If tan 2A.tan 3A=1, then cos 5A=

Using sin/cos=tan you get cos2Acos3A-sin2Asin3A=0
so cos(2A+3A)=0 and you have the answer.

tan 2A.tan 3A=1
=> sin(2A) sin(3A) / cos(2A) cos(3A) =1
=> cos(A)- Cos(5A) / cos(A)+ Cos(5A) =1

:::cos(A)*Cos(B) = 1/2 * (cos (a+b) + cos (a-b))
sin(A)*sin(B) = 1/2 * (cos (b-a) - cos (a+b))

=> cos(A)- Cos(5A)=cos(A)+ Cos(5A)
=> 2 cos(5A)=0
=> cos 5A =0