1. ## Trig Problem

Hi all.

I'm trying to show that...

1 - ( tan (x/2) )^2 = sec x + tan x

Any help will be appreciated.

Thanks.

2. ## expand the difference of squares

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

well the first thing you could do is expand the difference of squares

$\displaystyle (1-\tan \left(\frac{x}{2}\right))(1+\tan\left(\frac{x}{2}\ right)) =\sec{x} + \tan{x}$

you can probably see the rest...

3. How do I convert the x/2 to an x. Half/Double angle formulae?

4. Originally Posted by pollardrho06
Hi all.

I'm trying to show that...

1 - ( tan (x/2) )^2 = sec x + tan x
not an identity.

5. no wonder I couldn't resolve it.

6. OK. Let me start from the start...

I was working on the indefinite integral of sec x...

INT (sec x) dx...

I did this two different ways:

Method 1:

sec x * [ (sec x + tan x) / (sec x + tan x) ]

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leads to a simple substitution t = sec x + tan x...

and the integral evaluates to...

ln | sec x + tan x | + c

Method 2:

Using Weierstrass Substitution...u = tan (x/2)etc etc...

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I end up with the answer ln | 1 - ( tan (x/2) )^2 | + c

Now I'm trying to prove that both these answers (methods 1 and 2) are the same, hence the earlier post.

Where did I go wrong?

7. Originally Posted by pollardrho06
I end up with the answer ln | 1 - ( tan (x/2) )^2 | + c
Firstly, this is not right! Check your calculations. Secondly, the two integrals don't have to be equal. In the first substitution you have found that $\displaystyle I = \ln|\sec{x}+\tan{x}|+c$. What does the $\displaystyle c$ indicate? If, say, we had some function $\displaystyle f(x)$ and when we integrate, we find some anti-derivative $\displaystyle g(x_{0})$. Now, if for all the anti-derivatives we can find, $\displaystyle g(x_{1}), g(x_{2}), g(x_{3}),..., g(x_ {n})$, it's true that $\displaystyle g(x_{n}) = g(x_{0})$, for what purpose would there be for $\displaystyle c$ to serve? We use $\displaystyle c$ precisely because this is not the case! Consider the integral you have found in the first substitution $\displaystyle \ln|\sec{x}+\tan{x}|+c$ satisfies your $\displaystyle I$, for ANY constant $\displaystyle c$. $\displaystyle \ln|\sec{x}+\tan{x}|+\dfrac{\pi}{11}$ is valid for $\displaystyle I$ and so is $\displaystyle \ln|\sec{x}+\tan{x}|+{e^{\pi}}$. So what, for example, if the other substitution gives you $\displaystyle \ln|\sec{x}+\tan{x}|+{e^{\pi}}+k$, for some constant $\displaystyle k$? $\displaystyle k+e^{\pi}$ is a constant, and what we donated by $\displaystyle c$ was a constant, so both satisfy $\displaystyle I$. In general, any two or more anti-derivatives of some function F(x) don't have to be equal; rather, they have to differ by a constant.