1. Proving trigonometric Identities

Prove

tan(pi/4 + x) - tan(pi/4 - x) = 2tan2x

and

tan^2A - tan^2B = (sin^2A - sin^2B) / (cos^2A cos^B)

help on either would be very much appreciated. Thank you

2. Originally Posted by BeckyDWood
Prove

tan(pi/4 + x) - tan(pi/4 - x) = 2tan2x

and

tan^2A - tan^2B = (sin^2A - sin^2B) / (cos^2A cos^B)

help on either would be very much appreciated. Thank you
$\frac{\sin(\frac{\pi}{4}+x)}{\cos ({\pi}{4} +x)} - \frac{\sin (\frac{\pi}{4}-x) }{\cos (\frac{\pi}{4} -x) }$

$\frac{\sin(\frac{\pi}{4}+x)(\cos (\frac{\pi}{4} -x))-\sin (\frac{\pi}{4}-x)(\cos (\frac{\pi}{4} +x))}{\cos(\frac{\pi}{4}+x)(\cos (\frac{\pi}{4}-x)}$

$\sin (A-B) = \sin A \cos B - \sin B \cos A$

$\cos (A) \cos (B) = 1/2 (\cos(\frac{A+B}{2})+\cos (\frac{A-B}{2}))$

the numerator $\sin\left( \frac{\pi}{4} + x -(\frac{\pi}{4} -x )\right)$

the denominator

$1/2(\cos (\pi/4+x+\pi/4 -x)+\cos (\pi/4+x-(\pi/4-x)))$

3. Hello, Becky!

For the first one, we can use these identities:

. . $\tan(A \pm B) \:=\:\frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \qquad\qquad \frac{2\tan A}{1 - \tan^2A} \;=\;\tan2A$

Prove: . $\tan\left(\tfrac{\pi}{4} + x\right) - \tan\left(\tfrac{\pi}{4} - x\right) \;=\; 2\tan2x$

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

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

. . $= \;\;\frac{4\tan x}{1 - \tan^2\!x} \;\;=\;\;2\left(\frac{2\tan x}{1-\tan^2\!x}\right) \;\;=\;\;2\tan2x$