Just for fun!-VIII

Since, $\tan k^{o} \times \tan (90-k)^{o} = \tan k^{o} \times \cot k^{o} =1$

So, $\tan 1^{o} \tan 2^{o} \ldots \tan 89^{o} = (1 \times 1 \times \ldots \text{'44' times} \ldots 1) \times \tan 45^{o}$

But, $\tan 45^{o} = 1$ . So,

$\tan 1^{o} \tan 2^{o} \ldots \tan 89^{o} =\boxed{1}$

Method 1: $tan1°tan2°tan3°tan4°tan5°...tan89°\\ or\\ tan(90-89)°.tan(90-88)°....tan45°....tan88°.tan89°\\ or\\ cot89°.cot88°.....tan45°....tan88°.tan89°\\ \\ Now,\quad tan\theta \times cot\theta =1\\ \\ \Rightarrow \quad (1.1.1.....44times)\times tan45°\\ or\quad tan45°\quad =\quad 1$

Method 2: $tan1°tan2°tan3°tan4°tan5°...tan89°\\ or\\ \frac { sin1°sin2°sin3°....sin89° }{ cos1°cos2°cos3°....cos89° } \\ or\\ \frac { cos89°cos88°cos87°....cos3°cos2°cos1° }{ cos1°cos2°cos3°....cos89° } \\ Since\quad every\quad term\quad gets\quad cancelled\quad out,\\ so\quad the\quad answer\quad is\quad \boxed { 1 }$

= sin1.sin2.sin3..........sin89/ cos1.cos2......cos89

Now, cos89/sin1 = 1, cos88/sin2 = 1 Similarly the whole division = 1

(tan1 x tan89) x( tan2 xtan88)...............xtan45 = 1