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Find the value of the determinant with elements cot π, tan π, negative cot π, tan π.
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Remember, for a two-by-two matrix π΄ with elements π, π, π, π, 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.
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Thatβs ππ minus ππ.
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In this matrix, π is cot π, π is tan π, π is negative cot π, and π is tan π.
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Multiplying the top left and bottom right elements, thatβs π multiplied by π, and we get cot π tan π.
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Weβre then going to subtract the product of elements π and π.
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Thatβs tan π multiplied by negative cot π.
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Cot π tan π minus negative cot π tan π is cot π tan π plus cot π tan π.
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And that is of course two cot π tan π.
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Next, weβre going to need to recall some of our trigonometric identities to simplify this further.
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Remember, cot π is equal to one over tan π.
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So we can rewrite our expression for the determinant as two multiplied by one over tan π multiplied by tan π.
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And we can then cancel these two tan πs.
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And in doing so, we can see that the value of our determinant is two multiplied by one multiplied by one, which is two.
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The value of the determinant of the matrix given by cot π, tan π, negative cot π, tan π is two.