WEBVTT
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Simplify one minus tan π, squared, plus one plus tan π, squared.
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The first thing we do is expand the brackets; one minus tan π squared becomes one minus two tan π plus tan squared π; and one plus tan π all squared becomes one plus two tan π plus tan squared π.
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Now we can collect some like terms.
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We have two terms of one, so that becomes two.
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We have a minus two tan π and a plus two tan π, which cancel out; they form a zero pair.
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And finally we have to two tan squared πβs.
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Okay, so we have two plus two tan squared π now, which is certainly simpler than the expression we started with.
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But I think we can go further.
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Itβs possible that you will notice this is two times one plus tan squared π.
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And you would write it in this way because you remember that one plus tan squared π is equal to sec squared π.
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And so you could write this as two sec squared π.
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But what if you hadnβt thought to write two plus two tan squared π in that way?
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Well thereβs a more reliable method, which is to write everything in terms of sin π and cos π.
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So using the fact that tan π is sin π over cos π, we can rewrite tan squared π as sin π over cos π, squared.
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And so the whole expression becomes two plus two times sin squared π over cos squared π, which we can now write over a common denominator.
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So weβve used the common denominator of cos squared π, writing it is one fraction.
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So now it might be slightly easier to see which identity to apply.
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Weβve got two cos squared π plus two sine squared π, and we know the cos squared π plus sin squared π is one, which means as our numerator two cos squared π plus two sin squared π is just two.
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If you didnβt notice that immediately you might have noticed that sin squared π can be written in terms of cos squared π using this identity slightly rearranged.
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So you can subtract cos squared π from both sides and get sin squared π equals one minus cos squared π.
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And then substitute that in; you get the same numerator, two.
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The final thing is to write two over cos squared π as two times sec squared π, using the fact that sec π is just one over cos π.
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So thatβs our final answer.
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Of course we could have skipped a few lines of working by using the identity one plus tan squared π is equal to sec squared π.
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The advantage of writing everything in terms of cos π and sin π is that although you may take a few more steps to find the final answer, you only have to remember one identity to simplify, which is cos squared π plus sin squared π equals one rather than about six identities including one plus tan squared π equal sec squared π.