Question

If $\cos A=\frac{35}{37}$, find $\frac{\sec A+\tan A}{\sec A-\tan A}$

Answer 1

Given that $\cos A=\frac{35}{37}$. Let us consider $△ABC$ (see fig), $\angle B={90}^{\circ }$, with $AB=35$ and $AC=37$d. By Pythagoras theorem, we have
$A{C}^2=A{B}^2+B{C}^2$
${37}^2={35}^2+B{C}^2$
$B{C}^2={37}^2-{35}^2$
$=1369-1225=144$
$\therefore BC=\sqrt{144}=12$
$\sec A=\frac{AC}{AB}=\frac{37}{35},\tan A=\frac{BC}{AB}=\frac{12}{35}$
Now, $\sec A+\tan A=\frac{37}{35}+\frac{12}{35}=\frac{49}{35},\sec A-\tan A=\frac{37}{35}-\frac{12}{35}=\frac{25}{35}$
$\therefore \frac{\sec A+\tan A}{\sec A-\tan A}=\frac{\frac{49}{35}}{\frac{25}{35}}=\frac{49}{35}\times \frac{35}{25}=\frac{49}{25}$