DofotheroU

2021-08-12

Prove
$\mathrm{sec}\theta +\frac{1}{\mathrm{tan}\left(\theta \right)}=\frac{\mathrm{tan}\left(\theta \right)}{\mathrm{sec}\left(\theta \right)}-1$

i1ziZ

Prove:
$\frac{\mathrm{sec}\theta +1}{\mathrm{tan}\theta }=\frac{\mathrm{tan}\theta }{\mathrm{sec}\theta -1}$ Right hand side
$\frac{\mathrm{tan}\theta }{\mathrm{sec}\theta -1}=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta \left(\frac{1}{\mathrm{cos}\theta }-1\right)}=\frac{\mathrm{sin}\theta }{1-\mathrm{cos}\theta }$
Razionalize the function
$\frac{\mathrm{sin}\theta }{1-\mathrm{cos}\theta }×\frac{1+\mathrm{cos}\theta }{1+\mathrm{cos}\theta }=\frac{\mathrm{sin}\theta \left(1+\mathrm{cos}\theta \right)}{1-{\mathrm{cos}}^{2}\theta }$
$=\frac{\mathrm{sin}\theta +\mathrm{sin}\theta \mathrm{cos}\theta }{{\mathrm{sin}}^{2}\theta }=\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }$
$=\frac{1+\mathrm{cos}\theta }{\mathrm{sin}\theta }$
Also Left hand side
$\frac{\mathrm{sec}\theta +1}{\mathrm{tan}\theta }=\frac{\frac{1}{\mathrm{cos}\theta }+1}{\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }}=\frac{1+\mathrm{cos}\theta }{\mathrm{sin}\theta }$
which is equal to right hand

Jeffrey Jordon