Question

If $13\cos \theta -5=0$, find $\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }$.

Answer 1

Let $△ABC$ be a right angled triangle where $\angle B={90}^0$ and $\angle C=\theta$ as shown in the above figure:

Now it is given that $13\cos \theta -5=0$ or $\cos \theta =\frac{5}{13}$ and we know that, in a right angled triangle, $\cos \theta$ is equal to adjacent side over hypotenuse that is $\cos \theta =\frac{Adjacentside}{Hypotenuse}$, therefore, adjacent side $BC=5$ and hypotenuse $AC=13$.

Now, using pythagoras theorem in $△ABC$, we have

$A{C}^2=A{B}^2+B{C}^2\Rightarrow {13}^2=A{B}^2+5^2\Rightarrow 169=A{B}^2+25\Rightarrow A{B}^2=169-25=144\Rightarrow AB=\sqrt{144}=12$

Therefore, the opposite side $AB=12$.

We know that, in a right angled triangle,

$\sin \theta$ is equal to opposite side over adjacent side that is $\sin \theta =\frac{Oppositeside}{Hypotenuse}$

Here, we have opposite side $AB=12$, adjacent side $BC=5$ and the hypotenuse $AC=13$, therefore, $\sin \theta$ can be determined as follows:

$\sin \theta =\frac{Oppositeside}{Hypotenuse}=\frac{AB}{AC}=\frac{12}{13}$

Now, we find the following:

$\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{\frac{12}{13}+\frac{5}{13}}{\frac{12}{13}-\frac{5}{13}}=\frac{\frac{17}{13}}{\frac{7}{13}}=\frac{17}{7}$

Hence, $\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{17}{7}$.