# Question #b9e87

May 4, 2017

It all depends on where you wish to take this. One option is:

$\frac{\tan \left(x\right)}{\cos \left(x\right)} \text{ "=" } \sin \frac{x}{1 - {\sin}^{2} \left(x\right)}$

#### Explanation: SohCahToa

Soh$\to \sin \left(x\right) = \left(\text{opposite")/("hypotenuse}\right) \to \frac{b}{a}$

Cah$\to \cos \left(x\right) = \left(\text{adjacent")/("hypotenuse}\right) \to \frac{c}{a}$

Toa$\to \tan \left(x\right) = \left(\text{opposite")/("adjacent}\right) \to \frac{b}{c}$

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

Let the unknown term be $\beta$

Given equation: $\frac{\tan \left(x\right)}{\cos \left(x\right)} = \beta \text{ } \ldots \ldots \ldots E q u a t i o n \left(1\right)$

Write as: $\text{ } \tan \left(x\right) \times \frac{1}{\cos \left(x\right)} = \beta$

But $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$ so by substitution we have:

$\text{ } \frac{\sin \left(x\right)}{\cos \left(x\right)} \times \frac{1}{\cos \left(x\right)} = \beta$

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

This is written as:

$\text{ "sin(x)/cos^2(x)=beta" } \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(2\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As in the Pythagoras equation ${a}^{2} = {b}^{2} + {c}^{2}$
There is a comparable one involving ${\cos}^{2} \left(x\right)$

$\text{Pythagorean relationship} \to {\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

So $\text{ "cos^2(x)=1-sin^2(x)" } \ldots \ldots \ldots \ldots . E q u a t i o n \left(3\right)$

Using $E q u a t i o n \left(3\right)$ substitute for ${\cos}^{2} \left(x\right)$ in $E q u a t i o n \left(2\right)$

$\text{ "sin(x)/cos^2(x)=beta" "->" } \sin \frac{x}{1 - {\sin}^{2} \left(x\right)} = \beta$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus the finished relationship is:

$\text{ "(tan(x))/(cos(x))" "=" "beta" "=" } \sin \frac{x}{1 - {\sin}^{2} \left(x\right)}$

May 4, 2017

$\sin \frac{x}{\cos} ^ 2 \left(x\right)$ or $\sin \frac{x}{1 - {\sin}^{2} \left(x\right)}$

#### Explanation:

First of all, what are $\tan \left(x\right)$ and $\cos \left(x\right)$?

Both are trigonometric functions, but $\tan \left(x\right)$ is actually a function of both $\sin \left(x\right)$ and $\cos \left(x\right)$:

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

Substitute this in for the numerator in the original problem:

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

Remember that dividing by anything is the same as multiplying by the reciprocal of that thing. In other words, the following is mathematically valid:

$\frac{a}{b} = a \cdot \frac{1}{b}$

Applying this to our equation, we see that:

$\frac{\sin \frac{x}{\cos} \left(x\right)}{\cos} \left(x\right) = \sin \frac{x}{\cos} \left(x\right) \cdot \frac{1}{\cos} \left(x\right)$

Simplifying, we get:
$\sin \frac{x}{\cos} \left(x\right) \cdot \frac{1}{\cos} \left(x\right) = \sin \frac{x}{\cos} ^ 2 \left(x\right)$

If we want our answer to be in terms of $\sin \left(x\right)$, we could go further and use the identity: ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$

Replacing ${\cos}^{2} \left(x\right)$, we get:

$\sin \frac{x}{1 - {\sin}^{2} \left(x\right)}$