How do you prove $\frac{1}{sec x - tan x} - \frac{1}{sec x + tan x} = 2 tan x$ $1/(secx-tanx) - 1/(secx+tanx)=2tanx$ ?

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Answer 1 · 2016-05-11T18:13:50

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Use Property: ${sec}^{2} x = 1 + {tan}^{2} x$ $sec^2x=1+tan^2x$

Left Side: $= \frac{1}{sec x - tan x} - \frac{1}{sec x + tan x}$ $=1/(secx-tanx) -1/(secx+tanx)$

$= \frac{sec x + tan x - (sec x - tan x)}{(sec x - tan x) (sec x + tan x)}$ $=(secx+tanx-(secx-tanx))/((secx-tanx)(secx+tanx))$

$= \frac{sec x + tan x - sec x + tan x}{{sec}^{2} x - {tan}^{2} x}$ $=(secx+tanx-secx+tanx)/(sec^2x-tan^2x)$

$= \frac{2 tan x}{1}$ $=(2tanx)/1$

$= 2 tan x$ $=2tanx$

$=$ $=$ Right Side

How do you prove 1secx−tanx−1secx+tanx=2tanx 1/(secx-tanx) - 1/(secx+tanx)=2tanx?