# Question Video: Using the Pythagorean Identities to Evaluate a Trigonometric Function Involving Cofunction Identities given the Trigonometric Function and Quadrant of an Angle Mathematics

Find the value of sec (90Β° + π) given csc π = 17/8 where 0Β° < π < 90Β°.

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### Video Transcript

Find the value of sec 90 degrees plus π given that csc π equals 17 over eight, where π is between zero degrees and 90 degrees.

There are various ways that we can solve this question. We can use the fact that csc π is 17 over eight and π is between zero degrees and 90 degrees to find the value of π. And once we have this value of π, we can just substitute it into the expression sec 90 degrees plus π that we wanted to evaluate.

So letβs do this. Csc π is equal to 17 over eight; csc π is a one over sin π, and so we have that one over sin π is 17 over eight. Taking the reciprocal on both sides of the equation, or just rearranging the equation to make the subject sin π, we see that sin π is equal to eight over 17, and hence that π is the inverse sin or arcsin of eight over 17.

With our calculator in degree mode, we find that the π is equal to 28.072 dot dot dot degrees. This value that are calculator gives us is in the range that we want; itβs between zero degrees and 90 degrees. And so we donβt have to modify it in any way. We just substitute this value of π into sec 90 degrees plus π, remembering that sec π₯ is one over cos π₯ if our calculator doesnβt have a sec button to get a value of negative 17 over eight.

But perhaps your calculator doesnβt give you this exact value, negative 17 over eight, as a fraction or perhaps you canβt use a calculator at all. Luckily we can answer this question without using a calculator. We go back a few steps to where we discovered that sin π is equal to eight over 17, and we turn our attention to sec 90 degrees plus π.

Sec π₯ is one over cos π₯ and so sec 90 degrees plus π is one over cos 90 degrees plus π. We can use the angle sum formula for cosine, setting π₯ equal to 90 degrees and π¦ equal to π. 90 degrees is a special angle, so we know the values of cos 90 degrees and sin 90 degrees.

Cos 90 degrees is zero and sin 90 degrees is one. The zero cos π term disappears, and weβre left with one over negative one sin π, which is negative one over sin π. And we know that one over sin π is equal to csc π, so we find this is negative csc π. And weβre told in the question that csc π is 17 over eight, so the value of sec 90 degrees plus π is negative 17 over eight.

So that method preceded by simplifying sec 90 degrees plus π using an angle sum identity. There is a third method which Iβll show you for fun, which involves doing something slightly nonobvious. We write down the expression we wish to evaluate again, sec 90 degrees plus π, and we notice that it looks very much like the left-hand side of the cofunction identity sec 90 degrees minus π₯ equals csc π₯.

The only difference is that we have a 90 degrees plus something and not 90 degrees minus something. But we can fix that by writing 90 degrees plus π as 90 degrees minus negative π. We set π₯ equal to negative π in the identity to get csc negative π. But csc is an odd function, and if you donβt believe me, weβll prove this at the end of the video.

And so csc of negative π is negative csc π. Using the value of csc π weβre given in the question, we get a value of negative 17 over eight. That was a much quicker way of finding the value of sec 90 degrees plus π using the cofunction identity sec 90 degrees minus π₯ equals csc π₯ and the fact that csc is an odd function.

This last fact is something we should prove csc negative π₯ is equal to one over sin negative π₯ by definition and certainly sine is an odd function. We can prove this, but weβre not going to in this video. And so sine of negative π₯ is negative of sine π₯. Multiplying the numerator and denominator of this fraction by negative one, we get that this is equal to negative one over sine π₯. And one over sine π₯ is csc π₯, so we get that this is equal to negative csc π₯.

We have proved that csc negative π₯ is equal to negative csc π₯, and hence that the csc function is odd.

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