1-tan/1+tan = cot-1/cot+1,,,please prove?

Question

1-tan/1+tan = cot-1/cot+1,,,please prove?

Answer 1

replace tan = 1/cot then take LCM

1-1/cot / 1+1/cot = cot-1/cot+1

Answer 2

1- sin(z)/cos(z) / 1+sin(z)cos(z)

=1/sin(z) -1/cos(z) / 1/sin(z) + 1/cos(z)

=cos(z)-sin(z)/cos(z)sin(z)/ cos(z)+sin(z)/ cos(z)sin(z)

=cos(z)-sin(z)/ cos(z)+sin(z)

cot-1/ cot+1

cos(z)/sin(z) -1/ cos(z)/sin(z)+1

cos(z)(1/sin(z)-1/cos(z)) /cos(z)(sin(z)+ 1/cos(z))

cos(z)-sin(z)/ sin(z)cos(z)/ (cos(z)+sin(z)/sin(z)cos(z))

=cos(z)-sin(z)/ cos(z)+sin(z)

=cot(z)-1/cot(z)+1

Answer 3

GIVEN

1-tan/1+tan = cot-1/cot+1

The left hand side of the identity can be rewritten as

(1 - sin/cos)/(1 + sin/cos) =

[(cos - sin)/(cos)]/[(cos + sin)/(cos)] =

NOTE the "cos" will cancel out, hence

(cos - sin)/(cos + sin) = This is the left hand side of the identity

At this point, work on the right hand side of the identity, as in

cot-1/cot+1 = [(cos/sin) - 1]/[(cos/sin +1]

= [(cos - sin)sin]/[(cos + sin)/sin}

The "sin" will cancel out, hence

= (cos - sin)/(cos + sin) = this is the right hand side of the identity simplified

NOTE -- since we have established that the left hand side = the right hand side as in

(cos - sin)/(cos + sin) = (cos - sin)/(cos + sin) THEN the identity is proven.

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A second solution will be to simplify the left hand side of the equation as

(1 - 1/cot)/(1 + 1/cot) ---- since tan = 1/cot

Therefore,

{(cot - 1)/cot}/{(cot + 1) /cot}

The "cot" will cancel out, hence

(cot - 1) / (cot + 1) = (cot - 1) / (cot + 1) --- Identity proven

Answer 4

Cot x = 1/tan x

1- tan / 1+tan = (1/tan - 1)/(1/tan +1)

                         ((1-tan)/tan) / ((1+tan)/tan)=1-tan / 1 + tan

provided that tan <> 0 + n*pi/2