Prove Identity: cot(x)+1/cot(x)-1=1+tan(x)/1-tan(x)

Question

Prove from one side to another using trig identities. 

Answer 1

Proving this identity only requires one fact: cot(θ) = 1/(tan(θ)).

 

 [cot(θ) + 1]/[cot(θ) - 1] = [1/tan(θ) + 1]/[1/tan(θ) - 1]

 

numerator: 1 + 1/tan(θ) = tan(θ)/tan(θ) + 1/tan(θ) =

 

[1 + tan(θ)]/tan(θ)

 

denominator: 1/tan(θ) - 1 = 1/tan(θ) - tan(θ)/tan(θ) =

 

[1- tan(θ)]/tan(θ)

 

 

[[1 + tan(θ)]/tan(θ)] / [[1- tan(θ)]/tan(θ)] = 

 

[[1 + tan(θ)]/tan(θ)] * [tan(θ)/[1- tan(θ)]]

 

eliminating the tan(θ) in the numerator and denominator -> 

 

[1 + tan(θ)]/[1- tan(θ)].

 

Answer 2

Hi Jamie,

 

You must be missing extra parentheses, otherwise the identity doesn't hold.

 

With the extra parentheses/braces:

 

[ cot(x)+1 ] / [ cot(x)-1 ]

 

recall that cot(x) = 1 / tan(x)

 

[ cot(x)+1 ] / [ cot(x)-1 ] = [ 1 / tan(x) + 1 ] / [ 1 / tan(x) - 1 ] =

 

[ 1 / tan(x) + tan(x) / tan(x) ] / [ 1 / tan(x) - tan(x) / tan(x) ] =

 

[ ( 1 + tan(x) ) / tan(x) ] / [ ( 1 - tan(x) ) / tan(x) ] =

 

[ ( 1 + tan(x) ) / tan(x) ] * [ tan(x) / ( 1 - tan(x) ) ] =

 

[ 1 + tan(x) ] / [  1 - tan(x) ]   Q.E.D.