How do you prove $\frac{1 + tan x}{1 + cot x} = 2$ $(1+tan x) / (1+cot x) = 2$ ?

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Answer 1 · 2015-07-11T21:08:31

Let's suppose we were to even go through with this.

$\frac{1 + tan x}{1 + (\frac{1}{tan x})} \cdot \frac{tan x}{tan x}$ $(1+tanx)/(1+(1/tanx))*(tanx)/(tanx)$

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

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

$= (tanx+1)/(tanx+1) - cancel(1/(tanx+1)) + (tan^2x - 1)/(tanx+1) + cancel(1/(tanx+1))$

$= (tanx+1)/(tanx+1) + (tan^2x - 1)/(tanx+1)$

$= 1 + ((tanx - 1)cancel((tanx+1)))/cancel((tanx+1))$

$= 1 + tanx - 1$

$= color(blue)(tanx)$

So clearly, this is not true. This is equal to $tanx$ .

How do you prove (1+tan x) / (1+cot x) = 21+tanx1+cotx=2(1+tan x) / (1+cot x) = 2?