Question

if cos(A-B)=3/5 and tanA. tanB=2 then cosA cosB​

Answer 1

Answer:

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

So cosAcosB + sinAsinB = 3/5.....(.1)

Given tanA*tanB = 2

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

sinAsinB = 2 cosAcosB............(2)

Similarly ,

cosAcosB= ½(sinAsinB )........(.3)

Putting eqn.2 in eqn.1 we get,

cosAcosB + 2cosAcosB =3/5

3cosAcosB =3/5

cosAcosB =1/5......)4)

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

½(sinAsinB) + sinAsinB = 3/5

3/2(sinAsinB) = 3/5

sinAsinB = 2/5.....(5)

We know that

cos(A+B) = cosAcosB - sinAsinB...(6)

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

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

= -1/5. Proved.

Step-by-step explanation:

Answer 2

Answer:

For solution refer to the attachment

Step-by-step explanation:

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