hasil dari2 tan 37,5 / 1-tan 37,5 adalah....

Trigonometry.

Step 1. Determine sin 75° and cos 75°.
sin (α + β) = sin α cos β + cos α sin β
sin 75° = sin (45° + 30°)
            = sin 45° cos 30° + cos 45° sin 30°
            = 1/2 √2 (1/2 √3) + 1/2 √2 (1/2)
            = 1/4 (√6 + √2)

cos (α + β) = cos α cos β - sin α sin β
cos 75° = cos (45° + 30°)
             = cos 45° cos 30° - sin 45° sin 30°
             = 1/2 √2 (1/2 √3) - 1/2 √2 (1/2) 
             = 1/4 (√6 - √2)

Step 2. Calculate tan 37,5°.
We use tan (A / 2) = sin A / (1 + cos A)
[tex]\displaystyle \tan \left ( \frac{A}{2} \right )=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}\\ \tan 37,5^{\circ}=\frac{\sin 75^{\circ}}{1+\cos 75^{\circ}}\\ =\frac{\frac{\sqrt{6}+\sqrt{2}}{4}}{1+\frac{\sqrt{6}-\sqrt{2}}{4}}\\ =\frac{\sqrt{6}+\sqrt{2}}{4+\sqrt{6}-\sqrt{2}}[/tex]

Step 3. Calculate your problem.
[tex]\displaystyle \frac{2\tan 37,5^{\circ}}{1-\tan 37,5^{\circ}}\\ =\frac{2\left ( \frac{\sqrt{6}+\sqrt{2}}{4+\sqrt{6}-\sqrt{2}} \right )}{1-\frac{\sqrt{6}+\sqrt{2}}{4+\sqrt{6}-\sqrt{2}}}\\ =\frac{2\left ( \sqrt{6}+\sqrt{2} \right )}{4-2\sqrt{2}}\\ =\frac{2\sqrt{6}+2\sqrt{2}}{4-2\sqrt{2}}\frac{4+2\sqrt{2}}{4+2\sqrt{2}}\\ =1+\sqrt{2}+\sqrt{3}+\sqrt{6}[/tex]