Question

Question #b9e87

Answer 1

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

`(tan(x))/(cos(x))" "=" "sin(x)/(1-sin^2(x))`

Explanation:

SohCahToa

Soh`-> sin(x)=("opposite")/("hypotenuse") ->b/a`

Cah`->cos(x)=("adjacent")/("hypotenuse")->c/a`

Toa`->tan(x)=("opposite")/("adjacent")->b/c`

`tan(x)=(sin(x))/(cos(x))`
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Let the unknown term be `beta`

Given equation: `(tan(x))/(cos(x))=beta" ".........Equation(1)`

Write as: `" "tan(x)xx1/(cos(x))=beta`

But `tan(x)=(sin(x))/(cos(x))` so by substitution we have:

`" "(sin(x))/(cos(x))xx1/(cos(x))=beta`

`" "sin(x)/(cos(x))^2=beta`

This is written as:

`" "sin(x)/cos^2(x)=beta" "...............Equation(2)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As in the Pythagoras equation `a^2=b^2+c^2`
There is a comparable one involving `cos^2(x)`

`"Pythagorean relationship"->sin^2(x)+cos^2(x)=1`

So `" "cos^2(x)=1-sin^2(x)" ".............Equation(3)`

Using `Equation(3)` substitute for `cos^2(x)` in `Equation(2)`

`" "sin(x)/cos^2(x)=beta" "->" "sin(x)/(1-sin^2(x))=beta`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus the finished relationship is:

`" "(tan(x))/(cos(x))" "=" "beta" "=" "sin(x)/(1-sin^2(x))`

Answer 2

`sin(x)/cos^2(x)` or `sin(x)/(1-sin^2(x))`

Explanation:

First of all, what are `tan(x)` and `cos(x)`?

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

`tan(x)=sin(x)/cos(x)`

Substitute this in for the numerator in the original problem:

`tan(x)/cos(x)=(sin(x)/cos(x))/cos(x)`

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

`a/b=a*1/b`

Applying this to our equation, we see that:

`(sin(x)/cos(x))/cos(x)=sin(x)/cos(x)*1/cos(x)`

Simplifying, we get:
`sin(x)/cos(x)*1/cos(x)=sin(x)/cos^2(x)`

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

Replacing `cos^2(x)`, we get:

`sin(x)/(1-sin^2(x))`