Question

Cos^2x-sin^2x/sin^2x+sinxcosx=cotx-1

Answer 1

Answer:

$\bold{\frac{(cosx-sinx)}{(sinx)}}=\bold{\frac{cosx-sinx}{sinx}}$

Step-by-step explanation:

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

We're going to start by manipulating the left side of the equation and making it the same form as $cotx-1$.

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

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

Now simplify the denominator by expanding the $sin^2x$.

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

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

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

Cancel out the common factor $(cosx+sinx)$.

• $\bold{\frac{(cosx-sinx)}{(sinx)}}$

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

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

• $\frac{cosx}{sinx}-1$

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

To do this, multiply 1 by $\frac{sinx}{sinx}$. Now the expression should look like:

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

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

• $\bold{\frac{cosx-sinx}{sinx}}$

This looks exactly the same as what we manipulated the left side to be $(\frac{(cosx-sinx)}{(sinx)})$, just without parentheses. I put both expressions in bold. Therefore, this identity proves to be true as we just proved it.