If tan2A - 6tanA + 9 = 0, 0° < A < 90°, What is the value of 6cotA + 8 10 cosA?

Answer: Option C

tan²A - 6tanA + 9 = 0
tan²A - 3tanA - 3tanA + 9 = 0
tanA(tanA - 3) - 3(tanA - 3) = 0
(tanA - 3)(tanA - 3) = 0
tanA $$=\frac{3}{1}=\frac{P}{B}$$
H² = P² + B²
$$\begin{array}{rl}& H=\sqrt{9+1}=\sqrt{10}\\ & 6\cot A+8\sqrt{10}\cos A\\ & =6\left(\frac{B}{P}\right)+8\sqrt{10}\left(\frac{B}{H}\right)\\ & =6\left(\frac{1}{3}\right)+8\sqrt{10}\left(\frac{1}{\sqrt{10}}\right)\\ & =2+8\\ & =10\end{array}$$