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

c

Explanation:

plato

Answer 2

`\tan (A +B) = - 6(6)/(13)`

Explanation:

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