Prove identity: tanx-1/tanx+1= 1-cotx/1+cotx

to answer the question above, take the LHS. 
[(tan x - 1) / (tan x + 1)] = 

Remember that tan x = 1 / cot x. 
{[(1 / cot x) - 1] / [(1 / cot x) + 1]} = 

The LCD is cot x. Multiply as needed to get the common denominator for all terms. 
{[(1 / cot x) - 1(cot x / cot x)] / [(1 / cot x) + 1(cot x / cot x)]} = 
{[(1 / cot x) - (cot x / cot x)] / [(1 / cot x) + (cot x / cot x)]} = 

Then Simplify. 
[(1 - cot x) / cot x] / [(1 + cot x) / cot x] = 

Remember that (a / b) / (c / d) = (a / b) * (d / c). 
[(1 - cot x) / cot x] * [cot x / (1 + cot x)] = 
[(1 - cot x) / (1 + cot x )] = 
RHS
The answer is
 [(tan x - 1) / (tan x + 1)] = [(1 - cot x) / (1 + cot x )]