Question

If sin(A+B)=√3/2 and cos(A-B)=1/√2,find the value of A andB and the value of tan(A+B) and tan (A-B)

Answer 1

Hey dear your qsn ans i write following.

sin(A+B)=√3/2
sin(A+B)=sin60°
=> A+B=60° eqsn no(1)

and cos(A-B)=1/√2
=> cos(A+B)=cos45°
=> A+B=45° eqsn no(2)
now, eqsn no(1)+eqsn no(2)=>2A=105° so,A=52.5°
same, eqsn no(1)-eqsn no(2)=>2B=15°,so, B=7.5°

now ,
we know that,
$sin(a - b) = \sqrt{1 - cos {}^{2}(a - b) }$
sin(A-B)=√3/2
same typ...
$cos(a + b) = \sqrt{1 - sin {}^{2}(a + b) }$
cos(A+B)=1/2


there for ...
tan(A+B)=sin(A+B)/cos(A+B)
=(√3/2)/(1/2)
=√3.

tan(A-B)=sin(A-B)/cos(A-B)
=(√3/2)/(1/√2)
=√3/√2.
I hope you find your qsn ans.