Question

If $\sec \theta =\frac{13}{5}$, show that $\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }=3$

Answer 1

$\sec \theta =\frac{13}{5}\Rightarrow {\sec }^2\theta =\frac{169}{25}\Rightarrow 1+{\tan }^2\theta =\frac{169}{25}$

$\Rightarrow {\tan }^2\theta =\frac{169}{25}-1=\frac{144}{25}$

$\Rightarrow \tan \theta =\frac{12}{5}$

Now, $\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }=\frac{\cos \theta (2\tan \theta -3)}{\cos \theta (4\tan \theta -9)}=\frac{2\tan \theta -3}{4\tan \theta -9}$

$\Rightarrow \frac{2\times \frac{12}{5}-3}{4\times \frac{12}{5}-9}=\frac{\frac{24}{5}-3}{\frac{48}{5}-9}=\frac{\frac{9}{5}}{\frac{3}{5}}$

$=\frac{9}{5}\times \frac{5}{3}$

$=3$

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