tanの半角公式のような何かを見つけた

命題1
$0\lt\theta\lt\dfrac{\pi}{2}$ のとき
$\tan{\dfrac{\theta}{2}}= \sqrt{\dfrac{1}{\tan^2{\theta}}+1}-\sqrt{\dfrac{1}{\tan^2{\theta}}}$

命題2
$\dfrac{\pi}{2}\lt\theta\lt\pi$ のとき
$\tan{\dfrac{\theta}{2}}= \sqrt{\dfrac{1}{\tan^2{\theta}}+1}+\sqrt{\dfrac{1}{\tan^2{\theta}}}$

命題1の証明

$\tan$ の2倍角公式　 $\tan{\theta}=\dfrac{2\tan{\dfrac{\theta}{2}}}{1-\tan^2{\dfrac{\theta}{2}}}$ を使う

$\sqrt{\dfrac{1}{\tan^2{\theta}}+1}-\sqrt{\dfrac{1}{\tan^2{\theta}}}$
$=\sqrt{\left(\dfrac{1-\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}\right)^2+1}-\sqrt{\left(\dfrac{1-\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}\right)^2}$
$=\sqrt{\dfrac{1-2\tan^2{\dfrac{\theta}{2}}+\tan^4{\dfrac{\theta}{2}}}{4\tan^2{\dfrac{\theta}{2}}}+\dfrac{4\tan^2{\dfrac{\theta}{2}}}{4\tan^2{\dfrac{\theta}{2}}}}-\sqrt{\left(\dfrac{1-\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}\right)^2}$
$=\sqrt{\left(\dfrac{1+\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}\right)^2}-\sqrt{\left(\dfrac{1-\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}\right)^2}$
ここで, $0\lt\dfrac{\theta}{2}\lt\dfrac{\pi}{4}$ であるので, $1+\tan^2{\dfrac{\theta}{2}}\gt 0$ , $1-\tan^2{\dfrac{\theta}{2}}\gt 0$ , $2\tan{\dfrac{\theta}{2}}\gt 0$
ゆえに
$\dfrac{1+\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}-\dfrac{1-\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}$
$=\dfrac{2\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}$
$=\tan{\dfrac{\theta}{2}}$

命題2の証明

命題1の証明と同様にして
$\sqrt{\dfrac{1}{\tan^2{\theta}}+1}+\sqrt{\dfrac{1}{\tan^2{\theta}}}$
$=\sqrt{\left(\dfrac{1+\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}\right)^2}+\sqrt{\left(\dfrac{1-\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}\right)^2}$
ここで, $\dfrac{\pi}{4}\lt\dfrac{\theta}{2}\lt\dfrac{\pi}{2}$ であるので, $1+\tan^2{\dfrac{\theta}{2}}\gt 0$ , $1-\tan^2{\dfrac{\theta}{2}}\lt 0$ , $2\tan{\dfrac{\theta}{2}}\gt 0$
ゆえに
$=\dfrac{1+\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}+\dfrac{tan^2{\dfrac{\theta}{2}}-1}{2\tan{\dfrac{\theta}{2}}}$
$=\dfrac{2\tan^2{\dfrac{\theta}{2}}}{2\tan{\dfrac{\theta}{2}}}$
$=\tan{\dfrac{\theta}{2}}$