Proving basic/standard trigonometric identities How to prove the following trigonometric identities ?
1) If $\displaystyle \tan (\alpha) \cdot \tan(\beta) = 1 \text{ then } \alpha + \beta = \frac{\pi}{2}$
I tried to prove it by using the the formula for $\tan(\alpha + \beta)$ but ain't it valid only when $\alpha + \beta \neq \frac{\pi}{2}$ ?
2) $\displaystyle\sec\theta + \tan \theta = \frac{1}{ \sec\theta - \tan \theta}, \theta \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} $
For this one I tried substituting them with the sides of the triangle, but not successful to the final result.
These are not my homework, I am trying to learn maths almost on my own, so ...
 A: Added (and corrected twice):
$$\tan \alpha \tan \beta =1\Leftrightarrow \dfrac{\sin \alpha }{\cos \alpha }%
\dfrac{\sin \beta }{\cos \beta }=1$$
Multiplying by $\cos\alpha\cos\beta\ne 0$, gives
$$\sin \alpha \sin \beta -\cos \alpha\cos \beta =0 \Leftrightarrow \cos (\alpha +\beta )=0$$
This is equivalent to 
$$\alpha +\beta =\dfrac{\pi }{2}+n\pi,\qquad (\ast)$$
to which we still have to add the condition written above ($\cos\alpha\cos\beta\ne 0$), which means the constraint 
$$\alpha,\beta\ne\dfrac{\pi}{2}+n\pi,\qquad (\ast\ast)$$
where $n$ is an integer.
Note: In the original equation $\tan \alpha \tan \beta =1$, neither $\alpha$ nor $\beta$ can be zero.  The combined conditions $(\ast)$ and $(\ast\ast)$ assures that.

The identity $$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\qquad \theta \neq (2n+1)\dfrac{\pi}{2}$$
is equivalent to $$\sin ^{2}\theta +\cos ^{2}\theta =1.$$ 
Indeed, if 
$$\theta \neq (2n+1)\dfrac{\pi }{2}\Leftrightarrow \sin \theta \neq \pm 1 \Leftrightarrow \dfrac{\pm1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }\neq 0\Leftrightarrow \pm\sec \theta -\tan \theta \neq 0,$$
then
$$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\Leftrightarrow \left( \sec \theta +\tan \theta \right) \left( \sec \theta
-\tan \theta \right) =1$$
$$\Leftrightarrow \sec ^{2}\theta -\tan ^{2}\theta =1\Leftrightarrow \dfrac{1}{\cos ^{2}\theta }-\dfrac{\sin ^{2}\theta }{\cos^{2}\theta }=1\Leftrightarrow 1-\sin ^{2}\theta=\cos ^{2}\theta$$ $$\Leftrightarrow \sin ^{2}\theta +\cos ^{2}\theta =1.$$
A: When I need to prove trigonometric identities, I tend to use complex exponentials. For instance, to prove your first identity observe that
\begin{align}
\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} \quad \text{and} \quad \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2},
\end{align}
so
\begin{align}
\tan \theta = \frac{1}{i} \frac{e^{i\theta} - e^{-i\theta} }{e^{i \theta} + e^{-i \theta}}.
\end{align}
We calculate
\begin{align}
1 = \left( \frac{1}{i} \frac{e^{i\alpha} - e^{-i\alpha} }{e^{i \alpha} + e^{-i \alpha}} \right) \left(\frac{1}{i} \frac{e^{i\beta} - e^{-i\beta} }{e^{i \beta} + e^{-i \beta}} \right) = - \frac{(e^{2 i \alpha} - 1)(e^{2 i \beta} - 1)}{(e^{2 i \alpha} + 1)(e^{2 i \beta} + 1)}
\end{align}
Hence,
\begin{align}
(e^{2 i \alpha} + 1)(e^{2 i \beta} + 1) = - (e^{2 i \alpha} - 1)(e^{2 i \beta} - 1).
\end{align}
or $e^{2 i (\alpha + \beta)} + 1 = 0$, which implies that $\alpha + \beta = \frac{\pi}{2}$ by Euler's equation $e^{\pi i} + 1 = 0$, provided that we consider only angles in the fundamental region $[-\frac{\pi}{2}, \frac{\pi}{2}]$. A similar calculation works for your second identity.
A: 1) tan(a)tan(b) = 1 is equivilant to sin(a)sin(b) = Cos(a)cos(b) **using the fact that tan is sine divided by cosine.  Next use 
                    cos(a+b) = cos(a)cos(b) -sin(a)sin(b)
which implies      sin(a)sin(b)  = cos(a)cos(b) -cos(a+b)
Use this on the left hand side of ** to get 
                              cos(a+b) = 0

which implies a+b = Pi/2 or -Pi/2
A: For (2)
$$\sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}= \frac{1 + \sin \theta}{\cos \theta}$$ $$ = \frac{(1 + \sin \theta)\cdot (1 - \sin \theta)}{\cos \theta \cdot(1 - \sin \theta)}\text{ [Multiplying both sides by } (1 - \sin \theta)\text{]} $$
$$ = \frac{\cos^2 \theta}{ \cos \theta - \sin \theta \cdot \cos \theta} = \frac{1}{\sec \theta - \tan \theta} \text{ (Q.E.D) }$$
