Question

If $\sec \theta =\frac{13}{5},$ then what is the value of $\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }?$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

$\sec \theta =\frac{13}{5}$

$\sec \theta =\frac{H}{B}=\frac{13}{5}$
Using Pythagoras Theorem,
$H^2=P^2+B^2$
${13}^2=P^2+5^2$
$P^2=144$
$P=12$
Now, $\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }$

= $\frac{2\left(\frac{12}{13}\right)-3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right)-9\left(\frac{5}{13}\right)}$

= $\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$
= $3$