Question

How do you evaluate $\sec \left( {{\tan }^{-1}}8 \right)$ without a calculator?

Answer 1

We explain the function $\arctan \left( x \right)$. We express the inverse function of tan in the form of $\arctan \left( x \right)={{\tan }^{-1}}x$. We draw the graph of $\arctan \left( x \right)$ and the line $x=8$ to find the intersection point. Thereafter we take the sec ratio of that angle to find the solution. We also use the representation of a right-angle triangle with height and base ratio being 8 and the angle being $\theta$.

Complete step-by-step solution:
The internal part ${{\tan }^{-1}}8$ of $\sec \left( {{\tan }^{-1}}8 \right)$ is an angle. We assume ${{\tan }^{-1}}8=\theta$.
This gives in ratio $\tan \theta =8$. We know $\tan \theta =\dfrac{\text{height}}{\text{base}}$.
We can take the representation of a right-angle triangle with height and base ratio being 8 and the angle being $\theta$. The height and base were considered with respect to that particular angle $\theta$.

In this case we take $AB=x$ and keeping the ratio in mind we have $AC=8x$ as the ratio has to be 8.
Now we apply the Pythagoras’ theorem to find the length of BC. $B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$.
So, $B{{C}^{2}}={{x}^{2}}+{{\left( 8x \right)}^{2}}=65{{x}^{2}}$ which gives $BC=\sqrt{65}x$.
We need to find $\sec \left( {{\tan }^{-1}}8 \right)$ which is equal to $\sec \theta$.
This ratio gives $\sec \theta =\dfrac{\text{hypotenuse}}{\text{base}}$. So, $\sec \theta =\dfrac{BC}{AB}=\dfrac{\sqrt{65}x}{x}=\sqrt{65}$.
Therefore, $\sec \left( {{\tan }^{-1}}8 \right)$ is equal to $\sqrt{65}$.

Note: We can also apply the trigonometric image form to get the value of $\sec \left( {{\tan }^{-1}}8 \right)$.
It’s given that $\tan \theta =8$ and we need to find $\sec \theta$. We know $\sec \theta =\sqrt{1+{{\tan }^{2}}\theta }$.
Putting the values, we get $\sec \theta =\sqrt{1+{{\tan }^{2}}\theta }=\sqrt{1+{{8}^{2}}}=\sqrt{65}$.