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Solve the equation the determinant of negative cos π, sin π, csc π, csc π equals negative two given that π is greater than zero degrees and less than 90 degrees.
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Letβs begin by recalling what we mean by the determinant of a matrix, specifically a two-by-two matrix.
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The determinant of a general two-by-two matrix π, π, π, π is equal to ππ minus ππ.
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Using this definition, the determinant of the matrix negative cos π, sin π, csc π, csc π is negative cos π multiplied by csc π minus sin π multiplied by csc π.
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To help us progress from here, letβs firstly recall that csc π is equal to one over sin π.
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So letβs replace csc π with one over sin π.
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So we have negative cos π multiplied by one over sin π minus sin π multiplied by one over sin π equals negative two.
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Iβm going to rewrite this term as negative cos π over sin π and this term as sin π over sin π.
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Then we can see that sin π over sin π is just equal to one.
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And we can also rewrite cos π over sin π using an identity that we know.
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Because we know that sin π over cos π is equal to tan π, then cos π over sin π is equal to one over tan π.
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So we have that negative one over tan π minus one is equal to negative two.
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Letβs now simplify the equation weβve got.
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We can begin by adding one over tan π to both sides of this equation.
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That gives us negative one equals negative two add one over tan π.
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And then letβs add two to both sides, giving us one equals one over tan π.
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We can then multiply both sides by tan π.
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This gives us that tan π is equal to one, so now we just need to solve for π.
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We could do this by finding the inverse tan of one, which would give us 45 degrees.
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We could alternatively use the graph of tan to see that when tan of π is equal to one, π is equal to 45 degrees.
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Note that there will be more values where tan π is equal to one on this graph.
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In fact, the next one is at 180 add 45.
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But this is 225 and thatβs out of the range weβve been given for the question.
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So the answer is π equals 45 degrees.