Question Video: Simplifying Trigonometric Expressions Using Cofunction and Pythagorean Identities Mathematics

Simplify 1 + cotΒ²(3πœ‹/2 βˆ’ πœƒ)/1 + tanΒ²(3πœ‹/2 βˆ’ πœƒ).

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Video Transcript

Simplify one plus cot squared three πœ‹ over two minus πœƒ divided by one plus tan squared three πœ‹ over two minus πœƒ.

In order to answer this question, we will need to use a variety of trigonometric identities. In problems of this type, it is not always clear where we should start. And in this question, this is complicated further by the angle three πœ‹ over two minus πœƒ. As a result, we will begin by letting 𝛼 equal three πœ‹ over two minus πœƒ. This allows us to rewrite our expression as one plus cot squared 𝛼 divided by one plus tan squared 𝛼. The Pythagorean identities state that sin squared 𝛼 plus cos squared 𝛼 is equal to one. tan squared 𝛼 plus one is equal to sec squared 𝛼. And one plus cot squared 𝛼 is equal to csc squared 𝛼.

We notice that the numerator of our fraction is the same as the left-hand side of the third identity. This means we can rewrite this as csc squared 𝛼. The denominator of our expression is the same as the left-hand side of the second identity. We can therefore rewrite the expression as csc squared 𝛼 over sin squared 𝛼. Two of the reciprocal trigonometric identities state that csc 𝛼 is equal to one over sin 𝛼 and sec 𝛼 is equal to one over cos 𝛼. The second identity could also be rewritten as one over sec 𝛼 is equal to cos 𝛼.

Rewriting csc squared 𝛼 as one over sin squared 𝛼 and one over sec squared 𝛼 as cos squared 𝛼, we have one over sin squared 𝛼 multiplied by cos squared 𝛼. This is equal to cos squared 𝛼 over sin squared 𝛼, which in turn is equal to cot squared 𝛼. Replacing 𝛼 with three πœ‹ over two minus πœƒ, we have cot squared of three πœ‹ over two minus πœƒ. We might think this is our final answer. However, we can simplify this one stage further. And this can be done by considering related angles in the unit circle. We know that the point shown on the unit circle in the first quadrant has coordinates cos πœƒ, sin πœƒ. We know that three πœ‹ over two radians is equal to 270 degrees. The point with coordinates cos of three πœ‹ over two minus πœƒ, sin of three πœ‹ over two minus πœƒ is as shown.

We notice that the displacement in the negative π‘₯-direction of this triangle is the same as the displacement in the positive 𝑦-direction of our first triangle. This means that the cos of three πœ‹ over two minus πœƒ is equal to the negative of sin πœƒ. Likewise, the sin of three πœ‹ over two minus πœƒ is equal to negative cos πœƒ. Since cos πœƒ divided by sin πœƒ is equal to cot πœƒ, dividing these two equations we have cot of three πœ‹ over two minus πœƒ is equal to negative sin πœƒ over negative cos πœƒ. The right-hand side simplifies to tan πœƒ. cot squared three πœ‹ over two minus πœƒ is therefore equal to tan squared πœƒ. The initial expression one plus cot squared three πœ‹ over two minus πœƒ divided by one plus tan squared three πœ‹ over two minus πœƒ is tan squared πœƒ.

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