Zadanie 1
[tex]\text{L}=\dfrac{1+\cos x}{\sin x}+\dfrac{\sin x}{1+\cos x}=\dfrac{(1+\cos x)^2}{\sin x (1+\cos x)}+\dfrac{\sin^2 x}{\sin x(1+\cos x)}=\\\\\\=\dfrac{1+2\cos x+\cos^2 x+\sin^2 x}{\sin x(1+\cos x)}=\dfrac{1+2\cos x+1}{\sin x(1+\cos x)}=\\\\\\=\dfrac{2(1+\cos x)}{\sin x(1+\cos x)}=\dfrac{2}{\sin x}=\text{P}[/tex]
Zadanie 2
[tex]\text{L}=\dfrac{\text{tg }x}{\text{tg }x+\text{ctg }x}=\dfrac{\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}=\dfrac{\frac{\sin x}{\cos x}}{\frac{\sin^2 x}{\sin x \cos x}+\frac{\cos^2 x}{\sin x \cos x}}=\\\\\\=\dfrac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x \cos x}}=\dfrac{\sin ^2 x\cos x}{\cos x}=\sin^2 x=\text{P}[/tex]
Zadanie 3
[tex]\text{L}=\dfrac{\sin 2\alpha}{1+\cos 2\alpha}=\dfrac{2\sin\alpha\cos\alpha}{1+\cos^2\alpha-\sin^2\alpha}=\dfrac{2\sin\alpha\cos\alpha}{\cos^2\alpha+(1-\sin^2\alpha)}=\\\\\\=\dfrac{2\sin\alpha\cos\alpha}{\cos^2\alpha+\cos^2\alpha}=\dfrac{2\sin\alpha\cos\alpha}{2\cos^2\alpha}=\dfrac{\sin\alpha}{\cos\alpha}=\text{tg }\alpha=\text{P}[/tex]
