Solve the equation the determinant of negative cos 𝜃, sin 𝜃, csc 𝜃, csc 𝜃 equals negative two given that 𝜃 is greater than zero degrees and less than 90 degrees.
Let’s 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 𝑎, 𝑏, 𝑐, 𝑑 is equal to 𝑎𝑑 minus 𝑏𝑐. Using this definition, the determinant of the matrix negative cos 𝜃, sin 𝜃, csc 𝜃, csc 𝜃 is negative cos 𝜃 multiplied by csc 𝜃 minus sin 𝜃 multiplied by csc 𝜃.
To help us progress from here, let’s firstly recall that csc 𝜃 is equal to one over sin 𝜃. So let’s replace csc 𝜃 with one over sin 𝜃. So we have negative cos 𝜃 multiplied by one over sin 𝜃 minus sin 𝜃 multiplied by one over sin 𝜃 equals negative two. I’m going to rewrite this term as negative cos 𝜃 over sin 𝜃 and this term as sin 𝜃 over sin 𝜃. Then we can see that sin 𝜃 over sin 𝜃 is just equal to one. And we can also rewrite cos 𝜃 over sin 𝜃 using an identity that we know. Because we know that sin 𝜃 over cos 𝜃 is equal to tan 𝜃, then cos 𝜃 over sin 𝜃 is equal to one over tan 𝜃. So we have that negative one over tan 𝜃 minus one is equal to negative two.
Let’s now simplify the equation we’ve got. We can begin by adding one over tan 𝜃 to both sides of this equation. That gives us negative one equals negative two add one over tan 𝜃. And then let’s add two to both sides, giving us one equals one over tan 𝜃. We can then multiply both sides by tan 𝜃. This gives us that tan 𝜃 is equal to one, so now we just need to solve for 𝜃. 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 𝜃 is equal to one, 𝜃 is equal to 45 degrees. Note that there will be more values where tan 𝜃 is equal to one on this graph. In fact, the next one is at 180 add 45. But this is 225 and that’s out of the range we’ve been given for the question. So the answer is 𝜃 equals 45 degrees.