Szczegółowe wyjaśnienie:
[tex]\dfrac{\text{tg}\alpha(1+\text{ctg}^2\alpha)}{1+\text{tg}^2\alpha}=\dfrac{1-\sin^2\alpha}{\sin\alpha\cos\alpha}[/tex]
[tex]L=\dfrac{\text{tg}\alpha+\text{tg}\alpha\text{ctg}^2\alpha}{1+\text{tg}^2\alpha}=\dfrac{\text{tg}\alpha+\text{tg}\alpha\text{ctg}\alpha\text{ctg}\alpha}{1+\text{tg}^2\alpha}[/tex]
skorzystamy z [tex]\text{tg}x\cdot\text{ctg}x=1[/tex]
[tex]=\dfrac{\text{tg}\alpha+\text{ctg}\alpha}{1+\text{tg}^2\alpha}[/tex]
skorzystamy z [tex]\text{tg}x=\dfrac{\sin x}{\cos x},\ \text{ctg}x=\dfrac{\cos x}{\sin x}[/tex]
[tex]=\left(\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}\right):\left[1+\left(\dfrac{\sin\alpha}{\cos\alpha}\right)^2\right]\\=\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha}:\left(1+\dfrac{\sin^2\alpha}{\cos^2\alpha}\right)[/tex]
skorzystamy z jedynki trygonometrycznej [tex]\sin^2x+\cos^2x=1[/tex]
[tex]=\dfrac{1}{\sin\alpha\cos\alpha}:\left(\dfrac{\cos^2\alpha}{\cos^2\alpha}+\dfrac{\sin^2\alpha}{\cos^2\alpha}\right)=\dfrac{1}{\sin\alpha\cos\alpha}:\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}\\\\=\dfrac{1}{\sin\alpha\cos\alpha}:\dfrac{1}{\cos^2\alpha}=\dfrac{1}{\sin\alpha\cos\alpha}\cdot\dfrac{\cos^2\alpha}{1}=\dfrac{\cos\alpha}{\sin\alpha}=\text{ctg}\alpha[/tex]
[tex]P=\dfrac{1-\sin^2\alpha}{\sin\alpha\cos\alpha}[/tex]
skorzystamy z jedynki trygonometrycznej
[tex]\sin^2x+\cos^2x=1\to\cos^2\alpha=1-\sin^2\alpha[/tex]
[tex]=\dfrac{\cos^2\alpha}{\sin\alpha\cos\alpha}=\dfrac{\cos\alpha}{\sin\alpha}=\text{ctg}\alpha[/tex]
[tex]\huge\boxed{L=P}\qquad\qquad\blacksquare[/tex]