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) ]