### Video Transcript

Simplify one minus cos two π₯ over
one plus cos two π₯.

We write out the fraction
again. In order to simplify, it looks like
weβre going to have to rewrite cos two π₯. There are basically three ways to
rewrite cos two π₯: one involving both cos π₯ and sin π₯, another involving only cos
π₯, and the final one involving only sin π₯. Any one of these identities would
work, but we will use the first one. We might see later how we could
have made our lives easier for ourselves by choosing to rewrite cos two π₯ in other
ways, but for the moment we just replace cos two π₯ by cos squared π₯ minus sin
squared π₯.

We expand out the parentheses,
being careful to make the minus-minus sin squared π₯ in the numerator become a plus
sin squared π₯. What we have now doesnβt look any
simpler than what we started with, but we can apply another identity. sin squared π₯
plus cos squared π₯ is one; rearranging this, we get one minus cos squared π₯ equals
sin squared π₯ and one minus sin squared π₯ equals cos squared π₯. We notice that we have one minus
cos squared π₯ in the numerator of our fraction, so we can replace that by a sin
squared π₯. And we can combine this sin squared
π₯ we get with the other sin squared π₯ that comes from the numerator to get two sin
squared π₯.

Itβs a similar story in the
denominator, where there is a one and a minus sin squared π₯. Using one of our identities, we can
replace those two terms by cos π₯ squared. So we take this cos π₯ squared and
add it to the other cos π₯ squared that we get from the denominator of the fraction
and together they make two cos squared π₯. Now, it does look like we have
something simpler, but we can go further. Before we continue to simplify,
letβs note that we could have got to this step quicker by rewriting cos two π₯ in
some other way.

Had rewritten the cos two π₯ in the
numerator as one minus two sin squared π₯, we would have got the numerator two sin
squared π₯ straightaway after simplification. We wouldnβt have had to use the
further identity one minus cos squared π₯ equals sin squared π₯. This is because the identity sin
squared π₯ plus cos squared π₯ equals one is already built into the identity cos two
π₯ equals one minus two sin squared π₯. You can get this identity by
replacing the cos squared π₯ in the first identity by one minus sin squared π₯.

In a similar way had we rewritten
cos two π₯ as two cos squared π₯ minus one, the denominator would have become one
plus two cos squared π₯ minus one or just two cos squared π₯. We had to do some work to find this
denominator, applying the identity one minus sin squared π₯ equals cos squared
π₯. When had weβve been slightly more
clever about which cos two π₯ identity to apply, we could have got this straight
away.

In any case, we end up with the
fraction two sin squared π₯ over two cos squared π₯. We can cancel the common factor of
two in the numerator and the denominator. This leaves us with sin squared π₯
over cos squared π₯, which is the same as sin π₯ over cos π₯ squared. And as sin π₯ over cos π₯ is tan
π₯, this is tan squared π₯. And as we canβt simplify any
further, this is our final answer: one minus cos two π₯ over one plus cos two π₯ in
simplest form is tan squared π₯.