# Proving trigo identity

1. May 14, 2005

### whkoh

Prove that
$$\sec x + \tan x = \tan \left (\frac{\pi}{4} + \frac{x}{2}\right )$$

I've got to
$$\sec x + \tan x = \frac{1+\sin x}{\cos x}$$
and then I was stuck. Tried half angle but it didn't seem to work.

2. May 14, 2005

### Hurkyl

Staff Emeritus
There are two sides to the equation -- it sounds like you've only fiddled with the left hand side.

3. May 14, 2005

### arildno

Neat!
I've never seen that trig. identity before..

4. May 14, 2005

### whkoh

Well, manipulating RHS gives
$$\tan \left (\frac{\pi}{4} + \frac{x}{2} \right )$$
$$= \frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}$$
$$=\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}$$

and applying half angle to LHS gives
$$\frac{1+\sin x}{\cos x}$$
$$=\frac{1+2\sin\frac{x}{2}\cos\frac{x}{2}}{1-2\sin\frac{x}{2}\sin\frac{x}{2}}$$

Hmm.. how can $$\tan\frac{x}{2}$$ be equal to
$$2\sin\frac{x}{2}\cos\frac{x}{2}}$$
and
$$2\sin\frac{x}{2}\sin\frac{x}{2}}$$ at the same time?

5. May 14, 2005

### Curious3141

Here's help. :)

Try to manipulate the more complicated side to get the less complicated side. In this case, work on the RHS to get the LHS. Try not to work from both sides at once.

Let the respective sin, cos and tan trig ratios of x/2 be s, c and t. Let those of x be S, C and T. I'm doing this because I'm really fed up of clunky LaTex.

Taking it from where you left off,

RHS :

$$\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}$$
$$=\frac{(1 + t)^2}{(1-t)(1 + t)}$$
$$=\frac{1 + t^2 + 2t}{1 - t^2}$$
$$=\frac{1 + t^2 + 2t}{\frac{c^2 - s^2}{c^2}}$$
$$=\frac{(1 + t^2 + 2t)(c^2)}{C}$$
$$=\frac{c^2 + s^2 + 2sc}{C}$$
$$=\frac{1 + S}{C}$$
$$=\sec{x} + \tan{x}$$ (QED)

6. May 14, 2005

### Curious3141

BTW, the proof (and the original identity) fail for t = -1. In the proof, it's because I multiply the RHS by (1+t)/(1+t). In the orig. identity, the LHS becomes undefined while the RHS remains finite (so the failure is consistent).

7. May 14, 2005

### whkoh

8. May 15, 2005

### dextercioby

You know

$$\sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{1+\sin x}{\cos x}$$ (1)

Th RHS is

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

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

(Q.e.d.)

,pretty simple,right...?

Daniel.

Last edited: May 15, 2005
9. May 15, 2005

### arildno

I think he mentioned that identity in post 1, Daniel..

10. May 15, 2005

### dextercioby

Hehe,i thought he went backwards starting with the RHS.:tongue2:

Daniel.