If A and B are acute angle such that A+B and A−B satisfy the equation 4sin2θ−2√6sinθ+1=0, then-
On solving the equation sinθ=2√6−2√28 sinθ=√3−12√2 A−B=π15 A+B=14π15 A=π4 B=π6 Option A and B
If ∫α0dx1−cos α cos x=Asin α+B (α≠0).Then possible values of A and B are
If sinA = 1√10 and sinB = 1√5, where A and B are positive acute angles, then A+B=