How do you prove $\frac{tan (x) - 1}{tan (x) + 1} = \frac{1 - cot (x)}{1 + cot (x)}$ $(tan(x)-1)/(tan(x)+1)= (1-cot(x))/(1+cot(x))$ ?

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Answer 1 · 2015-04-19T20:14:42

Start off by cross multiplying

$(tan x - 1) (1 + cot x) = (tan x + 1) (1 - cot x)$ $(tanx-1)(1+cotx)=(tanx+1)(1-cotx)$

Expand each side using FOIL

$tan x + tan x cot x - 1 - cot x$ $tanx+tanxcotx-1-cotx$
$= tan x - tan x cot x + 1 - cot x$ $=tanx-tanxcotx+1-cotx$

Since $tan x$ $tanx$ and $cot x$ $cotx$ are reciprocals

$tan x cot x = 1$ $tanxcotx=1$

Now we can write

$tan x + 1 - 1 - cot x = tan x - 1 + 1 - cot x$ $tanx+1-1-cotx=tanx-1+1-cotx$

Simplifying each side

$tan x - cot x = tan x - cot x$ $tanx-cotx=tanx-cotx$

The right hand side and left hand side are the same

Answer 2 · 2015-04-19T21:21:50

Apr 19, 2015

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How do you prove tan(x)−1tan(x)+1=1−cot(x)1+cot(x)(tan(x)-1)/(tan(x)+1)= (1-cot(x))/(1+cot(x))?