rocedwrp

2021-08-13

If $\mathrm{sec}\theta +\mathrm{tan}\theta =x$, obtain the values of $\mathrm{sec}\theta ,\mathrm{tan}\theta$ and $\mathrm{sin}\theta$.

un4t5o4v

We have to use properties of trigonometry.
$\left(\mathrm{sec}\theta +\mathrm{tan}\theta \right)\frac{\left(\mathrm{sec}\theta -\mathrm{tan}\theta \right)}{\left(\mathrm{sec}\theta -\mathrm{tan}\theta \right)}=x$
$\to \frac{{\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta }{\mathrm{sec}\theta -\mathrm{tan}\theta }=x$
$\to \mathrm{sec}\theta -\mathrm{tan}\theta =\frac{1}{x}$
and $\mathrm{sec}\theta +\mathrm{tan}\theta =x$
$\mathrm{sec}\theta =\frac{1}{2}\left(x+\frac{1}{x}\right)$
$\mathrm{tan}\theta =\frac{1}{2}\left(x-\frac{1}{x}\right)$
$\to \frac{1}{\mathrm{cos}\theta }=\frac{1}{2}\left(\frac{1+{x}^{2}}{x}\right)$
$\to \mathrm{cos}\theta =\frac{2x}{1+{x}^{2}}$
$\mathrm{sin}\theta =\sqrt{1-\frac{4{x}^{2}}{{\left(1+{x}^{2}\right)}^{2}}}=\sqrt{\frac{{\left({x}^{2}-1\right)}^{2}}{{\left({x}^{2}+1\right)}^{2}}}$
$\mathrm{sin}\theta =\frac{{x}^{2}-1}{{x}^{2}+1}$

Jeffrey Jordon