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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

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

The key observation is, both products have the same number of terms.

More details on how to simplify the middle term inside the product:

$$(1+\tan r^\circ)(1+\cot (45^\circ + r^\circ))$$ $$ = (1+\tan r^\circ) \left( 1+{1 \over \tan (45^\circ + r^\circ)} \right) $$ $$ = (1+\tan r^\circ) \left( 1+{1 - \tan 45^\circ \times \tan r^\circ \over \tan 45^\circ + \tan r^\circ} \right) $$ $$ = (1+\tan r^\circ) \left( 1+{1 - \tan r^\circ \over 1 + \tan r^\circ} \right) $$ $$ = (1+\tan r^\circ) \left( {2 \over 1 + \tan r^\circ} \right) $$ $$ = 2 $$

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