Question

tan(A+B)=√3. , tan+A-B)=0 then find sin(A+B). and sin(A-B)?

Answer 1

tan(60)=3^1/2 tan(0)=0 given tan(a+b) =3^1/2 so a+b =60.......(1) tan(a-b) = o so a-b=0..........(2) from (1) and (2) a=b=30 so sin(a+b) = 3^1/2 /2 sin(a-b) =0

Answer 2

Answer:

tan (A+B) = ✓3 ( tan60°=✓3)

tan(A+B)= tan60°

A+B= 60°

tan(A-B) = 0 ( tan0°= 0)

tan(A-B)= tan0°

A-B = 0

A= B

A+B=60

B+B= 60

B= 30°

A = 30°

sin (A+B) = 30+30

sin(A+B)= 60°

sin(A+B)=√3/2