in this video we're going to be going through 14 verifying trigonometric identity problems I want to challenge you to see how many of these can you get right on your own put the answer down in the comment section below let's jump into the first one we've got cosecant squared theta minus 1 divided by cosecant squared theta and we want to show or prove or verify that it equals cosine squared theta so the first thing you want to do when you tackle these identities is you want to look at which side is like larger or more expanded or complicated and then we'll work with that side and we'll condense it down to the side that's more condensed or compact so in this case it clearly seems to be the left side that's more expanded so let's start there we've got cosecant squared theta minus 1 whenever I see the squared okay trig functions I always think of these three Pythagorean trig identities right here it doesn't always have to be these but oftentimes it's helpful to use these so here we have cosecant squared if we subtract 1 from both sides see how cosecant squared minus 1 equals cotangent squared so we can go ahead and make that replacement so we're gonna replace cosecant squared theta minus 1 with cotangent squared okay and this is all over cosecant squared theta and we're trying to verify or prove that it equals cosine squared theta so the next thing I want to do is when I look at cotangent squared you can see that cotangent equals cosine over sine but if cotangent is squared then we're going to square this side as well it's going to be cosine squared over sine squared let's go ahead and replace that numerator with cosine squared theta over sine squared theta and then cosecant squared theta we know that cosecant is 1 over sine but if this is squared this will be 1 over sine squared so let's replace that so when you're thinking about identity is really you know these quantities they're identical we're just interchanging one with the other so we're not really changing the problem we're just changing the way that it looks now here what we're going to do is we've got cosine squared over sine squared 1 over sine squared let's go ahead and multiply the numerator and the denominator by sine squared theta because anything divided by itself this is 1 right and so what happens is this science Squared's are going to cancel and we've got cosine squared over 1 which equals cosine squared theta which is what we were trying to prove that it's equal to it and you got okay number two see if you can do this one on your own remember when you were in geometry and you had to do those two column proofs this is kind of what this is like it's just that we're doing with trig functions so here when we look at the left side and the right side which one you think is more expanded well definitely the right side you can see we've got two fractions here and here we just have one term here so what we want to do is we want to combine these together into one term right the nice thing about these is that we know what the answer is that's the good news the bad news is there's not one set you know way of doing each one of these problems but as long as you make an identical substitution you know you're gonna have the same quantity it's just in a different form so definitely want to practice a lot of these to get good at them but here the first thing I noticed is that you know just like an algebra you know how to get a common denominator we want to get a common denominator for both these fractions this one has a 1 minus cosine theta but it doesn't have a 1 plus cosine theta and this has a 1 plus cosine theta but it doesn't have a 1 minus cosine theta so what I'm gonna do here is I'm going to multiply by what it's missing so here I'm gonna multiply this by 1 plus cosine theta and remember whatever you do to the denominator you have to do that to the numerator okay because that's like multiplying by 1 here we're missing a 1 minus cosine theta so we're gonna multiply the numerator by 1 minus cosine theta okay so now let's go ahead and simplify a little bit here we have a binomial times a binomial what you want to do is you want to multiply this together so you have 1 times 1 which is 1 you have cosine theta and negative cosine theta which cancel and then we have cosine theta times negative cosine theta which is negative cosine squared theta and now what I'm going to do is when I combine these together since they have the same denominator now we're gonna put it all over this common denominator in the numerator we have 1 plus cosine theta and then over here we have 1 minus cosine theta okay now keep in mind you want to keep one eye on the left side one eye on the right side we know where we want to go to so we want to make sure we're heading in that right direction but at this point we're just trying to simplify our condense as much as we can so here you can see I've got a cosine theta and a negative cosine theta those cancel here we've got 1 plus 1 which is 2 over 1 minus cosine squared theta now remember when I said you see these ones with the second degree the squared ones I always think of these three Pythagorean trig identities well that's what we have here we've got 1 if I subtract cosine squared theta from both sides of this equation it's still going to be an identity it's still going to be equivalent but now we have sine squared equals 1 minus cosine squared so let's replace this denominator with what it equals which is sine squared theta ok now we're almost there you notice the sign is in the denominator right so you can see that this is actually sine squared so you can think of this as 2 times 1 okay 2 times 1 is still 2 but look one over sine squared we know is going to be cosecant squared so this is going to equal 2 cosecant squared theta and that's what we were trying to prove there was equal to and you got our 3 see if you can do this one we've got cotangent squared theta plus 1 times sine squared theta minus 1 and we're trying to show that equals a negative cotangent squared theta so when I was learning this in school myself you know my teachers they would say if you don't know what to do just do something just make some type of substitution if it's the wrong one meaning it you know maybe it gets more complicated you can always back up a couple steps and start over and try a different substitution but don't stay stuck you definitely want to try some type of substitution and try to move forward you're gonna get better as you do these more and more and you'll start to recognize what's going to be an easier substitution as you get experience so in this one again we want to look at which side is more expanded which side is more condensed right so what do you think it looks definitely like the left side is more expanded so let's start there now notice we've got those second-degree trig functions right there squared again like we've been talking about we use these Pythagorean trig identities a lot it's good to memorize them or you know at least write them down and have them in front of you so let's take a look what we have here we've got cotangent squared theta plus 1c cotangent squared squared theta plus 1 we know that equals cosecant squared theta let's replace that with cosecant squared theta okay now over here we have sine squared theta minus 1 now I'll follow me on this one see if I subtract 1 to the left side of the equation and if I subtract cosine squared theta to the right side sine squared theta minus 1 actually equals negative cosine squared theta so we're going to replace this with negative cosine squared theta okay now let's take a look what we have so far it's a little bit more condensed cosecant squared we know is 1 over sine squared so let's go ahead and make that replacement 1 over sine squared theta and then over here we've got negative cosine squared theta okay let's just leave that as a negative cosine squared theta another little hint is if you ever get stuck all these trig functions are basically made up of sines and cosines so if you get really stuck you can always convert back to sines and cosines those are the basic building blocks and you can figure it out from there but I try not to do that you know at the very beginning if I can avoid it I try to use some of the other identities first but that's always the last kind of bet that you can use so we've got negative cosine squared theta over sine squared theta and now let's take a look what we have we cosine over sine is cotangent but these are both squared this is going to be cotangent square because both sides are squared and we have a negative so it's going to be negative cotangent squared theta and you can see we've got to match so we proved it okay number four see if you can do this one cosecant theta plus cotangent theta equals sine of theta over one minus cosine theta how would you do that one well first of all I wanted to say that you know congratulations most people probably bailed already on this video because it seems too challenging but I want to congratulate you for sticking with it it definitely the more you do the better you're gonna get at it and you know I'm here to cheer you on you know that's what my whole channel is really devoted to you know in athletics you know how they have cheerleaders that are cheering on that the athletes well you guys are mathletes right and so what we want to do here is we want to improve our math skills so in this one what do you think is the better side to work with you really can't make a mistake you know but and this one I'm going to start with the right side and we're going to try and work with that so here what I'm gonna do is I'm gonna use a little technique called multiplying by the conjugate so conjugate is you change the sign in between this these two terms of this binomial so I'm gonna multiply by one plus cosine theta and one plus cosine theta and you're probably saying Mario how did you know how to do that right part of it comes with experience when you see these come up over and over again you'll recognize this as a valid technique so let's go ahead and multiply the denominators together first we've got 1 times 1 which is 1 again the inside and outside cancel we have negative cosine theta and positive cosine theta and then the last term's give us negative cosine squared theta and the numerator I'm just gonna leave this factored as sine theta and 1 plus cosine theta because sometimes what happens is you're gonna be able to get some terms that cancel in the numerator and denominator when I look at 1 minus cosine squared theta I see that squared so I'm thinking Pythagorean trig identities possibly right 1 minus cosine squared theta will leave us with sine squared theta so I'm going to replace this denominator with sine squared theta I'm gonna leave the numerator the same sine theta times 1 plus cosine theta now notice we've got 1 sine theta here to sign you know sine squared theta here are basically sine times sine so what you can do is you can think of one of these signs canceling with one of these sines and we just have sine theta left over in the denominator so now let's see what we have left we've got 1 plus cosine theta over sine theta what I'm going to do here is I'm going to actually split this up into two fractions I'm going to say 1 over sine theta plus cosine theta over sine theta see I'm just like splitting it up into two parts so let's write that down 1 over sine theta plus cosine theta over sine theta and one of our sine theta we know is cosecant theta right right here so that's going to be cosecant theta these are called the reciprocal identities and cosine over sine theta we know that that is cotangent theta and you can see we've got a match so we verified or R approved okay number 5 how would you do this one we've got cotangent sin - 4 theta equals cotangent squared theta times cosecant squared theta minus cotangent squared they okay so that was a mouthful so I think here I'll start with the right sign it looks like a little bit more expanded and the technique I'm going to use here on this one is I'm going to actually see this group here and see this group here see how they have a cotangent squared theta in common well let's go ahead and factor that out as the greatest common factor so cotangent squared theta and then we're left with cosecant squared theta minus one a nice thing about factoring you can check your work if you distribute back in you should get back to your original now notice how we have this cosecant squared theta when I have this squared I again think of the Pythagorean trig identities cosecant squared minus one would leave us with cotangent squared on the left there so let's go ahead and replace this with cotangent squared theta and this is x cotangent squared theta and we've got cotangent x cotangent when you multiply just like with an algebra right the rules of exponents you add the exponents so it's gonna be cotangent to the fourth theta which is what we have here and we got it so some of the problems that we're doing are a little bit more challenging some of them are a little bit more easier I try to get a nice mix in here but I'd say they're you know kind of in the middle at about the medium level of difficulty number six how would you do this one we've got 1 plus cosine theta over sine theta plus sine theta over one plus cosine theta equals two cosecant theta so definitely the left side seems more expanded let's start with that one the key here would be really to combine these into one fraction with a common denominator you can see there's like two terms here there's only one term here so what's the common denominator well we have to multiply by what's missing right so here you've got one plus cosine theta we need a sine theta so we're gonna multiply top and bottom by sine theta so that's gonna make this sine squared theta okay and it's all over the common denominator which is going to be sine theta one plus cosine theta over here we've got a sine theta but we're missing a one plus cosine theta so we're gonna have to multiply one plus cosine theta times one plus cosine theta which is one plus cosine theta the quantity squared okay because the same thing twice and this is added so now let's go ahead and simplify this out if we foil this out one plus cosine theta times one plus cosine theta we get one plus two cosine theta plus cosine squared theta plus sine squared theta all over our common denominator sine theta times one plus cosine theta now cosine squared plus sine squared that's your basic Pythagorean trig identity that's going to equal one so we can replace this here with one and notice we also have one in one here so let's go ahead and combine those together to two so we've got two plus two cosine theta over sine theta one plus cosine theta now notice we can factor out a two here in the numerator so this is two times one plus cosine theta all divided by sine theta one plus cosine theta and see the numerator and denominator that 1 plus cosine theta cancels we have two over sine theta which is equal to two cosecant theta and we've got a match and you've proved it okay let's try some number seven and number eight these are a little bit easier we've got cotangent of PI over 2 minus theta times cotangent of theta we want to show that it equals one when you see this pi over 2 minus theta or 90 minus theta these are called KO functions now let's go down here to the bottom it's cofunctions cotangent of PI over 2 minus theta is equal to tangent of theta so let's go ahead and replace this with tangent of theta times the cotangent of theta now we know that cotangent and tangent are reciprocals of one another so we can replace cotangent with one over tangent and then tangent of theta we can just think of that as tangent of theta over one and these are gonna cancel on the diagonal their numerator denominator we get 1 over 1 which is 1 and we've proved it 1 equals 1 okay try number 8 now negative sine of negative theta divided by negative cosine of negative theta we want to show that equals negative tangent theta now when you see these negative angles these are what are called even and odd identities and that's these ones right here so the sine of negative theta is the same as sine of positive theta but it's multiplied by negative 1 okay that's this is called an odd function because it's the opposite right but we have this negative here so we've got negative sine of negative theta we said is negative sine positive theta okay all divided by negative cosine of negative theta so cosine of negative theta is the same as cosine of positive theta so this is going to be cosine of positive theta multiplied by this negative here in front and now you can see a negative times a negative that's gonna be a positive a positive divided by a negative is going to be a negative and sine divided by cosine is tangent so we've got negative tangent of theta equals negative tangent of theta and we proved it okay number nine see if you can do this one you've got cosecant to the fourth theta times cosecant theta cotangent theta minus cosecant squared theta times cosecant theta cotangent theta and we want to show it equals cosecant cubed theta times cotangent q theta so this one's a real large one definitely the left side is more expanded than the right side and what I'm going to do on this one is factor out the greatest common factor so you can see that this whole thing is a group this whole thing is a group we want to see what they have in common let's start off with what's in front here we've got cosecant squared theta cosecant to the fourth theta so we can factor out a cosecant squared theta also look what's in parentheses you got a cosecant theta cotangent theta so let's also factor out a cosecant theta cotangent theta and if we do that we're gonna be left with this cosecant squared cosecant squared theta minus one okay because we factored out this whole thing here so we're just left with one here we factored out cosecant squared so we were just left with cosecant squared when we divided this out of here so now look what we've got we've got cosecant squared theta minus one if we subtract one from both sides of this Pythagorean trig identity you can see we're going to be left with cotangent squared theta and now let's see what we've got we've got cosecant squared times cosecant to the first that's going to give us cosecant cubed here we have cotangent to the first times cotangent to the second that gives us cotangent cubed and we've got it so that was an easy one once you've factored out the greatest common factor number ten how would you do this one one plus cosine theta times one minus cosine of negative you know we want to show that equals sine squared theta so definitely easier to start with on the left side it's more expanded you can see we've got a negative angle so we're thinking of our even and odd identities the cosine of negative theta we know is the same as cosine of positive theta okay so this whole thing is going to be cosine positive theta and we still have one minus and this over here is one plus cosine theta and now if we foil this together one times one is one cosine theta and negative cosine theta those cancel and then cosine theta times negative cosine theta is negative cosine squared theta you can see we've got that squared cosine so we can think of our Pythagorean trig identities 1 minus cosine squared equals sine squared and we proved it ok number 11 how would you do this one sine theta times tangent theta all over 1 minus cosine theta minus 1 equals secant theta all right so on this one you can see the left side is more expanded we want to condense it down to secant theta but what I have here is I've got two groups two terms separated by this minus sign so what we really want to do is combine them into one group by getting a common denominator Loosli are common denominators going to be 1 minus cosine theta so I'm going to multiply the numerator and denominator by 1 minus cosine theta so that gives us 1 minus cosine theta over 1 minus cosine theta ok and because we had a common denominator just can extend this fraction here sine theta and tangent theta is sine theta over cosine theta right here tangent theta so now what I'm going to do is I'm going to do a little bit of simplifying here I'm gonna distribute the negative and sine theta is like sine theta over one so I'm gonna multiply the numerators and denominators so I get sine squared theta over 1 times cosine is cosine theta minus 1 ok and then a negative times a negative gives you plus cosine theta all over the common denominator 1 minus cosine theta and keep in mind we're trying to show that this equals secant theta so now what's interesting is we have this fraction within a larger for action so what we want to do is want to get rid of that complex fraction I'm going to do that by multiplying the numerator by cosine theta and the denominator by cosine theta because that's like multiplying by one so if I distribute cosine to each of these terms in the numerator here are the cosines are going to cancel and that's just gonna leave us with sine squared theta cosine theta times negative one is negative cosine theta cosine theta times cosine theta is cosine squared theta and then the denominator I'm just gonna leave this as cosine theta times one minus cosine theta and the reason I'm leaving a factor is because we might get some cancellation numerator and denominator now notice how we have sine squared plus cosine squared and that's our basic Pythagorean trig identity that equals one so this gives us one minus cosine theta over cosine theta times one minus cosine theta so you can see that this 1 minus cosine theta and the numerator denominator cancel out and we just have 1 over cosine theta which equals secant theta which is what we were trying to prove see if you can do number 12 and 13 here we've got cosecant squared theta minus tangent squared of PI over 2 minus theta and we're trying to show that equals 1 so definitely the left side is more expanded notice we've got this pi over 2 minus theta we want to refer to our cofunction identities and you can see over here tangent of PI over 2 minus theta is equal to cotangent of theta but because this is tangent squared then this is going to be cotangent squared it's like we're squaring both sides so this is going to be cotangent squared theta okay cotangent squared theta and this is cosecant squared theta minus cotangent squared theta and again when you have these ones to the second power I like to think of the Pythagorean trig identities cosecant squared if we subtract the cotangent squared over that's going to leave us with 1 so you can see one equals one that was kind of a quick and short one let's try number 13 now 1 over sine theta plus 1 plus 1 over sine theta minus 1 we want to show that equals negative 2 tangent theta secant theta all right how would you do that one well notice how we have a 1 group here we've got two groups here we really want to combine these into one fraction by getting a common denominator so what I do is I multiply by what I'm missing I've got a sine theta plus one but I need a sine theta minus one so I want to multiply the numerator and denominator by sine theta minus one so that's gonna make this sine theta minus one over our common denying our common denominator sine theta plus one times sine theta minus one which when you multiply that out it comes out two sine squared theta minus one and then here we're missing a sine theta plus one so I'm going to multiply the numerator by sine theta plus one all over our common denominator sine squared theta minus one now we can see the negative one in the possible and cancel sine theta plus sine theta is two sine theta over sine squared theta minus one now what's sine squared theta minus one equal to well if we go to our Pythagorean trig identities if we subtract one to the left and subtract the cosine squared to the right sine squared minus one equals negative cosine squared so now we have two sine theta over negative cosine squared theta okay so how we're going to work with this one well what you can do is you can split this up you can say this is two sine theta over now cosine squared is really like you've got a cosine theta times another cosine theta okay so see how that gives you back the negative cosine squared theta but one over cosine theta equals secant theta and sine over cosine is tangent and two over negative one is negative two so you proved it last one number fourteen remember to put down in the comments below how many of these were able to get right let's see if we can solve this one tangent of theta over cosecant theta we want to show equals secant theta minus cosine theta now remember there's no right or wrong way to do these you just want to stick with one side if you decide you're gonna work with the right show it equals the left or start with the left to show that equals the right either way I'm going to start with the left on this one tangent of theta is equal to sine of theta over cosine theta okay and cosecant theta is one over sine theta now remember when you divide by a fraction it's like multiplying by the reciprocal so if we take this one over sign and flip it and multiply it by the numerator that's going to give us sine squared theta over cosine theta now sine squared theta if we go to our Pythagorean tree identity here we can subtract the sine squared to the other side whoops I said that wrong we're here subtract the cosine squared to the other side sine squared is equal to one minus cosine squared so we're gonna replace sine squared with 1 minus cosine squared all divided by cosine theta now what I can do is I can split this up into two fractions 1 over cosine theta which is equal to secant theta and I can split this up over here minus cosine squared divided by cosine is just gonna give us cosine theta and we proved it if you want to see more examples follow me over to that video right there we'll do some more trig identities I'll see you in the next video