WEBVTT
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Simplify tan π sin π over sec π.
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To solve this problem, weβre actually gonna have to use a couple of trigonometric identities.
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The first is that sec π, just short for secant π, is actually the reciprocal function of cos π.
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So therefore, we can say that sec π is equal to one over cos π.
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And the second identity that weβre gonna use is actually this one that says that tan π is equal to sin π over cos π.
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And this one is a very common one that you will use in a lot of different problems.
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First thing weβre gonna do to simplify our expression, is actually substitute the values that we have from our trig identities into our expression.
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First, we can see that weβve substituted in sin π over cos π for tan π.
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And now, weβve substituted in one over cos π for sec π.
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So we now have this expression that says that sin π over cos π multiplied by sin π all over one over cos π is equal to our original expression of tan π sin π over sec π.
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Okay, well this is a bit clumsy.
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So letβs see if I can simplify this a bit.
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So it can help us to simplify it.
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So Iβve made it easier to actually understand.
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We have sin π over cos π, then multiplied by sin π.
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And then because we were dividing by one over cos π, weβve actually found the reciprocal.
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And actually, itβs cos π over one.
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And instead of dividing, weβre multiplying.
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So now, this is much easier to deal with.
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And we could now fully simplify.
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Well, as we can see here, Iβve tidied it up.
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Havenβt quite fully simplified yet, but itβs equal to sin squared π cos π all over cos π.
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Okay, so one more step.
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So then if we divide our sin squared π cos π by cos π, weβre just left with sin squared π.
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So we could say that if we simplify tan π sin π over sec π fully, then weβre left with sin squared π.
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I just want to quickly recap what we did.
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Just gonna remind you of this kind of problem.
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First of all, we had a look at some of our trigonometric identities.
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And we picked these two here to help us with this problem.
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Next, we used these to substitute back into our expression, to actually eliminate some of the trig ratios we actually had within it.
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So that gave us this expression over here.
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We then simplified the expression we had.
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And it gave us our final answer of sin squared π.