"MATEMĀTIKA 11. KLASEI"
No identitātēm $\mathrm{tg}\mathrm{\alpha }=\frac{\mathrm{sin}\mathrm{\alpha }}{\mathrm{cos}\mathrm{\alpha }}$ un $\mathrm{ctg}\mathrm{\alpha }=\frac{\mathrm{cos}\mathrm{\alpha }}{\mathrm{sin}\mathrm{\alpha }}$ seko, ka

1) $\mathrm{tg}\mathrm{\alpha }\cdot \mathrm{ctg}\mathrm{\alpha }=1$

2) $\mathrm{tg}\mathrm{\alpha }=\frac{1}{\mathrm{ctg}\mathrm{\alpha }};\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathrm{ctg}\mathrm{\alpha }=\frac{1}{\mathrm{tg}\mathrm{\alpha }}$

Izmantojot arī pamatidentitāti${\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }=1$, var iegūt jaunas sakarības.

Dalot identitātes ${\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }=1$ abas puses ar sinusa kvadrātu, ja ${\mathrm{sin}}^{2}\mathrm{\alpha }\ne 0$, iegūst formulu

$\frac{{\mathrm{sin}}^{2}\mathrm{\alpha }}{{\mathrm{sin}}^{2}\mathrm{\alpha }}+\frac{{\mathrm{cos}}^{2}\mathrm{\alpha }}{{\mathrm{sin}}^{2}\mathrm{\alpha }}=\frac{1}{{\mathrm{sin}}^{2}\mathrm{\alpha }}$ jeb
$1+{\mathrm{ctg}}^{2}\mathrm{\alpha }=\frac{1}{{\mathrm{sin}}^{2}\mathrm{\alpha }}$, kur $\mathrm{\alpha }\ne \mathrm{\pi }n,\phantom{\rule{0.147em}{0ex}}n\in \mathrm{ℤ}$

Dalot identitātes ${\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }=1$ abas puses ar kosinusa kvadrātu, ja ${\mathrm{cos}}^{2}\mathrm{\alpha }\ne 0$, iegūst formulu

$\frac{{\mathrm{sin}}^{2}\mathrm{\alpha }}{{\mathrm{cos}}^{2}\mathrm{\alpha }}+\frac{{\mathrm{cos}}^{2}\mathrm{\alpha }}{{\mathrm{cos}}^{2}\mathrm{\alpha }}=\frac{1}{{\mathrm{cos}}^{2}\mathrm{\alpha }}$ jeb
${\mathrm{tg}}^{2}\mathrm{\alpha }+1=\frac{1}{{\mathrm{cos}}^{2}\mathrm{\alpha }}$, kur $\mathrm{\alpha }\ne \frac{\mathrm{\pi }}{2}+\mathrm{\pi }n,\phantom{\rule{0.147em}{0ex}}n\in \mathrm{ℤ}$
Piemērs:
Vienkāršo izteiksmi $\frac{1}{{\mathit{cos}}^{2}x}-{\mathit{tg}}^{2}x$
Risinājums:
$\begin{array}{l}\frac{1}{{\mathit{cos}}^{2}x}-{\mathit{tg}}^{2}x=\\ ={\mathit{tg}}^{2}x+1-{\mathit{tg}}^{2}x=1\end{array}$
Abas sakarības ir atrodamas Matemātika II formulu lapā.