Question Video: Determinants of Matrices Involving Trigonometric Functions Mathematics • 10th Grade

Video Transcript

Evaluate the determinant of negative four eight sec 𝜃 negative sec 𝜃 two tan squared 𝜃.

Remember these lines represent the determinant of the matrix. For the matrix 𝐴, given by 𝑎, 𝑏, 𝑐, 𝑑, its determinant is calculated by finding the product of elements 𝑎 and 𝑑 and subtracting the product of 𝑏 and 𝑐. Let’s apply this to the matrix we’ve been given.

The determinant of 𝐴 is negative four times two tan squared 𝜃 minus eight sec 𝜃 times negative sec 𝜃. We can simplify our expression to get negative eight tan squared 𝜃 add eight sec squared 𝜃. We are next going to factorize our expression to give us the determinant of 𝐴 equals eight multiplied by six squared 𝜃 minus tan squared 𝜃.

Then, in order to take this further, we do need to recall our trigonometric identities. The one we’re interested in today is tan squared 𝜃 add one equals sec squared 𝜃. Let’s rearrange this formula to get sec squared 𝜃 minus tan squared 𝜃 equals one. We’ve done this so we can replace sec squared 𝜃 minus tan squared 𝜃 with one in our expression, leaving us with the determinant of 𝐴 equals eight times one which is eight.

The determinant of our matrix is eight.