三角関数の公式(証明)
$3$\displaystyle{{\tan }^{2}{\theta}+1=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}}$ $3$\displaystyle{ $ \cos {\theta}\neq 0 $ }$

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<証明>

$3$\displaystyle{{\sin }^{2}{\theta}+{\cos }^{2}{\theta}=1}$ の両辺を $3$\displaystyle{{\cos }^{2}{\theta}}$ で割ると，

$3$\displaystyle{\frac{\hspace{2} { \sin }^{2}{\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}+\frac{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} \sin {\theta} \hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}=\tan {\theta}}$ を上式に代入し，

$3$\displaystyle{{\tan }^{2}{\theta}+=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}}$

の関係式が得られる。

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