Question Video: Solving Trigonometric Equations Involving Determinants | Nagwa Question Video: Solving Trigonometric Equations Involving Determinants | Nagwa

# Question Video: Solving Trigonometric Equations Involving Determinants Mathematics • First Year of Secondary School

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Solve the equation |βcos π, sin π and csc π, csc π| = β2 given that 0Β° < π < 90Β°.

02:45

### Video Transcript

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.

Letβs begin by recalling what we mean by the determinant of a matrix, specifically a two-by-two matrix. The determinant of a general two-by-two matrix π, π, π, π is equal to ππ minus ππ. 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.

To help us progress from here, letβs firstly recall that csc π is equal to one over sin π. So letβs replace csc π with one over sin π. So we have negative cos π multiplied by one over sin π minus sin π multiplied by one over sin π equals negative two. Iβm going to rewrite this term as negative cos π over sin π and this term as sin π over sin π. Then we can see that sin π over sin π is just equal to one. And we can also rewrite cos π over sin π using an identity that we know. 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. So we have that negative one over tan π minus one is equal to negative two.

Letβs now simplify the equation weβve got. We can begin by adding one over tan π to both sides of this equation. That gives us negative one equals negative two add one over tan π. And then letβs add two to both sides, giving us one equals one over tan π. We can then multiply both sides by tan π. This gives us that tan π is equal to one, so now we just need to solve for π. We could do this by finding the inverse tan of one, which would give us 45 degrees.

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. Note that there will be more values where tan π is equal to one on this graph. In fact, the next one is at 180 add 45. But this is 225 and thatβs out of the range weβve been given for the question. So the answer is π equals 45 degrees.

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