Prove that identity tan2x - tanx = tanx sec2x
I managed to start i think but am not sure
(2tan x / 1-tan^2 x) - tan x
but i don't know what else
you may want to start with..
$\displaystyle \tan x \sec 2x = \frac{\tan x}{\cos 2x} = \frac{\tan x}{2\cos^2 x - 1} = \frac{\tan x}{\frac{2}{\sec^2 x}-1}$
$\displaystyle = \frac{\tan x}{\frac{2 - \sec^2 x}{\sec^2 x}} = \frac{\tan x \sec^2 x}{2-\sec^2 x}$
this seems manageable..
EDIT: from what you got,
$\displaystyle \frac{2\tan x}{1-\tan^2 x} - \tan x = \frac{2\tan x - \tan x + \tan^3 x}{1-\tan^2 x}$
hi!gracey (sweet name)
your starting is absolutly fine.
(2tan x / 1-tan^2 x) - tan x
now what you should really do is take tanx common
tanx[(2/(1-tan^2x)-1]
now use
tan^2x=(1-cos2x)/1+cos2x
after substituting we get
tanx[2/(1-{(1-cos2x)/1+cos2x})-1]
=tanx[{(1+cos2x)/cos2x}-1]
=tanx[1/cos2x]
=tanxsec2x
hence proved
Hello, gracey!
A variation of nikhil's solution . . .
Prove the identity: .$\displaystyle \tan2x - \tan x \:=\: \tan x\sec2x$
I managed to start i think, but am not sure
. . $\displaystyle \frac{2\tan x}{1-\tan^2\!x} - \tan x $
Factor: .$\displaystyle \tan x\left(\frac{2}{1-\tan^2\!x}-1\right) \;=\;\tan x\left(\frac{1+\tan^2\!x}{1-\tan^2\!x}\right) $
. . . . $\displaystyle = \;\tan x\left(\frac{\sec^2\!x}{1-\tan^2\!x}\right) \;=\;\tan x\left(\frac{\frac{1}{\cos^2\!x}}{1 - \frac{\sin^2\!x}{\cos^2\!x}}\right)$
. . . . $\displaystyle = \;\tan x\left(\frac{1}{\cos^2\!x - \sin^2\!x}\right) \;=\;\tan x\left(\frac{1}{\cos2x}\right)$
. . . . $\displaystyle = \;\tan x\sec2x$