Question Video: Determinants of Matrices Involving Trigonometric Functions | Nagwa Question Video: Determinants of Matrices Involving Trigonometric Functions | Nagwa

Question Video: Determinants of Matrices Involving Trigonometric Functions Mathematics • First Year of Secondary School

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Evaluate |βˆ’4, 8 sec πœƒ and βˆ’sec πœƒ, 2 tanΒ² πœƒ|.

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Video Transcript

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

Remember these lines represent the determinant of the matrix. For the matrix 𝐴, given by π‘Ž, 𝑏, 𝑐, 𝑑, its determinant is calculated by finding the product of elements π‘Ž 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 πœƒ minus eight sec πœƒ times negative sec πœƒ. We can simplify our expression to get negative eight tan squared πœƒ add eight sec squared πœƒ. We are next going to factorize our expression to give us the determinant of 𝐴 equals eight multiplied by six squared πœƒ minus tan squared πœƒ.

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 πœƒ add one equals sec squared πœƒ. Let’s rearrange this formula to get sec squared πœƒ minus tan squared πœƒ equals one. We’ve done this so we can replace sec squared πœƒ minus tan squared πœƒ 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.

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