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# 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

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Solution

## (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.

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