A geometry problem by Waxim Channa

We have:

$\tan1^{\circ}=\cot89^{\circ}=\dfrac{1}{\tan89^{\circ}}\Rightarrow \tan1^{\circ}\tan89^{\circ}=1$

$\tan5^{\circ}=\cot85^{\circ}=\dfrac{1}{\tan85^{\circ}}\Rightarrow \tan5^{\circ}\tan85^{\circ}=1$

$\tan15^{\circ}=\cot75^{\circ}=\dfrac{1}{\tan75^{\circ}}\Rightarrow \tan15^{\circ}\tan75^{\circ}=1$

So, $\tan1^{\circ}\tan5^{\circ}\tan15^{\circ}\tan75^{\circ}\tan85^{\circ}\tan89^{\circ}=\boxed{1}$

since,

tan (90-x) =1/tanx

..therefore, tan 89 =tan (90-1) =1/tan 1

tan _ 85 =tan (90-5) =1/tan 5 _

tan 75 =tan (90-15) =1/tan 15

so that, (Tan1 $\times$ tan5 $\times$ tan15 $\times$ tan75 $\times$ tan85 $\times$ tan89) =

(Tan1 $\times$ tan5 $\times$ tan15 x $\frac{1}{tan15}$ x $\frac{1}{tan5}$ x $\frac{1}{tan15}$ =

$\boxed{1}$

There is no tan89° term in the question

Couple the terms which give a sum of 90 degrees. So, let us consider the first pair: tan(1°)*tan(89°)

We know that tan(90°)=(tan(1°)+tan(89°)) / (1-tan(1°) * tan(89°))

The denominator must be zero, hence tan(1°) * tan(89°) is 1.

Similarly for the pairs: tan(5°) * tan(85°) and tan(15°) * tan(75°).

So, we effectively get 1 * 1 * 1, hence the answer is 1.