How to simplify $\sec\tan^{-1}\pi$? How to simplify $\sec\tan^{-1}\pi$?
I've never dealt with inverse trigonometry before. Thanks in advance.
 A: Here is a right triangle.

In any right triangle, by definition, $\tan x = \frac ba$ and $\sec x = \frac ca$.
Also, of course, $a^2+b^2 = c^2$.
Since $\tan x =\frac ba $, we have $x = \tan^{-1}\frac ba$ and then:  $$\begin{align}\sec x & =  \frac ca \\ \sec\left(\tan^{-1} \frac ba\right) & = \frac ca \\ & = \frac {\sqrt{a^2+b^2}}a.\end{align}$$
Now take $b=\pi$ and $a=1$ and we have $$
\sec\tan^{-1} \frac \pi1 =  \frac {\sqrt{1^2+\pi^2}}1.$$

Also, observe: since this works for any right triangle, we can also take $a=3, b=4$ and obtain $$\sec\tan^{-1} \frac 43 =  \frac 53$$
without actually knowing what $\tan^{-1} \frac43$ is, and without even being able to calculate it.  
A: Picture a right triangle whose horizontal leg is 1 and whose vertical leg (extending upwards from the right side of the horizontal leg) is $\pi$. Then the angle opposite the vertical leg is $\tan^{-1}\pi$.
The secant of this angle is the hypotenuse $\sqrt{1+\pi^2}$ divided by the horizontal leg, but the latter is just 1, so
$$\sec\tan^{-1}\pi=\sqrt{1+\pi^2}$$
(And yes, the domain of $\tan^{-1}$ does contain $\pi$ – indeed, the whole real line too. It's its range that doesn't contain $\pi$.)
A: We know from this that$$\sec\arctan x=\sqrt{1+x^2}$$
Then $$\sec\arctan \pi=\sqrt{1+\pi^2}$$
A: Remember the identity, $$ sec^2(x) = 1+tan^2(x)$$
You know your $tan(x)=\pi$, so your $$sec(x) = \sqrt {1+\pi ^2} $$ 
The above identity is found by dividing $$ sin^2(x) + cos^2(x)=1$$ by $ cos^2(x)$ 
A: Recall the following trigonometric identity:

$$\sec\theta=\frac{1}{\cos\theta} \ $$

Thus:

$$\sec(\arctan(\pi))=\frac{1}{\cos(\arctan(\pi))}\ $$

Now, recall the following inverse trigonometric identity:

$$\cos(\arctan(x))=\frac{1}{\sqrt{1+x^{2}}}$$

Therefore, if $\\ x=\pi, \ $we have:

$$\cos(\arctan(\pi))=\frac{1}{\sqrt{1+\pi^{2}}}$$

Thus:

$$\sec(\arctan(\pi))=\frac{1}{\cos(\arctan(\pi))}\ =\frac{1}{\frac{1}{\sqrt{1+\pi^{2}}}} $$

Or simply:

$$\sec(\arctan(\pi))=\sqrt{1+\pi^{2}}\ $$

