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

2 Answers
Oct 18, 2017

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

Oct 18, 2017

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

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

Explanation:

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