# How to prove the following identities cos(30°)-cos(60°)\cos(30°)+sin(60°)=tan(15°)?

Aug 12, 2018

A Proof is given in The Explanation.

#### Explanation:

$\textcolor{red}{\text{The Correct Problem is to prove that,}}$

$\textcolor{red}{\frac{\cos {30}^{\circ} - \cos {60}^{\circ}}{\cos {30}^{\circ} + \cos {60}^{\circ}} = \tan {15}^{\circ}}$.

$\text{The Expression} = \frac{\cos {30}^{\circ} - \cos {60}^{\circ}}{\cos {30}^{\circ} + \cos {60}^{\circ}}$,

$= \frac{- 2 \sin \left(\frac{{30}^{\circ} + {60}^{\circ}}{2}\right) \sin \left(\frac{{30}^{\circ} - {60}^{\circ}}{2}\right)}{2 \cos \left(\frac{{30}^{\circ} + {60}^{\circ}}{2}\right) \cos \left(\frac{{30}^{\circ} - {60}^{\circ}}{2}\right)}$,

$= \frac{- 2 \sin {45}^{\circ} \sin \left(- {15}^{\circ}\right)}{2 \cos {45}^{\circ} \cos \left(- {15}^{\circ}\right)}$,

$= \frac{\frac{1}{\sqrt{2}} \cdot \sin {15}^{\circ}}{\frac{1}{\sqrt{2}} \cdot \cos {15}^{\circ}}$,

$= \tan {15}^{\circ}$, as desired!