Question

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

Answer 1

Hint: 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 $$\tan \theta ={\displaystyle \frac{\text{height}}{\text{base}}}$$.
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, $B{C}^2=x^2+{\left(8x\right)}^2=65x^2$ 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 $$\sec \theta ={\displaystyle \frac{\text{hypotenuse}}{\text{base}}}$$. So, $$\sec \theta ={\displaystyle \frac{BC}{AB}}={\displaystyle \frac{\sqrt{65}x}{x}}=\sqrt{65}$$.
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 $\sec \theta =\sqrt{1+{\tan }^2\theta }=\sqrt{1+8^2}=\sqrt{65}$.