# Thread: prove this identity

1. ## prove this identity

2sin(x)/[(cos(x)+sin(x)] = tan(2x) + 1 - sec(2x)

First, I turn tan(2x) into [sin(2x) / cos(2x)] then I turn sec(2x) into 1/cos(2x)

So, I end up with... [sin(2x) / cos(2x)] + 1 - (1/(cos2x))

I multiple and get the GCD of cos(2x)..

[sin(2x) + cos(2x) - 1] / cos(2x)

Now things get kind of fuzzy.. I probably didn't do this the best way thus far but I can't really advance much further.. I need help, please.

2. Originally Posted by zodiacbrave
2sin(x)/[(cos(x)+sin(x)] = tan(2x) + 1 - sec(2x)

First, I turn tan(2x) into [sin(2x) / cos(2x)] then I turn sec(2x) into 1/cos(2x)

So, I end up with... [sin(2x) / cos(2x)] + 1 - (1/(cos2x))

I multiple and get the GCD of cos(2x)..

[sin(2x) + cos(2x) - 1] / cos(2x)

Now things get kind of fuzzy.. I probably didn't do this the best way thus far but I can't really advance much further.. I need help, please.
Start with right side with the identy

$tan(2x)=\frac{2tanx}{1-tan^2x}$ and $sec(2x)=\frac{1}{cos(2x)}=\frac{1+tan^2x}{1-tan^2x}$

Then the right side reduces to

$\frac{2tanx+2}{1-tan^2x}$

Chage tan to terms of sin and cos and you should be able to get there.

3. Hello, zodiacbrave!

Another approach . . .

Prove: . $\frac {2\sin x}{\cos x+\sin x} \:=\: \tan(2x) + 1 - \sec(2x)$
Multiply the left side by: . $\frac{\cos x - \sin x}{\cos x - \sin x}$

$\frac{2\sin x}{\cos x + \sin x}\cdot\frac{\cos x - \sin x}{\cos x - \sin x}$

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

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

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

. . $=\;\tan(2x) + 1 - \sec(2x)$

4. Soroban's LATEXing technique is superb and very instructive, thanks MHF expert