Question

What is the tangent ratio at $15$ degrees?

Answer 1

$\tan {15}^o=2-\sqrt{3}$
Explanation:
Recell that ${15}^o$ can be written as ${60}^o-{45}^o$.
So, $\tan {15}^o=\tan ({60}^o-{45}^o)$. We have to expand this using the difference formula for tangent. The sum / difference formulas are shown in the following table.
$\tan (\alpha -\beta )=\frac{\tan \left(\alpha \right)-\tan \left(\beta \right)}{1+\tan \left(\alpha \right)\tan \left(\beta \right)}$
$\therefore \tan (60-45)=\frac{\tan \left(60\right)-\tan \left(45\right)}{1+\tan \left(60\right)\tan \left(45\right)}$

$\therefore \tan \left(15\right)=\frac{\sqrt{3}-1}{1+\left(\sqrt{3}\right)\left(1\right)}$

$\therefore \tan \left(15\right)=\frac{\sqrt{3}-1}{1+\sqrt{3}}$

$\therefore \tan \left(15\right)=\frac{\sqrt{3}-1}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}$

$\therefore \tan \left(15\right)=\frac{(\sqrt{3}-1)(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}$

$\therefore \tan \left(15\right)=\frac{\sqrt{3}-3-1+\sqrt{3}}{1-3}$

$\therefore \tan \left(15\right)=\frac{2\sqrt{3}-4}{-2}$

$\therefore \tan \left(15\right)=\frac{2(\sqrt{3}-2)}{-2}$

$\therefore \tan \left(15\right)=-(\sqrt{3}-2)$

$\therefore \tan \left(15\right)=2-\sqrt{3}$