<s>[INST] 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
 [/INST] Generalization:
$$\dfrac{(1+\tan y\tan z)(\tan z+\tan(y-z))}{1+\tan^2z} =\cdots =\dfrac{\cos(y-z)\sin(z+y-z)\cos^2z}{\cos y\cos z\cos z\cos(y-z)} =\tan y$$
$$\text{ if }\cos(y-z)\cos z\ne0$$
Here $y=60^\circ$
Then $z=1^\circ, 2^\circ$
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