`cotAcotB=2,cos(A+B)=3/5=>sinAsinB=`

cot A cot B=2,cos(A+B)=(3)/(5)rArr sin A sin B=

Cot A Cot B = 2, Cos (A+B) = 3/5 => sinA.sin B = (i) 2/5 (ii) 1/5 (iii) 4/5 (iv) 3/5

If 2cos A+3cos B+5cos C=0=2sin A+3sin B+5sin C then 3A+27cos3B+125cos3C=k cos(A+B+C) then k is

If cos(A-B)=(3)/(5) andtan Atan B=2, then cos A cos B=(1)/(5)(b)sin A sin B=-(2)/(5)cos A cos B=-(1)/(5)(d)sin A sin B=-(1)/(5)

A and B are positive acute angles satisfying the equation 3cos^(2)A+2cos^(2)B=4(3sin A)/(sin B)=(2cos B)/(cos A) then A+2B is

Using vector method prove that cos(A-B)=cosAcosB+sinAsinB

If sin(A+B)=sinAcosB+cosAsinB and cos(A-B)=cosAcosB+sinAsinB , find the values of (i) sin75^(@) and (ii) cos15^(@) .

If A=60^(@) and B=30^(@) , verify that : (i)sin(A+B)=sinAcosB+cosAsinB (ii) cos(A+B)=cosAcosB-sinAsinB

Prove that: (sin(A-B))/(sinAsinB)=(sinAcosB-cosAsinB)/(sinAsinB) (sinAcosB)/(sinAsinB)-(cosAsinB)/(sinCsinA) =cotB-cotA-cotC =0 = RHS Hence Proved.