Now, recall that: $\cos^2(x) - \sin^2(x) = \cos(2x)$

$= \dfrac{2}{\cos(2x)}$

$= 2\sec(2x)$

Shown!
• Nov 21st 2010, 10:07 AM
TheCoffeeMachine
Quote:

Originally Posted by Also sprach Zarathustra
tan(pi/4 + x) =cotx
+
tan(pi/4 - x)=-cotx

=0

hmm.. or maybe I am fool!

You are confusing it with $\tan(\frac{\pi}{2} \pm x) = \mp\cot(x)$.
• Nov 21st 2010, 07:41 PM
TheCoffeeMachine
Quote:

Originally Posted by doodofnood
I know that I must use compound angle formula for tan, but I'm still having heaps of trouble.

If you want to avoid the compound angle identity for the tangent function, express everything in terms of
sine and cosine and use the symmetric relation $\cos\left(\frac{\pi}{4}-t\right) = \sin\left(\frac{\pi}{4}+t\right)$. It will be less involving.

Spoiler:
$\displaystyle \tan\left(\frac{\pi}{4} + x\right) + \tan\left(\frac{\pi}{4} - x\right) = \frac{\sin\left(\frac{\pi}{4}+x\right)}{\cos\left( \frac{\pi}{4}+x\right)}+\frac{\sin\left(\frac{\pi} {4}-x\right)}{\cos\left(\frac{\pi}{4}-x\right)}$
Spoiler:
$\displaystyle = \frac{\sin^2\left(\frac{\pi}{4}+x\right)+\cos^2\le ft(\frac{\pi}{4}+x\right)}{\cos^2\left(\frac{\pi}{ 4}+x\right)} = \frac{1}{\frac{1}{2}\cos{2x}} = \frac{2}{\cos{2x}} = 2\sec{2x}.$

• Nov 21st 2010, 10:16 PM
Soroban
Hello, doodofnood!

Quote:

$\tan(\frac{\pi}{4} + x) + \tan(\frac{\pi}{4} - x) \:=\: 2\sec2x$

$\tan(\frac{\pi}{4} + x) + \tan(\frac{\pi}{4} - x) \;=\;
\dfrac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4}\tan x} + \dfrac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4}\tan x}$

. . $\displaystyle =\;\frac{1+\tan x}{1 - \tan x} + \frac{1-\tan x}{1+\tan x} \;=\;\frac{(1+\tan x)^2 + (1-\tan x)^2}{(1-\tan x)(1 + \tan x)}$

. . $\displaystyle =\;\frac{1 + 2\tan x + \tan^2\!x + 1 - 2\tn x + \tan^2\!x}{1-\tan^2\!x} \;=\;\frac{2+2\tan^2\!x}{1-\tan^2\!x}$

. . $\displaystyle =\;\frac{2(1+\tan^2\!x)}{1-\tan^2\!x} \;=\;\frac{2\sec^2\!x}{1-\tan^2\!x}$

$\displaystyle \text{Multiply by }\frac{\cos^2\!x}{\cos^2\!x}\!:\;\;\frac{2}{\cos^2 \!x - \sin^2\!x} \;=\;\frac{2}{\cos2x} \;=\;2\sec2x$