How do you verify the identity $\frac{tan x}{sec x - 1} = \frac{sec x + 1}{tan x}$ $(tan x)/(Sec x-1)=(sec x+1)/ (Tan x)$ ?

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Answer 1 · 2015-06-16T21:21:02

Verify the identity

Use trig identity: ${tan}^{2} x = {sec}^{2} - 1 = (sec - 1) (sec + 1)$ $tan^2 x = sec^2 - 1 = (sec - 1)(sec + 1)$
Therefor:

$\frac{tan x}{sec - 1} = \frac{sec + 1}{tan x}$ $(tan x)/(sec - 1) = (sec + 1)/(tan x)$

How do you verify the identity tanxsecx−1=secx+1tanx(tan x)/(Sec x-1)=(sec x+1)/ (Tan x)?