2 tan (pi/2) / 1- tan^2 (pi/2)

tan(2θ) = 2*tan(θ)/(1- tan²(θ))

 

Taking the limit as θ→pi/2 we have 2*tan(pi/2)/(1- tan²(pi/2)) = tan(pi) = 0.

 

(Geno, I think you got a sign wrong: should be 1 + tan² = sec² in your fourth line)

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Tan(π/2) is undefined; thus making the expression undefined.

However, if you are willing to play with the expression, you can do this:

tan(π/2)  =  sin(π/2) / cos(π/2)

1 - tan²(π/2)  =  sec²(π/2)  =  1 / cos²(π/2) 

So:  2·tan(π/2) / [1 - tan²(π/2) ]  =  [ 2·sin(π/2) / cos(π/2) ] / [ 1 / cos²(π/2) ]

=  [ 2·sin(π/2) / cos(π/2) ] · [ 1 / cos²(π/2) ]

= 2·sin(π/2)·cos(π/2)                               Since  sin(2x) = 2xin(x)cos(x):

= sin(2 · π/2)  =  sin(π)  =  0