If 6tanA(tanA + 1) = 5 - tanA, given that 0 < A < $$\frac{\pi }{2}$$ what is the value of (sinA + cosA)?

6tanA(tanA + 1) = 5 - tanA
6tan²A + 6tanA + tanA = 5
6tan²A + 7tanA = 5
6tan²A + 7tanA - 5 = 0
6tan²A + 10tanA - 3tanA - 5 = 0
2tanA(3tanA + 5) - 1(3tanA + 5) = 0
(2tanA - 1)(3tanA + 5) = 0
tanA = $$\frac{1}{2}$$ and tanA = $$ - \frac{5}{2}$$
taking tanA $$ = \frac{1}{2} = \frac{P}{B}$$
H = √5
$$\eqalign{ & \therefore \sin A + \cos A \cr & = \frac{P}{H} + \frac{B}{H} \cr & = \frac{1}{{\sqrt 5 }} + \frac{2}{{\sqrt 5 }} \cr & = \frac{3}{{\sqrt 5 }} \cr} $$