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

Solution (By Examveda Team)

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
$$\eqalign{ & H = \sqrt {9 + 1} = \sqrt {10} \cr & 6\cot A + 8\sqrt {10} \cos A \cr & = 6\left( {\frac{B}{P}} \right) + 8\sqrt {10} \left( {\frac{B}{H}} \right) \cr & = 6\left( {\frac{1}{3}} \right) + 8\sqrt {10} \left( {\frac{1}{{\sqrt {10} }}} \right) \cr & = 2 + 8 \cr & = 10 \cr} $$