Help for tangent

Question

Is this equality correct? how can I get to it starting from the trigonometry basics?

$$tan(\alphaº/2)=\frac{\sqrt{1+tan²(\alpha)}-1}{tan(\alpha)}$$

where $\alpha$ is an angle

Answer 1

$\tan \frac {\alpha}{2} \\ \frac {\sin \frac {\alpha}{2}}{\cos \frac {\alpha}{2}}\\ \frac {\sqrt {1-\cos\alpha}}{\sqrt {1+\cos\alpha}}\\ \frac {(\sqrt {1-\cos\alpha})^2}{\sqrt {1-\cos^2\alpha}}\\ \frac {1-\cos\alpha}{\sin \alpha}\\ \frac {\sec\alpha-1}{\tan \alpha}\\ \frac {\sqrt {1+\tan^2\alpha}-1}{\tan \alpha}\\ $

But I played a little bit fast and loose with the trig there.

In particular $\sec\alpha = \sqrt {1+\tan^2\alpha}$ only when $\sec \alpha \ge 0$

And this identity is not true if $\alpha \in (\frac {\pi}{2}, \frac {3\pi}{2})$

Answer 2

From the tangent double angle formula we have \begin{eqnarray*} \tan \alpha =\frac{ \tan( \alpha/2)}{1- \tan^2 (\alpha/2)}. \end{eqnarray*} Let $T=\tan \alpha$ and $t = \tan( \alpha/2)$ \begin{eqnarray*} Tt^2+2t-T=0. \end{eqnarray*} Now solve this as a quadratic in $t$ and your formula follows.

Answer 3

$$\dfrac{\sqrt{1+\tan^2 \alpha}-1}{\tan \alpha} $$

$\sqrt{1+\tan^2\alpha}= \sec\alpha \space \forall \alpha \notin \left({\dfrac{\pi}{2}, {\dfrac{3\pi}{2}}}\right)$

$$\\= \dfrac{\sec\alpha-1}{\tan \alpha}\\= \dfrac{1-\cos\alpha}{\sin \alpha }\\$$Now use half angle identitities,$$=\dfrac{1-{( 1- 2\sin^2\alpha/2)}}{2\sin \left(\dfrac{\alpha}{2}\right)\cos \left(\dfrac{\alpha}{2}\right)}\\= \color{blue}{\tan \alpha /2}$$