Question

If sec $\theta ==\frac{13}{5}$ find the value of $\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }$

• A. 1
• B. 3
• C. 5
• D. 7

Answer 1

The correct option is B 3

Let $BC=13,AB=5$
Given,
sec$\theta =\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{BC}{AB}=\frac{13}{5}$

By Pythagoras theorem,
$AC=\sqrt{(B{C}^2-A{B}^2)}$
$AC=\sqrt{({13}^2-5^2)}$
$AC=\sqrt{(169-25)}$
$AC=\sqrt{144=12}$

So,
$\sin \theta =\frac{AC}{BC}=\frac{12}{13}$ $\cos \theta =\frac{AB}{BC}=\frac{5}{13}$

Hence,
$=\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }$
$=\frac{2\times \left(\frac{12}{13}\right)-3\times \left(\frac{5}{13}\right)}{4\times \left(\frac{12}{13}\right)-9\times \left(\frac{5}{13}\right)}$
$=\frac{\frac{(24-15)}{13}}{\frac{(48-45)}{13}}$
$=\frac{9}{3}$
$=3$