Odpowiedź i szczegółowe wyjaśnienie:
[tex]a)\ \dfrac{1}{1-cosx}+\dfrac{1}{1+cosx}=\dfrac{2}{sin^2x}\\\\P=\dfrac{2}{sin^2x}\\\\L=\dfrac{1}{1-cosx}+\dfrac{1}{1+cosx}=\dfrac{1\cdot(1+cosx)}{(1-cosx)(1+cosx)}+\dfrac{1\cdot(1-cosx)}{(1-cosx)(1+cosx)}=\\\\\\=\dfrac{1+cosx}{1-cos^2x}+\dfrac{1-cosx}{1-cos^2x}=\dfrac{1+cosx+1-x}{1-cos^2x}=\dfrac{2}{sin^2x}\\\\L=P[/tex]
[tex]b)\ tg^2x-sin^2x=tg^2x\cdot sin^2x\\\\P=tg^2x\cdot sin^2x\\L=\dfrac{sin^2x}{cos^2x}-sin^2x=\dfrac{sin^2x}{cos^2x}-\dfrac{sin^2xcos^2x}{cos^2x}=\dfrac{sin^2x-sin^2xcos^2x}{cos^2x}=\\\\=\dfrac{sin^2x(1-cos^2x)}{cos^2x}=\dfrac{sin^2x}{cos^2x}(1-cos^2x)=tg^2x\cdot sin^2x\\\\L=P[/tex]
[tex]c)\ \dfrac{tgx}{1-tg^2x}\cdot\dfrac{ctg^2x-1}{ctgx}=1\\\\P=1\\L=\dfrac{tgx}{1-tg^2x}\cdot\dfrac{ctg^2x-1}{ctgx}=\dfrac{tgxctg^2x-tgx}{ctgx-ctgx tg^2x}=\dfrac{\frac{1}{ctgx}ctg^2x-\frac{1}{ctgx}}{ctgx-ctgx\cdot\frac{1}{ctg^2x}}=\\\\=\dfrac{ctgx-\frac{1}{ctgx}}{ctgx-\frac{1}{ctgx}}=1\\\\L=P\\[/tex]
[tex]d)\ \dfrac{sin^2x}{sinx-cosx}+\dfrac{sinx+cosx}{1-tg^2x}=sinx+cosx\\\\P=sinx+cosx\\\\L=\dfrac{sin^2x}{sinx-cosx}+\dfrac{sinx+cosx}{1-tg^2x}=\\\\\dfrac{sin^2x(1-tg^2x)}{(sinx-cosx)(1-tg^2x)}+\dfrac{(sinx+cosx)(sinx-cosx)}{(sinx-cosx)(1-tg^2x)}=\\\\\\=\dfrac{sin^2x(1-tg^2x)+sin^2x-cos^2x}{(sinx-cosx)(1-tg^2x)}=\\\\\\=\dfrac{(1-cos^2x)(1-\frac{sin^2x}{cos^2x})+sin^2x-cos^2x}{(sinx-cosx)(1-tg^2x)}=\\\\\\=\dfrac{1-\frac{sin^2x}{cos^2x}-cos^2x+sin^2x+sin^2x-cos^2x}{(sinx-cosx)(1-tg^2x)}=\\[/tex]
[tex]=\dfrac{\frac{cos^2x}{cos^2x}-\frac{sin^2x}{cos^2x}+sin^2x-cos^2x+sin^2x-cos^2x}{(sinx-cosx)(1-tg^2x)}=\\\\\\=\dfrac{\frac{cos^2x-sin^2x}{cos^2x}+sinx^2x-cos^2x+sin^2x-cos^2x}{(sinx-cosx)(\frac{cos^2x}{cos^2x}-\frac{sin^2x}{cos^2x})}=\\\\\\=\dfrac{\frac{-sin^2x+cos^2x}{cos^2x}+sin^2x-cos^2x+sin^2x-cos^2x}{(sinx-cosx)(\frac{cos^2x-sin^2x}{cos^2x})}=\\\\\\=\dfrac{-\frac{sin^2x-cos^2x}{cos^2x}+sin^2x-cos^2x+sin^2x-cos^2x}{(sinx-cosx)(\frac{cos^2x-sin^2x}{cos^2x})}=\\[/tex]
[tex]=\dfrac{(sin^2x-cos^2)(-\frac{1}{cos^2}+1+1)}{(sinx-cosx)(\frac{cos^2x-sin^2x}{cos^2x})}=\\\\\\=\dfrac{(sinx-cosx)(sinx+cosx)(2-\frac{1}{cos^2x})}{(sinx-cosx)(\frac{cos^2x-sin^2x}{cos^2x})}=\\\\\\=\dfrac{(sinx+cosx)(\frac{2cos^2x}{cos^2x}-\frac{sin^2x+cos^2x}{cos^2x})}{\frac{cos^2x-sin^2x}{cos^2x}}=\\\\\\=\dfrac{(sinx+cosx)(\frac{2cos^2x-sin^2x-cos^2x}{cos^2x})}{\frac{cos^2x-sin^2x}{cos^2x}}=\\\\\\=\dfrac{(sinx+cosx)(\frac{cos^2x-sin^2x}{cos^2x})}{\frac{cos^2x-sin^2x}{cos^2x}}=\\\\=sin+cosx\\[/tex]
L=P
