Question

If $\cos \theta =\frac{5}{13}$ and $\theta$ is acute, find the value of $\frac{5\tan \theta +12\cot \theta }{5\tan \theta -12\cot \theta }$.

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 $\cos \theta =\frac{5}{13}$ 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,

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

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

$\tan \theta =\frac{Oppositeside}{Adjacentside}=\frac{AB}{BC}=\frac{12}{5}$

$\cot \theta =\frac{1}{\tan \theta }=\frac{1}{\frac{12}{5}}=1\times \frac{5}{12}=\frac{5}{12}$

Now, we find the following:

$\frac{5\tan \theta +12\cot \theta }{5\tan \theta -12\cot \theta }=\frac{5\left(\frac{12}{5}\right)+12\left(\frac{5}{12}\right)}{5\left(\frac{12}{5}\right)-12\left(\frac{5}{12}\right)}=\frac{12+5}{12-5}=\frac{17}{7}$

Hence, $\frac{5\tan \theta +12\cot \theta }{5\tan \theta -12\cot \theta }=\frac{17}{7}$.