$\frac{1-tanA}{1+tanA}=\frac{tan3°tan15°tan30°tan75°tan87°}{tan27°tan39°tan51°tan60°tan63°}$, then the value of cot A is :

2 

$\frac{1-tanA}{1+tanA}=\frac{tan3°tan15°tan30°tan75°tan87°}{tan27°tan39°tan51°tan60°tan63°}$

We know,

tanA × tanB = 1  iff  A + B = 90°

So, tan3° × tan87° = 1

tan15° × tan75° = 1

tan27° × tan63° = 1

tan39° × tan51° = 1

tan30° = \(\frac{1 }{√3}\)

tan60° = √3

Now,

\(\frac{1-tanA }{1+tanA}\) = \(\frac{1/√3 }{√3}\)

\(\frac{1-tanA }{1+tanA}\) = \(\frac{1 }{3}\)

3 - 3tanA = 1 +  tanA

tanA = \(\frac{1 }{2}\)

cotA = 2