Find #A+B# when #tanA+tanB=1# and #cosA*cosB=(3^(1/2))/2#.?

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Answer 1 · 2017-10-18T07:31:28

Given #tanA+tanB=1# and #cosA*cosB=(sqrt3)/2#,

we are to find out #A+B=?#

Now
#tanA+tanB=1#

#=>sinA/cosA+sinB/cosB=1#

#=>(sinAcosB+cosAsinB)/(cosAcosB)=1#

#=>sinAcosB+cosAsinB=cosAcosB#

#=>sin(A+B)=sqrt3/2=sin(pi/3)#

#=>A+B=npi+(-1)^npi/3" where " n in ZZ#

Answer 2 · 2017-10-18T07:36:41

In the interval #[0,2pi)#, we have #A+B=pi/3# or #(2pi)/3#

and general solution is #A+B=npi+(-1)^npi/3#

#tanA+tanB=sinA/cosA+sinB/cosB=1#

or #(sinAcosB+cosAsinB)/(cosAcosB)=1#

or #sin(A+B)/(3^(1/2)/2)=1#

or #sin(A+B)=sqrt3/2=sin(pi/3)#

and #A+B=pi/3# or #(2pi)/3# in the interval #[0,2pi)#

and general solution is #A+B=npi+(-1)^npi/3#