If sec $ A = \frac{17}{8}$, given that A < 90o, what us the value of the following? $\frac{34sinA +15cotA}{68cosA-16tanA}$
Answer & explanation
Correct answer: option 2
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