# Verification: Why does $\frac{dy}{dx}$ of $\tan(x)$ = $\sec^2x$?

So, I once read somewhere that if: $y = \tan x, \quad\frac{dy}{dx}=\sec^2x$.

I think I have found the proof, but I am not so sure.

Recall the identity:

$$\tan x \equiv \frac{\sin x}{\cos x}$$

Recall the quotient rule:

(where $u$ and $v$ are each functions:) $$y=\frac{u}{v}, \quad \frac{dy}{dx}=\frac{v * \frac{du}{dx}-u* \frac{dv}{dx}}{v^2}$$

I did:

$u=\sin x$

$v = \cos x$

$\frac{du}{dx} = \cos x$

$\frac{dv}{dx} = -\sin x$

I believe that if you substitute all these values in, you get:

$$\frac{\cos^2x + \sin^2x}{\cos^2x}$$

Recall the identity:

$$\sin^2x + \cos^2x \equiv 1$$

This is the case on the top so:

$$\frac{dy}{dx}=\frac{1}{\cos^2x}$$

On the same webpage that I found these identities in, I also found that:

$$\sec^2x \equiv\frac{1}{\cos^2x}$$

This means that $$\frac{dy}{dx}=\sec^2x \equiv\frac{1}{\cos^2x}$$

But why does $\frac{1}{\cos^2x} \equiv \sec^2x$? I want a proof that the $\sec \leftarrow \rightarrow \cos$ identity works. From my research, not much has touched on this, so how is it provable that $\frac{dy}{dx}$ of $\tan x = \sec x^2$?

P.S. I have a hunch that it is something that is taken for granted, and that we don't necessarily have to know this.

P.P.S. Could you check this proof for me, and state if I have missed any steps?

• This is true by definition: $$\sec{x}\equiv \frac{1}{\cos{x}}$$ – projectilemotion May 4 '17 at 15:37
• The definition of $\sec$ is $\frac1\cos$. – Arthur May 4 '17 at 15:38
• @projectilemotion , I know, but is there a proof? – VortexYT May 4 '17 at 15:38
• @simplest_mathematics $\sec(x) = \frac{1}{\cos(x)}$ by definition -- you may be interested to read math.stackexchange.com/questions/1399781/… – Marcus Andrews May 4 '17 at 15:39
• @simplest_mathematics, $\sec(x)$ just means $\frac{1}{\cos(x)}$, so it's not something you can prove, that's just what it means. Basically at some point in history people just decided it was inconvenient to always be writing $\frac{1}{\cos(x)}$ so they gave it a new name. – DMcMor May 4 '17 at 15:40

But why does $\frac{1}{\cos^2x} \equiv \sec^2x$? I want a proof that the $\sec \leftarrow \rightarrow \cos$ identity works.

There is nothing to prove because this is a definition. We simply give another name to $\tfrac{1}{\cos x}$, we call it $\sec x$. You can avoid using $\sec x$ by replacing every occurence by $\tfrac{1}{\cos x}$, if you prefer. Using it allows to rewrite some formulas in a shorter and/or more elegant way.

Compare it to giving the name $\tan x$ to the quotient $\tfrac{\sin x}{\cos x}$: you can't prove this either, at least not if this is how you define $\tan x$.

$\sec x = \frac 1{\cos x}$ by definition.

However, there is a nice geometric proof that $\frac {d}{dx} \tan x = \sec^2 x$ that is worth knowing. In the figure we have two right triangles. The base is 1.

Lets call the angle at the center of the unit circle $x, h, x+h$

The heights of the right triangles are $\tan x$ and $\tan (x+h)$ hypoteni are $\sec x$ and $\sec (x+h)$

The areas are $\frac 12 \tan x$ and $\frac 12 \tan (x+h)$

The small triangle with angle $h$ has area $\frac 12 \tan (x+h) - \frac 12 \tan x$

The area of this triangle is less than the area of the section of the circle with radius $\sec (x+h)$ and greater than the section of the circle of radius $\sec x$

These areas are $\frac 12 (\sec^2 (x+h)) h$ and $\frac 12 (\sec^2 x) h$

So,

$\frac 12 (\sec^2 x) h \le \frac 12 (\tan (x+h) - \tan x) \le \frac 12 (\sec^2 (x+h)) h$

or

$(\sec^2 x)\le \frac {\tan (x+h) - \tan x}{h} \le \sec^2 (x+h)$

Now take the limit as h goes to 0.

$\frac {d}{dx} \tan x = \sec^2 x$