Udowodnij tożsamość trygonometryczną [latex]frac{tgx - ctgx }{tgx+ctgx} = frac{tg^{2}x-1 }{tg^{2}x+1 } [/latex]

[latex]frac{tg x-ctg x}{tgx+ctgx}=frac{tg^2x-1}{tg^2x+1} \Z: tgx+ctgx eq0 Rightarrow xinmathbb{R}ackslash {frac{pi}{2}+kpi, kpi}, kinmathbb{C} \Dowod: \L=frac{tgx-ctgx}{tgx+ctgx}=frac{frac{sin x}{cos x}-frac{cos x}{sin x}}{frac{sin x}{cos x}+frac{cos x}{sin x}}=frac{frac{sin^2x-cos^2x}{sin xcos x}}{frac{sin^2x+cos^2x}{sin xcos x}}=frac{sin^2x-cos^2x}{sin^2x+cos^2x}=sin^2x-cos^2x [/latex] [latex]\P=frac{tg^2x-1}{tg^2x+1}=frac{frac{sin^2x}{cos^2x}-1}{frac{sin^2x}{cos^2x}+1}=frac{frac{sin^2x}{cos^2x}-frac{cos^2x}{cos^2x}}{frac{sin^2x}{cos^2x}+frac{cos^2x}{cos^2x}}=frac{frac{sin^2x-cos^2x}{sin^2xcos^2x}}{frac{sin^2x+cos^2x}{sin^2xcos^2x}}= \=frac{sin^2x-cos^2x}{sin^2x+cos^2x}=sin^2x-cos^2x \left.egin{matrix} \ L=sin^2x-cos^2x \P=sin^2x-cos^2x end{matrix} ight|Rightarrow L=P[/latex]