Question

If sin A = 3/5 and tan B = 15/8 and angles A and B are in Quadrant I, find the value of tan(A + B)

Answer 1

Answer:

c

Step-by-step explanation:

plato

Answer 2

Answer:

$\tan (A +B) = - 6\frac{6}{13}$

Step-by-step explanation:

$\sin \: A = \frac{3}{5}...(given) \\ \\ \because\cos A = \pm \sqrt{1 - { \sin }^{2}\: A } \\ \\ \ \therefore\cos A = \pm \sqrt{1 - \bigg( \frac{3}{5} \bigg) ^{2}} \\ \\ \ \therefore\cos A = \pm \sqrt{1 - \frac{9}{25}} \\ \\ \therefore\cos A = \pm \sqrt{ \frac{25 - 9}{25} } \\ \\ \therefore\cos A = \pm \sqrt{ \frac{16}{25}} \\ \\ \therefore\cos A = \pm { \frac{4}{5}} \\ \\ \because \: \angle A \: is \: in \: quadrant \: I \\ \\ \therefore \: \cos A = { \frac{4}{5}} \\ \\ \tan \: A = \frac{ \sin \: A}{\cos \: A} \\ \\ \tan \: A = \frac{ \frac{3}{5} }{ \frac{4}{5} } \\ \\ \tan \: A = \frac{3}{4} \\ \\ \because \tan (A +B) = \frac{ \tan A + \tan B }{1 - \tan A \tan B } \\ \\ \therefore \tan (A +B) = \frac{ \frac{3}{4} + \frac{15}{8} }{1 - \frac{3}{4} \times \frac{15}{8} } \\ \\ \therefore \tan (A +B) = \frac{ \frac{6}{8} + \frac{15}{8} }{1 - \frac{45}{32}} \\ \\ \therefore \tan (A +B) = \frac{ \frac{21}{8} }{\frac{32 - 45}{32}} \\ \\ \therefore \tan (A +B) = \frac{ \frac{21}{8} }{\frac{ - 13}{32}} \\ \\ \therefore \tan (A +B) = \frac{21}{8} \times \bigg(- \frac{32}{13 } \bigg)\\ \\ \therefore \tan (A +B) = - \frac{84}{13} \\ \\ \therefore \tan (A +B) = - 6\frac{6}{13}$