Question

If $cosec\theta =13/12$, then evaluate $\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }$.

Answer 1

Given,

$\csc \theta =\frac{13}{12}$

$\Rightarrow \sin \theta =\frac{12}{13}$

$\Rightarrow \cos \theta =\frac{5}{13}$

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

$=\frac{2\times \frac{12}{13}-3\times \frac{5}{13}}{4\times \frac{12}{13}-9\times \frac{5}{13}}$

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

$=3$