Proving Trig Identities?

**ccarnessali** · January 7th 2009, 07:52 AM

I cannot understand this at all. how do i prove:

1-sinx/1+sinx = (tanx-secx)^2

**red_dog** · January 7th 2009, 08:04 AM

$(\tan x-\sec x)^2=\left(\frac{\sin x}{\cos x}-\frac{1}{\cos x}\right)^2=\left(\frac{\sin x-1}{\cos x}\right)=$

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

**ccarnessali** · January 7th 2009, 08:19 AM

thank you so much! this proving trig identities stuff is killing me. i cannot understand it.

**Soroban** · January 7th 2009, 08:24 AM

Hello, ccarnessali!

We can go the other way, too . . .

$\frac{1-\sin x}{1+\sin x} \:=\: (\tan x-\sec x)^2$

Multiply the left side by: $\frac{1-\sin x}{1-\sin x}$

. . $\frac{1-\sin x}{1+\sin x}\cdot {\color{blue}\frac{1-\sin x}{1-\sin x}}$

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

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

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

. . $= \;\left(\frac{1}{\cos x}\right)^2 - 2\left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) + \left(\frac{\sin x}{\cos x}\right)^2$

. . $= \;\sec^2\!x - 2\sec x\tan x + \tan^2\!x$

. . $= \;(\sec x - \tan x)^2$

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