Question

If $\sec \theta =\frac{26}{10}$, find $\frac{3\cos \theta +4\sin \theta }{4\cos \theta -2\sin \theta }$

Answer 1

$\sec \theta =\frac{26}{10}$

$\Rightarrow {\sec }^2\theta ={\left(\frac{26}{10}\right)}^2=\frac{676}{100}$

$\Rightarrow 1+{\tan }^2\theta =\frac{676}{100}$

$\Rightarrow {\tan }^2\theta =\frac{676}{100}-1=\frac{576}{100}$

$\Rightarrow \tan \theta =\sqrt{\frac{576}{100}}=\frac{24}{10}$

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

$=\frac{3+4\times \frac{24}{10}}{4-2\times \frac{24}{10}}$

$=\frac{3+\frac{48}{5}}{4-\frac{24}{5}}$

$=\frac{\frac{(15+48)}{5}}{\frac{(20-24)}{5}}$

$=\frac{63}{-4}$

Hence, the answer is $\frac{63}{-4}.$