If secA = $$\frac{{17}}{8},$$ given that A < 90°, what is the value of the following?
$$\frac{{34\sin A + 15\cot A}}{{68\cos A - 16\tan A}}$$

Solution (By Examveda Team)

$$\eqalign{ & \sec A = \frac{{17 \to H}}{{8 \to B}},\,A < {90^ \circ } \cr & {L^2} = {H^2} - {B^2} \cr & {L^2} = 289 - 64 \cr & {L^2} = 225 \cr & L = 15 \cr & {\text{Then}}, \cr & \frac{{34\sin A + 15\cot A}}{{68\cos A - 16\tan A}} \cr & = \frac{{34 \times \frac{{15}}{{17}} + 15 \times \frac{8}{{15}}}}{{68 \times \frac{8}{{17}} - 16 \times \frac{{15}}{8}}} \cr & = \frac{{30 + 8}}{{32 - 30}} \cr & = \frac{{38}}{2} \cr & = 19 \cr} $$