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

Evaluate |β4, 8 sec π and βsec π, 2 tanΒ² π|.

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### Video Transcript

Evaluate the determinant of negative four eight sec π negative sec π two tan squared π.

Remember these lines represent the determinant of the matrix. For the matrix π΄, given by π, π, π, π, its determinant is calculated by finding the product of elements π and π and subtracting the product of π and π. Letβs apply this to the matrix weβve been given.

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

Then, in order to take this further, we do need to recall our trigonometric identities. The one weβre interested in today is tan squared π add one equals sec squared π. Letβs rearrange this formula to get sec squared π minus tan squared π equals one. Weβve done this so we can replace sec squared π minus tan squared π with one in our expression, leaving us with the determinant of π΄ equals eight times one which is eight.

The determinant of our matrix is eight.