If 6tanA(tanA + 1) = 5 - tanA, given that 0 < A < /2 what is the value of (sinA + cosA)?

Answer: Option D

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
$$\begin{array}{rl}& \therefore \sin A+\cos A\\ & =\frac{P}{H}+\frac{B}{H}\\ & =\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}\\ & =\frac{3}{\sqrt{5}}\end{array}$$