# Solve the equation frac{1-sin x}{1+sin x}=(sec x-tan x)^{2}

Solve the equation $\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}=\left(\mathrm{sec}x-\mathrm{tan}x{\right)}^{2}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

hosentak
$\left(\mathrm{sec}\left(x\right)-\mathrm{tan}\left(x\right){\right)}^{2}=\left(\frac{1}{\mathrm{cos}\left(x\right)}-\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}{\right)}^{2}$
$=\left(\frac{1-\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}{\right)}^{2}$
$=\frac{\left(1-\mathrm{sin}\left(x\right){\right)}^{2}}{co{s}^{2}\left(x\right)}$
$=\frac{\left(1-\mathrm{sin}\left(x\right){\right)}^{2}}{1-si{n}^{2}\left(x\right)}$
$=\frac{-\left(1-\mathrm{sin}\left(x\right){\right)}^{2}}{\left(\mathrm{sin}\left(x\right)+1\right)\left(\mathrm{sin}\left(x\right)-1\right)}$
$=\frac{1-\mathrm{sin}\left(x\right)}{\mathrm{sin}\left(x\right)+1}$
$=\frac{1-\mathrm{sin}\left(x\right)}{1+\mathrm{sin}\left(x\right)}$
Therefore
$\frac{1-\mathrm{sin}\left(x\right)}{1+\mathrm{sin}\left(x\right)}=\left(\mathrm{sec}\left(x\right)-\mathrm{tan}\left(x\right){\right)}^{2}$