Question

If cos(A-B) = 3/5 and tanA*tanB = 2 then

(a) cosA*cosB= 1/5 (c) cos(A+B)= -1/5

(b) sinA*sinB= 2/5 (d) sinA*sinB=4/5

Answer 1

(a) cosA*cosB= 1/5 (c) cos(A+B)= -1/5

(b) sinA*sinB= 2/5

a)

cos(A - B) = cosA . cosB + sinA . sinB.

Now, tanA.tanB = 2

sinA / cosA . sinB / cosB = 2

Therefore, sinA + sinB = 2.cosA.cosB

Now we have, cos(A - B) = cosA . cosB + sinA . sinB.

3/5 = cosA . cosB + 2. cosA . cosB

3/5 = 3 cosA . cosB

Therefore we have the answer, cosA. cosB = 1/5

b)

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

c)

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.