Question

If $12\csc \theta =13$, find the value of $\frac{2\sin \theta -9\cos \theta }{4\sin \theta -9\cos \theta }$.

Answer 1

$\because 12cosec\theta =13\Rightarrow cosec\theta =\frac{13}{12}⟶\left(1\right)weknowthatcosecx=\frac{h}{p}⟶\left(2\right)comparing\left(1\right)and\left(2\right)wegeth=13kandp=12k\Rightarrow b=\sqrt{h^2-p^2}=\sqrt{{\left(13k\right)}^2-{\left(12k\right)}^2}=5k\therefore \sin \theta =\frac{p}{h}=\frac{12}{13}and\cos \theta =\frac{b}{h}=\frac{5}{13}now,\frac{2\sin \theta -9\cos \theta }{4\sin \theta -9\cos \theta }=\frac{2.\frac{12}{13}-9.\frac{5}{13}}{4.\frac{12}{13}-9.\frac{5}{13}}=-7$