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 …
Best Answer
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
$$\sec \theta +\tan \theta =\dfrac{1}{\sec \theta -\tan \theta }\Leftrightarrow \left( \sec \theta +\tan \theta \right) \left( \sec \theta -\tan \theta \right) =1$$
$$\Leftrightarrow \sec ^{2}\theta -\tan ^{2}\theta =1\Leftrightarrow \dfrac{1}{\cos ^{2}\theta }-\dfrac{\sin ^{2}\theta }{\cos^{2}\theta }=1\Leftrightarrow 1-\sin ^{2}\theta=\cos ^{2}\theta$$ $$\Leftrightarrow \sin ^{2}\theta +\cos ^{2}\theta =1.$$