If $tan^2 A - 6 tan A+ 9 = 0$, 0 <A < 90°, What is the value of $6 cot A + 8\sqrt{10} cos A$?

10

tan²A - 6tanA + 9 = 0

tan²A - 3tanA - 3tanA + 9 = 0

tanA ( tanA - 3 ) - 3 ( tanA - 3 ) = 0

So , tanA - 3 = 0

tanA = 3

P = 3 , B = 1

Using pythagoras theorem ,

P² + B² = H²

3² + 1² = H²

H² = 10

H = √10

Now , 6cotA + 8√10cosA

= 6 × \(\frac{1}{3}\)  + 8√10 × \(\frac{1}{√10}\)

= 2 + 8

= 10