Find the value of the determinant
with elements cot 𝜃, tan 𝜃, negative cot 𝜃, tan 𝜃.
Remember, for a two-by-two matrix
𝐴 with elements 𝑎, 𝑏, 𝑐, 𝑑, its determinant can be found by subtracting the
product of the top right and bottom left element from the product of the top left
and bottom right. That’s 𝑎𝑑 minus 𝑏𝑐. In this matrix, 𝑎 is cot 𝜃, 𝑏 is
tan 𝜃, 𝑐 is negative cot 𝜃, and 𝑑 is tan 𝜃. Multiplying the top left and bottom
right elements, that’s 𝑎 multiplied by 𝑑, and we get cot 𝜃 tan 𝜃.
We’re then going to subtract the
product of elements 𝑏 and 𝑐. That’s tan 𝜃 multiplied by
negative cot 𝜃. Cot 𝜃 tan 𝜃 minus negative cot 𝜃
tan 𝜃 is cot 𝜃 tan 𝜃 plus cot 𝜃 tan 𝜃. And that is of course two cot 𝜃
Next, we’re going to need to recall
some of our trigonometric identities to simplify this further. Remember, cot 𝜃 is equal to one
over tan 𝜃. So we can rewrite our expression
for the determinant as two multiplied by one over tan 𝜃 multiplied by tan 𝜃. And we can then cancel these two
tan 𝜃s. And in doing so, we can see that
the value of our determinant is two multiplied by one multiplied by one, which is
The value of the determinant of the
matrix given by cot 𝜃, tan 𝜃, negative cot 𝜃, tan 𝜃 is two.