# 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

$$\prod^{10}_{r=1}\bigg(1+\tan r^\circ\bigg)\cdot \prod^{55}_{r=46}\bigg(1+\cot r^\circ\bigg)$$ $$\prod^{10}_{r=1}\bigg( 1+\tan r^\circ \bigg)\cdot \prod^{10}_{r=1}\bigg( 1+\cot (45^\circ + r^\circ) \bigg)$$ $$= \prod^{10}_{r=1}\bigg( (1+\tan r^\circ)(1+\cot (45^\circ + r^\circ)) \bigg)$$ $$= \prod^{10}_{r=1} 2$$ $$= 2^{10}$$
$$(1+\tan r^\circ)(1+\cot (45^\circ + r^\circ))$$ $$= (1+\tan r^\circ) \left( 1+{1 \over \tan (45^\circ + r^\circ)} \right)$$ $$= (1+\tan r^\circ) \left( 1+{1 - \tan 45^\circ \times \tan r^\circ \over \tan 45^\circ + \tan r^\circ} \right)$$ $$= (1+\tan r^\circ) \left( 1+{1 - \tan r^\circ \over 1 + \tan r^\circ} \right)$$ $$= (1+\tan r^\circ) \left( {2 \over 1 + \tan r^\circ} \right)$$ $$= 2$$