Trig Identities

**casey_k** · November 7th 2008, 06:23 PM

(hi) I'm having trouble with the longer problems.

(sin x + cos x)/(sin x-cos x) = sec² x + 2 tan x/ sec² - 2 tan x

**Soroban** · November 7th 2008, 08:16 PM

Hello, casey_k!

$\frac{\sin x + \cos x}{\sin x-\cos x} \:=\: \frac{\sec^2\!x + 2\tan x}{\sec^2\!x- 2\tan x}$ . . . . This is not an identity.

Is the first fraction squared? . $\left(\frac{\sin x + \cos x}{\sin x - \cos x}\right)^2 \;=\;\frac{\sec^2\!x + 2\tan x}{\sec^2\!x - 2\tan x}$

The left side is: . $\frac{\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}} + \:2\sin x\cos x}{\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} - \:2\sin x\cos x} \;=\; \frac{1 + 2\sin x\cos x}{1 - 2\sin x\cos x}$

Divide top and bottom by $\cos^2\!x\!:\;\;\frac{\dfrac{1}{\cos^2\!x} + \dfrac{2\sin x\cos x}{\cos^2\!x}} {\dfrac{1}{\cos^2\!x} - \dfrac{2\sin x\cos x}{\cos^2\!x}}$

. . . . $=\;\;\frac{\sec^2\!x + 2\,\dfrac{\sin x}{\cos x}}{\sec^2\!x - 2\,\dfrac{\sin x}{\cos x}} \;\;=\;\;\frac{\sec^2\!x + 2\tan x}{\sec^2\!x - 2\tan x}<br /> <br />$

**casey_k** · November 7th 2008, 08:25 PM

Yes, the left side is squared. I forgot to add the square.

Thank you so much for your help!

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