### Video Transcript

Simplify one minus tan π, squared, plus one plus tan π, squared.

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.

Now we can collect some like terms. We have two terms of one, so that becomes two. We have a minus two tan π and a plus two tan π, which cancel out; they form a zero pair. And finally we have to two tan squared πβs.

Okay, so we have two plus two tan squared π now, which is certainly simpler than the expression we started with. But I think we can go further. Itβs possible that you will notice this is two times one plus tan squared π. And you would write it in this way because you remember that one plus tan squared π is equal to sec squared π.

And so you could write this as two sec squared π. But what if you hadnβt thought to write two plus two tan squared π in that way? Well thereβs a more reliable method, which is to write everything in terms of sin π and cos π. 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.

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. So weβve used the common denominator of cos squared π, writing it is one fraction.

So now it might be slightly easier to see which identity to apply. 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.

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. So you can subtract cos squared π from both sides and get sin squared π equals one minus cos squared π. And then substitute that in; you get the same numerator, two.

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. So thatβs our final answer.

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.

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.