Question

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

Answer 1

$\cos \alpha =\frac{5}{13}$, then $\sin \alpha =\frac{12}{13}$ [Using ${\sin }^2\alpha +{\cos }^2\alpha =1$]........(1).

Now,

$\frac{2\sin \alpha -3\cos \alpha }{4\sin \alpha -9\cos \alpha }$

$=\frac{2.\frac{12}{13}-3.\frac{5}{13}}{4.\frac{12}{13}-9.\frac{5}{13}}$[ Using values of $\sin \alpha$, $\cos \alpha$]

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

$=\frac{9}{3}$

$=3$.