### Video Transcript

Solve the equation the determinant
of negative cos π, sin π, csc π, csc π equals negative two given that π is
greater than zero degrees and less than 90 degrees.

Letβs begin by recalling what we
mean by the determinant of a matrix, specifically a two-by-two matrix. The determinant of a general
two-by-two matrix π, π, π, π is equal to ππ minus ππ. Using this definition, the
determinant of the matrix negative cos π, sin π, csc π, csc π is negative cos π
multiplied by csc π minus sin π multiplied by csc π.

To help us progress from here,
letβs firstly recall that csc π is equal to one over sin π. So letβs replace csc π with one
over sin π. So we have negative cos π
multiplied by one over sin π minus sin π multiplied by one over sin π equals
negative two. Iβm going to rewrite this term as
negative cos π over sin π and this term as sin π over sin π. Then we can see that sin π over
sin π is just equal to one. And we can also rewrite cos π over
sin π using an identity that we know. Because we know that sin π over
cos π is equal to tan π, then cos π over sin π is equal to one over tan π. So we have that negative one over
tan π minus one is equal to negative two.

Letβs now simplify the equation
weβve got. We can begin by adding one over tan
π to both sides of this equation. That gives us negative one equals
negative two add one over tan π. And then letβs add two to both
sides, giving us one equals one over tan π. We can then multiply both sides by
tan π. This gives us that tan π is equal
to one, so now we just need to solve for π. We could do this by finding the
inverse tan of one, which would give us 45 degrees.

We could alternatively use the
graph of tan to see that when tan of π is equal to one, π is equal to 45
degrees. Note that there will be more values
where tan π is equal to one on this graph. In fact, the next one is at 180 add
45. But this is 225 and thatβs out of
the range weβve been given for the question. So the answer is π equals 45
degrees.