# How to prove (costheta+cosbeta)/(sintheta-sinbeta)=(sintheta+sinbeta)/(costheta-cosbeta ?

Jun 3, 2018

We use that ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

#### Explanation:

By cross multiplication and using that
$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$
we get

${\cos}^{2} \left(\theta\right) - {\cos}^{2} \left(\beta\right) = {\sin}^{2} \left(\theta\right) - {\sin}^{2} \left(\beta\right)$

Jun 4, 2018

$L H S = \frac{\cos \theta + \cos \beta}{\sin \theta - \sin \beta}$

$= \frac{\cos \theta + \cos \beta}{\sin \theta - \sin \beta} \times \frac{\cos \theta - \cos \beta}{\cos \theta - \cos \beta}$

$= \frac{\cos \theta \left(\cos \theta - \cos \beta\right) + \cos \beta \left(\cos \theta - \cos \beta\right)}{\left(\sin \theta - \sin \beta\right) \cdot \left(\cos \theta - \cos \beta\right)}$

$= \frac{{\cos}^{2} \theta \cancel{- \cos \theta \cdot \cos \beta} \cancel{+ \cos \theta \cdot \cos \beta} - {\cos}^{2} \beta}{\left(\sin \theta - \sin \beta\right) \cdot \left(\cos \theta - \cos \beta\right)}$

$= \frac{\cancel{1} - {\sin}^{2} \theta \cancel{- 1} + {\sin}^{2} \beta}{\left(\sin \theta - \sin \beta\right) \cdot \left(\cos \theta - \cos \beta\right)}$

$= \frac{\left(\sin \beta + \sin \theta\right) \cancel{\left(\sin \beta - \sin \theta\right)}}{- \cancel{\left(\sin \beta - \sin \theta\right)} \cdot \left(\cos \theta - \cos \beta\right)}$

$= \frac{\sin \theta + \sin \beta}{\cos \beta - \cos \theta} = R H S$