Answer:
[tex]\bold{\frac{(cosx-sinx)}{(sinx)}}=\bold{\frac{cosx-sinx}{sinx}}[/tex]
Step-by-step explanation:
[tex]\frac{cos^2x-sin^2x}{sin^2x+sinxcosx}=cotx-1[/tex]
We're going to start by manipulating the left side of the equation and making it the same form as [tex]cotx-1[/tex].
Start by applying the difference of two squares formula to the numerator, like so:
- [tex]\frac{(cosx+sinx)(cosx-sinx)}{sin^2x+sinxcosx}[/tex]
Now simplify the denominator by expanding the [tex]sin^2x[/tex].
- [tex]\frac{(cosx+sinx)(cosx-sinx)}{(sinx)(sinx)+sinxcosx}[/tex]
The denominator can even be further simplified since both addends (when added together = a sum) have the common factor of [tex]sinx[/tex]. Factor it out.
- [tex]\frac{(cosx+sinx)(cosx-sinx)}{(sinx)(sinx+cosx)}[/tex]
Cancel out the common factor [tex](cosx+sinx)[/tex].
- [tex]\bold{\frac{(cosx-sinx)}{(sinx)}}[/tex]
Since this is the furthest simplified that the left side can be manipulated, let's see if can try to manipulate the right side to also look like [tex]\frac{(cosx-sinx)}{(sinx)}[/tex].
Start by expressing [tex]cotx-1[/tex] with [tex]sinx[/tex] and [tex]cosx[/tex], since we know that cotangent is simply [tex]\frac{x}{y} \rightarrow\frac{cosx}{sinx}[/tex].
- [tex]\frac{cosx}{sinx}-1[/tex]
We can simplify this expression to look like our expression we found by manipulating the left side [tex](\frac{(cosx-sinx)}{(sinx)})[/tex] by making the 1 have a common denominator of [tex]sinx[/tex].
To do this, multiply 1 by [tex]\frac{sinx}{sinx}[/tex]. Now the expression should look like:
- [tex]\frac{cosx}{sinx}-\frac{sinx}{sinx}[/tex]
Since they have a common denominator we can write the expression under one fraction, like so:
- [tex]\bold{\frac{cosx-sinx}{sinx}}[/tex]
This looks exactly the same as what we manipulated the left side to be [tex](\frac{(cosx-sinx)}{(sinx)})[/tex], just without parentheses. I put both expressions in bold. Therefore, this identity proves to be true as we just proved it.
