Simplifying trigonometry function by substitution I want to simplify the following expression: $\frac{{\left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + {{\tan }^2}{1^{\circ}}} \right)\left( {1 + {{\tan }^2}{2^{\circ}}} \right)}}$
My approach is as follow
$T = \left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$
$\Rightarrow \tan {58^{\circ}}\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right) = \sqrt 3  - \tan {2^{\circ}}$
$\Rightarrow \tan {59^{\circ}}\left( {1 + \sqrt 3 \tan {2^0}} \right) = \sqrt 3  - \tan {1}$
$\tan {60^{\circ}} = \frac{{\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)}}{{\left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)}} \Rightarrow \left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)$
$\tan {60^{\circ}} = \frac{{\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)}} \Rightarrow \left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)$
$T = \frac{{\sqrt 3  - \tan {1^{\circ}}}}{{\tan {{59}^{\circ}}}} \times \frac{{\sqrt 3  - \tan {2^{\circ}}}}{{\tan {{58}^{\circ}}}} \times \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right) \times \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$
After this step I am confused

$$
\tan 59^{\circ} = \frac{\sqrt{3}-\tan 1^{\circ}}{1+\sqrt{3} \tan 1^{\circ}}
$$
$$
\tan 59^{\circ}+\tan 1^{\circ} = \frac{\sqrt{3}(1+\tan^2 1^{\circ})}{1+\sqrt{3} \tan 1^{\circ}},
$$so we have
$$
(1+\sqrt{3} \tan 1^{\circ})(\tan 59^{\circ}+\tan 1^{\circ})=\sqrt{3}(1+\tan^2 1^{\circ})
$$A similar argument works for $\tan 2^{\circ}$ and $\tan 58^{\circ}$. So the answer is $3$.
