How to find the values of: a) cos (A-B) b) tan (A+B) while angle is obtuse with sin A = 3/5 and angle B is acute with sin B = 12/13 ?

Jul 27, 2018

Below

Explanation:

If angle A is obtuse, then using a right-angled triangle,

$\sin A = \frac{3}{5}$
$\cos A = - \frac{4}{5}$ (cos is negative in the second quadrant)
$\tan A = - \frac{3}{4}$ (tan is negative in the second quadrant)

If angle B is acute, then using a right-angled triangle,

$\sin B = \frac{12}{13}$
$\cos B = \frac{5}{13}$
$\tan B = \frac{12}{5}$

a)
$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$
$= \left(- \frac{4}{5}\right) \left(\frac{5}{13}\right) + \left(\frac{3}{5}\right) \left(\frac{12}{13}\right)$
$= - \frac{4}{13} + \frac{36}{65}$
$= \frac{16}{65}$

b)
$\tan \left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
$= \frac{- \frac{3}{4} + \frac{12}{5}}{1 - \left(- \frac{3}{4}\right) \left(\frac{12}{5}\right)}$
$= \frac{\frac{33}{20}}{1 + \frac{9}{5}}$
$= \frac{33}{20} \times \frac{5}{14}$
$= \frac{33}{56}$