How do you verify $\frac{1 + tan x}{1 - tan x} = \frac{cos x + sin x}{cos x - sin x}$ $(1+tanx)/(1-tanx) = (cosx+sinx)/(cosx-sinx)$ ?

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How do you verify $\frac{1 + tan x}{1 - tan x} = \frac{cos x + sin x}{cos x - sin x}$ $(1+tanx)/(1-tanx) = (cosx+sinx)/(cosx-sinx)$ ?

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Answer 1 · 2016-03-09T10:27:14

see explanation

Explanation:

manipulate the left side

$\Rightarrow \frac{1 + tan x}{1 - tan x} = \frac{1 + \frac{sin x}{cos x}}{1 - \frac{sin x}{cos x}}$ $rArr (1+tanx)/(1-tanx) = (1+sinx/cosx)/(1-sinx/cosx)$

rewrite numerator / denominator as single fractions.

$= \frac{\frac{cos x + sin x}{cos x}}{\frac{cos x - sin x}{cos x}}$ $=((cosx+sinx)/cosx)/((cosx-sinx)/(cosx)$

multiply the fractions by ' inverting' the fraction on denominator.

$rArr (cosx+sinx)/cancel(cosx) xx (cancel(cosx))/(cosx-sinx)$

$= (cosx+sinx)/(cosx-sinx) = " right side "$

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How do you verify $\frac{1 + tan x}{1 - tan x} = \frac{cos x + sin x}{cos x - sin x}$ $(1+tanx)/(1-tanx) = (cosx+sinx)/(cosx-sinx)$ ?

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Explanation:

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1 Answer

Explanation:

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How do you verify $\frac{1 + tan x}{1 - tan x} = \frac{cos x + sin x}{cos x - sin x}$ $(1+tanx)/(1-tanx) = (cosx+sinx)/(cosx-sinx)$ ?