Question

cosA=3/5, tanB = 5/12 , A lies in first and
B lies in third quadrant find sin(A+B), cos (A+B)​

Answer 1

QUESTION :

If cosA = 3/5 and sinB = 5/13 , then the value of [(tanA + tanB) ÷ (1 - tanA.tanB)] is

[NOTE : A and B lies in Quadrant I]

Answer:

63/16 or 3.9375

Step-by-step explanation:

cosA = 3/5

sin B = 5/13

cos^2A + sin^2A = 1

sin^2A = 1 - (9/25)

sin^2A = 16/25

sinA = +4/5 or -4/5

But SinA = +4/5 (Hence, it is in Quadrant I)

cos^2 B = 1 - sin^2 B

cos^2 B = 1 - 25/169

cos^2 B = 144/169

cosB = +12/13 or -12/13

But Cos B = +12/13 (Hence, it is in quadrant I)

(tanA + tanB) / (1 - tanA.tanB)

= [(sinA/cosA) + (sinB/cosB)] ÷ [1 - (sinA/cosA) × (sinB/cosB)]

= [{(4/5) ÷ (3/5)} + {(5/13) ÷ (12/13)}] / [1 - {(4/5) ÷ (3/5)} × {(5/13) ÷ (12/13)}]

= [(4/3) + (5/12)] ÷ [1 - (20/36)]

= (21/12) ÷ (4/9)

= 189/48

= 63/16 or 3.9375