Simplify $\frac{tan x + sec x - 1}{tan x - sec x + 1}$ $(tanx+secx-1)/(tanx-secx+1)$ ?

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Simplify $\frac{tan x + sec x - 1}{tan x - sec x + 1}$ $(tanx+secx-1)/(tanx-secx+1)$ ?

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Answer 1 · 2016-04-29T10:18:28

Please see the proof below.

Explanation:

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

Multiplying numerator and denominator by $tan x + sec x + 1$ $tanx+secx+1$

= $\frac{tan x + sec x - 1}{tan x - sec x + 1} \times \frac{tan x + sec x + 1}{tan x + sec x + 1}$ $(tanx+secx-1)/(tanx-secx+1)xx(tanx+secx+1)/(tanx+secx+1)$

= $\frac{{tan}^{2} x + tan x sec x + tan x + tan x sec x + {sec}^{2} x + sec x - tan x - sec x - 1}{{tan}^{2} x + tan x sec x + tan x - tan x sec x - {sec}^{2} x - sec x + tan x + sec x + 1}$ $(tan^2x+tanxsecx+tanx+tanxsecx+sec^2x+secx-tanx-secx-1)/(tan^2x+tanxsecx+tanx-tanxsecx-sec^2x-secx+tanx+secx+1)$

= $\frac{{tan}^{2} x + 2 tan x sec x + {sec}^{2} x - 1}{{tan}^{2} x + 2 tan x - {sec}^{2} x + 1}$ $(tan^2x+2tanxsecx+sec^2x-1)/(tan^2x+2tanx-sec^2x+1)$

As ${sec}^{2} x = {tan}^{2} x + 1$ $sec^2x=tan^2x+1$ , above is equal to

= $\frac{{tan}^{2} x + 2 tan x sec x + {tan}^{2} x + 1 - 1}{{tan}^{2} x + 2 tan x - {tan}^{2} x - 1 + 1}$ $(tan^2x+2tanxsecx+tan^2x+1-1)/(tan^2x+2tanx-tan^2x-1+1)$

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

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

= $tan x + sec x$ $tanx+secx$

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Simplify $\frac{tan x + sec x - 1}{tan x - sec x + 1}$ $(tanx+secx-1)/(tanx-secx+1)$ ?

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Explanation:

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Simplify $\frac{tan x + sec x - 1}{tan x - sec x + 1}$ $(tanx+secx-1)/(tanx-secx+1)$ ?