Question

cos^2x-sin^2x/sin^2x+sinxcosx=cotx-1​

Answer 1

$\large{\sf{\frac{cos^2x-sin^2x}{sin^2x+sinxcosx}=cotx-1}}$

We're going to star by manipulation the left side of the equation and making it the same from as $\sf{cotx-1}$

Start by applying the difference of two squares formula to the numerator,like so :

• $\large{\bf{\frac{(cosx+sinx)(cosx-sinx)}{sin^2x+sinxcosx}}}$

Now simply the denominator by expanding the $\sf{sin^2x}$

• $\large{\bf{\frac{(cosx+sinx)(cosx-sinx)}{(sinx)(sinx)+sinxcosx}}}$

The denominator can even be further simplified since both addends (when added together = a sum) have the common factor of sinxsinx . Factor it out.

• $\large{\bf{\frac{(cosx+sinx)(cosx-sinx)}{(sinx)(sinx+cosx)}}}$

Cancel out the common factor$\sf{ (cosx+sinx)(cosx+sinx) .}$

• $\large{\bf{\frac{(cosx-sinx)}{(sinx)}}}$

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 $\sf{\frac{(cosx-sinx)}{(sinx)}}$

Start by expressing$\sf{ cotx-1}$with $\sf{sinx}$and$\sf{ cosx}$ , since we know that cotangent is simply$\sf{ \frac{x}{y} \rightarrow\frac{cosx}{sinx}}$

• $\large{\bf{\frac{cosx}{sinx}-1 }}$

We can simplify this expression to look like our expression we found by manipulating the left side $\sf{(\frac{(cosx-sinx)}{(sinx)})}$) by making the 1 have a common denominator of$\sf{ sinx}$ .

To do this, multiply 1 by$\frac{sinx}{sinx}$

Now the expression should look like:

• $\large{\bf{\frac{cosx}{sinx}-\frac{sinx}{sinx}}}$

Since they have a common denominator we can write the expression under one fraction, like so:

• $\large{\bf{\frac{cosx-sinx}{sinx}}}$

This looks exactly the same as what we manipulated the left side to be$\sf{ (\frac{(cosx-sinx)}{(sinx)})}$, just without parentheses. Therefore, this identity is true.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\sf\large{\red{★proved★}}$