If sec $ A = \frac{17}{8}$, given that A < 90^o, what us the value of the following? $\frac{34sinA +15cotA}{68cosA-16tanA}$

19

secA = \(\frac{17}{8}\)

By using pythagoras theorem ,

P² + B² = H²

P² + 8² = 17²

P = 15

34sinA + 15cotA = 34 ( \(\frac{15}{17}\) ) + 15 ( \(\frac{8}{15}\) )

= 30 + 8 = 38

68cosA - 16 tanA = 68 ( \(\frac{8}{17}\) )  - 16 ( \(\frac{15}{8}\) 

= 32 - 30 = 2

Now , $\frac{34sinA +15cotA}{68cosA-16tanA}$

By putting values of numerator of denominator

= \(\frac{38}{2}\)

= 19