If tan(A+B) =√3 and sin(A-B) = 12 Where A and B are acute angles.
Then, Cos2A + Sin2B =
Tan (A+B)=√3
And we know that tan 60∘ =√3
So, A+B =60∘………………………………………………….(eq. 1)
and, sin(A-B)=12
We know that sin 30∘=12
So, A-B=30∘……………………………………………………(eq. 2)
Adding eq.1 and eq.2, we get
2A=90∘
A=45 & B= 15∘