Product of Trigonometric function 
The value of $$\prod^{10}_{r=1}\bigg(1+\tan r^\circ\bigg)\cdot \prod^{55}_{r=46}\bigg(1+\cot r^\circ\bigg)$$

Attempt: $\displaystyle \prod^{10}_{r=1}\bigg(1+\tan r^\circ\bigg)=(1+\tan 1^\circ)(1+\tan 9^\circ)\cdots \cdots (1+\tan 4^\circ)(1+\tan 6^\circ)\tan 5^\circ$
from $\tan(A+B) = \tan 10^\circ\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B} = \tan 10^\circ$
could some help me to solve it, thanks

$$ \prod^{10}_{r=1}\bigg(1+\tan r^\circ\bigg)\cdot \prod^{55}_{r=46}\bigg(1+\cot r^\circ\bigg) $$
$$ \prod^{10}_{r=1}\bigg(   1+\tan r^\circ   \bigg)\cdot \prod^{10}_{r=1}\bigg( 
  1+\cot (45^\circ + r^\circ)   \bigg)$$
$$ = \prod^{10}_{r=1}\bigg(   (1+\tan r^\circ)(1+\cot (45^\circ + r^\circ))   \bigg) $$
$$ = \prod^{10}_{r=1}   2   $$
$$ = 2^{10} $$
The key observation is, both products have the same number of terms.
More details on how to simplify the middle term inside the product:
$$(1+\tan r^\circ)(1+\cot (45^\circ + r^\circ))$$
$$ = (1+\tan r^\circ) \left( 1+{1 \over \tan (45^\circ + r^\circ)} \right) $$
$$ = (1+\tan r^\circ) \left( 1+{1 - \tan 45^\circ \times \tan r^\circ \over \tan 45^\circ + \tan r^\circ} \right) $$
$$ = (1+\tan r^\circ) \left( 1+{1 - \tan r^\circ \over 1 + \tan r^\circ} \right) $$
$$ = (1+\tan r^\circ) \left( {2 \over 1 + \tan r^\circ} \right) $$
$$ = 2 $$
