Question

Cos(A-B)=3/5 and tanA.tanB=2 Then......

Answer 1

Given that cos (A-B) = 3/5

So cosAcosB + sinAsinB = 3/5.....eqn.1

Given tanA*tanB = 2

So (sinA/cosA)*(sinB/cosB) = 2

→ sinAsinB = 2 cosAcosB............eqn.2

Similarly , cosAcosB= ½(sinAsinB )........eqn.3

Putting eqn.2 in eqn.1 we get,

cosAcosB + 2cosAcosB =3/5

3cosAcosB =3/5

cosAcosB =1/5......eqn.4

Putting eqn.3 in eqn. 1 , we get,

½(sinAsinB) + sinAsinB = 3/5

3/2(sinAsinB) = 3/5

sinAsinB = 2/5.....eqn.5

We know that cos(A+B) = cosAcosB - sinAsinB.....eqn.6

Putting eqn.4 and eqn.5 in eqn.6 we get,

cos(A-B) = 1/5- 2/5

= -1/5. Proved.

Answer 2

We are given that,
tan A ×tan B = 2
(sin A × sin B)/(cos A ×cos B)=2
Applying componendo and dividendo rule
sin A sin B + cos A cos B /sin A sin B - cos A cos B =3/1
cos (A-B)/cos(A-B)= 3
cos(A-B) /3= cos (A+B)
cos (A+B) = (-3/5)×3=-1/5
cos (A+B) =-1/5