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3.1.4: Піфагорійська ідентичність
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$\begin{aligned} \dfrac{\sin ^2 x}{\sin ^2 x}+\dfrac{\cos ^2 x}{\sin ^2 x}&=\dfrac{1}{\sin ^2 x}\\ 1+\cot ^2 x&=\csc ^2 x \end{aligned}$ \ dfrac {\ sin x (\ csc x-\ sin x)} {1-\ sin x} &=\ dfrac {\ ... $\begin{aligned} \dfrac{\sin ^2 x}{\sin ^2 x}+\dfrac{\cos ^2 x}{\sin ^2 x}&=\dfrac{1}{\sin ^2 x}\\ 1+\cot ^2 x&=\csc ^2 x \end{aligned}$ \ dfrac {\ sin x (\ csc x-\ sin x)} {1-\ sin x} &=\ dfrac {\ sin x\ cdot\ csc x-\ sin ^ {2} x} {1-\ sin x}\\\ $\begin{aligned} (\sec ^2 x)(1−\sin ^2 x)−(\dfrac{\sin x}{\csc x}+\dfrac{\cos x}{\sec x})&=\sec ^2 x\cdot \cos ^2 x−(\sin ^2 x+\cos ^2 x) \\ &=1−1 \\&=0\end{aligned}$