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