### Video Transcript

Simplify one plus cot squared three
π over two minus π divided by one plus tan squared three π over two minus π.

In order to answer this question,
we will need to use a variety of trigonometric identities. In problems of this type, it is not
always clear where we should start. And in this question, this is
complicated further by the angle three π over two minus π. As a result, we will begin by
letting πΌ equal three π over two minus π. This allows us to rewrite our
expression as one plus cot squared πΌ divided by one plus tan squared πΌ. The Pythagorean identities state
that sin squared πΌ plus cos squared πΌ is equal to one. tan squared πΌ plus one is equal to
sec squared πΌ. And one plus cot squared πΌ is
equal to csc squared πΌ.

We notice that the numerator of our
fraction is the same as the left-hand side of the third identity. This means we can rewrite this as
csc squared πΌ. The denominator of our expression
is the same as the left-hand side of the second identity. We can therefore rewrite the
expression as csc squared πΌ over sin squared πΌ. Two of the reciprocal trigonometric
identities state that csc πΌ is equal to one over sin πΌ and sec πΌ is equal to one
over cos πΌ. The second identity could also be
rewritten as one over sec πΌ is equal to cos πΌ.

Rewriting csc squared πΌ as one
over sin squared πΌ and one over sec squared πΌ as cos squared πΌ, we have one over
sin squared πΌ multiplied by cos squared πΌ. This is equal to cos squared πΌ
over sin squared πΌ, which in turn is equal to cot squared πΌ. Replacing πΌ with three π over two
minus π, we have cot squared of three π over two minus π. We might think this is our final
answer. However, we can simplify this one
stage further. And this can be done by considering
related angles in the unit circle. We know that the point shown on the
unit circle in the first quadrant has coordinates cos π, sin π. We know that three π over two
radians is equal to 270 degrees. The point with coordinates cos of
three π over two minus π, sin of three π over two minus π is as shown.

We notice that the displacement in
the negative π₯-direction of this triangle is the same as the displacement in the
positive π¦-direction of our first triangle. This means that the cos of three π
over two minus π is equal to the negative of sin π. Likewise, the sin of three π over
two minus π is equal to negative cos π. Since cos π divided by sin π is
equal to cot π, dividing these two equations we have cot of three π over two minus
π is equal to negative sin π over negative cos π. The right-hand side simplifies to
tan π. cot squared three π over two minus π is therefore equal to tan squared
π. The initial expression one plus cot
squared three π over two minus π divided by one plus tan squared three π over two
minus π is tan squared π.