tan2A + 5 sec A = 13, where 0 < A < 90°. Solve for A (in degrees).

tan²A + 5 sec A = 13

( Sec²A - 1 ) + 5 SecA = 13

Sec²A + 5SecA = 14

Sec²A + 7SecA - 2SecA - 14= 0

SecA ( SecA + 7 ) -2 ( SecA + 7 ) = 0

( SecA - 2 ) ( SecA + 7 ) = 0

Either ( SecA - 2 = 0 ) OR ( SecA + 7 = 0  ).

SecA = - 7 is not possible because 0 < A < 90°

So , SecA = 2

⇒  SecA = Sec60°

A = 60°