# How do you prove: secx - cosx = sinx tanx?

Feb 11, 2016

Using the definitions of $\sec x$ and $\tan x$, along with the identity
${\sin}^{2} x + {\cos}^{2} x = 1$, we have

$\sec x - \cos x = \frac{1}{\cos} x - \cos x$

$= \frac{1}{\cos} x - {\cos}^{2} \frac{x}{\cos} x$

$= \frac{1 - {\cos}^{2} x}{\cos} x$

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

$= \sin x \cdot \sin \frac{x}{\cos} x$

$= \sin x \tan x$

Feb 11, 2016

First convert all terms into $\sin x$ and $\cos x$.
Second apply fraction sum rules to the LHS.
Lastly we apply the Pythagorean identity: ${\sin}^{2} x + {\cos}^{2} x = 1$

#### Explanation:

First in questions of these forms it's a good idea to convert all terms into sine and cosine: so, replace $\tan x$ with $\sin \frac{x}{\cos} x$
and replace $\sec x$ with $\frac{1}{\cos} x$.

The LHS, $\sec x - \cos x$ becomes $\frac{1}{\cos} x - \cos x$.
The RHS, $\sin x \tan x$ becomes $\sin x \sin \frac{x}{\cos} x$ or ${\sin}^{2} \frac{x}{\cos} x$.

Now we apply fraction sum rules to the LHS, making a common base (just like number fraction like 1/3 +1/4 => 4/12 + 3/12 = 7/12).
LHS=$\frac{1}{\cos} x - \cos x \implies \frac{1}{\cos} x - {\cos}^{2} \frac{x}{\cos} x \implies \frac{1 - {\cos}^{2} x}{\cos} x$.

Lastly we apply the Pythagorean identity: ${\sin}^{2} x + {\cos}^{2} x = 1$! (one of the most useful identities for these types of problems).
By rearranging it we get $1 - {\cos}^{2} x = {\sin}^{2} x$.
We replace the $1 - {\cos}^{2} x$ in the LHS with ${\sin}^{2} x$.

LHS = $\frac{1 - {\cos}^{2} x}{\cos} x \implies \frac{{\sin}^{2} x}{\cos} x$ which is equal to the modified RHS.

Thus LHS= RHS Q.E.D.

Note this general pattern of getting things into terms of sine and cosine, using the fraction rules and the Pythagorean identity, often solves these types of questions.

Sep 7, 2016

If we so desire, we can also modify the right-hand side to match the left-hand side.

We should write $\sin x \tan x$ in terms of $\sin x$ and $\cos x$, using the identity $\textcolor{red}{\tan x = \sin \frac{x}{\cos} x}$:

$\sin x \tan x = \sin x \left(\sin \frac{x}{\cos} x\right) = {\sin}^{2} \frac{x}{\cos} x$

Now, we use the Pythagorean identity, which is ${\sin}^{2} x + {\cos}^{2} x = 1$. We can modify this to solve for ${\sin}^{2} x$, so: $\textcolor{red}{{\sin}^{2} x = 1 - {\cos}^{2} x}$:

${\sin}^{2} \frac{x}{\cos} x = \frac{1 - {\cos}^{2} x}{\cos} x$

Now, just split up the numerator:

$\frac{1 - {\cos}^{2} x}{\cos} x = \frac{1}{\cos} x - {\cos}^{2} \frac{x}{\cos} x = \frac{1}{\cos} x - \cos x$

Use the reciprocal identity color(red)(secx=1/cosx:

$\frac{1}{\cos} x - \cos x = \sec x - \cos x$

Sep 8, 2016

It's really this simple...

#### Explanation:

Using the identity $\tan x = \sin \frac{x}{\cos} x$, multiply the $\sin x$ onto the identity to get:

$\sec x - \cos x = {\sin}^{2} \frac{x}{\cos} x$

Then, multiply $\cos x$ through the equation to yield:

$1 - {\cos}^{2} x = {\sin}^{2} x$

Considering that $\sec x$ is the inverse of $\cos x$.
Finally, using the trigonometric identity $1 - {\cos}^{2} x = {\sin}^{2} x$, the final answer would be:

${\sin}^{2} x = {\sin}^{2} x$