2倍角の公式の導出( $3$\displaystyle{\tan 2{\alpha}}$ 編)

$3$\displaystyle{\begin{array}{lll}\tan 2{\alpha}=\tan $ {\alpha}+{\alpha} $ & \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}=\frac{\hspace{2} \tan {\alpha}+\tan {\alpha} \hspace{2}}{\hspace{2} 1- \tan {\alpha}\tan {\alpha} \hspace{2}}& \vspace{6}\\ \text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}\text{\hspace{5}}=\frac{\hspace{2} 2\tan {\alpha} \hspace{2}}{\hspace{2} 1- { \tan }^{2}{\alpha} \hspace{2}}& \vspace{6}\\ \end{array}}$

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