# How do you find tan(1/2sin^-1(1/2))?

Jun 5, 2018

rarrtan(1/2(sin^(-1)(1/2))

$= \tan \left({30}^{\circ} / 2\right) \mathmr{and} \tan \left({150}^{\circ} / 2\right)$

$= \tan {15}^{\circ} \mathmr{and} \tan {75}^{\circ}$

Now, $\rightarrow \tan {15}^{\circ}$

$= \tan \left({45}^{\circ} - {30}^{\circ}\right)$

$= \frac{\tan {45}^{\circ} - \tan {30}^{\circ}}{1 + \tan {30}^{\circ} \tan {45}^{\circ}}$

$= \frac{1 - \left(\frac{1}{\sqrt{3}}\right)}{1 + \frac{1}{\sqrt{3}}}$

$= \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$

$= \frac{{\left(\sqrt{3}\right)}^{2} - 2 \sqrt{3} + {1}^{2}}{{\left(\sqrt{3}\right)}^{2} - {1}^{2}} = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$

Similary, $\rightarrow \tan {75}^{\circ}$

$= \tan \left({45}^{\circ} + {30}^{\circ}\right)$

$= \frac{\tan {45}^{\circ} + \tan {30}^{\circ}}{1 - \tan {30}^{\circ} \tan {45}^{\circ}}$

$= \frac{1 + \left(\frac{1}{\sqrt{3}}\right)}{1 - \frac{1}{\sqrt{3}}}$

$= \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$= \frac{{\left(\sqrt{3}\right)}^{2} + 2 \sqrt{3} + {1}^{2}}{{\left(\sqrt{3}\right)}^{2} - {1}^{2}} = \frac{4 + 2 \sqrt{3}}{2} = 2 + \sqrt{3}$