# If csc A= -5/3 and csc B = -13/5 with angle A and angle B in quadrant 3. What is value of cos(A-B)?

Jul 23, 2016

$\cos \left(A - B\right) = \frac{63}{65}$.

#### Explanation:

We start by taking note that angles $A \mathmr{and} B$ are in Q_(III, so that, $\cos A \mathmr{and} \cos B$ both will be $- v e$.

$\csc A = - \frac{5}{3} \Rightarrow \sin A = - \frac{3}{5} \Rightarrow \cos A = - \sqrt{1 - {\sin}^{2} A} = - \sqrt{1 - \frac{9}{25}} = - \frac{\sqrt{16}}{25} = - \frac{4}{5}$

Similarly, $\csc B = - \frac{13}{5} \Rightarrow \cos B = - \frac{12}{13}$

Therefore, reqd. value$= \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

$= \left(- \frac{4}{5}\right) \left(- \frac{12}{13}\right) + \left(- \frac{3}{5}\right) \left(- \frac{5}{13}\right)$

$\frac{48}{65} + \frac{15}{65} = \frac{63}{65}$.