 Welcome all, we are going to resume with the session that we could not have last Wednesday unfortunately because of some technical glitches. So today we will be resuming with the session on trigonometry according to our fast track revision plan. So those who are attending the session, I would request you all to type in your names in the chat box so that I know who are attending the session. Alright, so let's begin with today's session with the very first problem of the day. Good afternoon, Sai. So here the first problem of the day says cos alpha by cos a plus sin alpha by sin a is equal to cos beta by cos a plus sin beta by sin a equal to 1 where alpha beta do not differ by an even multiple of pi. Then I'll find the value of cos alpha cos beta by cos square a plus sin alpha sin beta by sin square a. Good afternoon to all of you who joined the session. Welcome to the live session. Alright, so we'll follow our wait till five people respond and once I get the response from five people only then I'll be starting the discussion of every problem. Because in that case what will happen, it will not be a one-sided affair. Else I'll be just solving it and new people will be just seeing the solution that will not be helpful. I want the response to come from at least five of you. Those who give me the first response would be getting a shout out from me. Hello, good afternoon Rohan. Good afternoon. Welcome. Cheers. I request people who have joined in the session to please type in your names in the chat box so that I know who all are attending the session. Alright, so Atmesh is the first one to respond this and he says option two is correct. Atmesh, I'm assuming you're saying option two, not the answer two. Alright guys, so we'll begin with this slightly tricky problem to solve especially when it's the first problem it becomes even more trickier. If you look at these two expressions, they resemble a lot, right? It resembles to that extent where you can say that alpha and beta are the roots of, are the roots of this equation cos theta by cos a plus sin theta by sin a equal to one. Right? Okay. Now, how should I use this particular information? Now if you focus on the required expression, you see something like symmetrical terms right? Cos alpha cos beta by cos square a, which you can actually write it as cos alpha cos beta by cos a cos a, okay? And this term is like sin alpha sin beta by sin a sin a, okay? Now in my opinion, if I would try to get a equation, so I would try to get a equation whose roots are these two terms, right? And I'll also try to get an equation whose roots are these two terms, okay? So let us see how do I transform this equation in order to get another equation whose roots are first these two terms. Let's focus on these two first. I'll come back to these two little later on, okay? So what I'll do here, I will write sin theta on the other side, so it becomes one minus sin theta by sin a, okay? I'll square both the sides, I'll square both the sides. Let's see what do I get from here? So this becomes cos theta cos square theta by cos square a is equal to 1 minus 2 sin theta by sin a plus sin square theta by sin square a, okay? Now let me bring the cos square on the other side. Let me bring, let me first write cos square as 1 minus sin square theta over here. Now actually I'm trying to get these two first. So I'm trying to get these two first rather than these two. So I'm trying to get this from here, okay? So please see the steps very, very carefully. Now what I will do is I will take the sin square cos square a term on the right hand side. So I can write this as 1 by cos square a is equal to 1 minus 2 sin theta by sin a plus sin square theta 1 by sin square a plus 1 by cos square a. So that gives me 1 by cos square a is 1 minus 2 sin theta sin a and on this side I'll get sin square theta. If I take the LCM I get 1 by sin square a cos square a, okay? Now what I will do? I'll multiply throughout with cos square a. When I multiply throughout with cos square a this is what I get minus 2 sin theta by sin a into cos square a and here I will get sin square theta by sin square a, okay? Now treat this as a quadratic where sin theta by sin a is taken as your x. So basically it becomes a quadratic of this nature. x square minus 2x cos square a, okay? And of course this term will also get multiplied with cos square a which I forgot cos square a over here, okay? And you will have minus of sin square a when I bring the 1 on the other side. If I bring this 1 on the other side I will get a minus of sin square a, correct? Now can I say that the roots of this equation will be the roots of this equation will be sin alpha by sin a and sin beta by sin a, correct? Because I have assumed in general sin theta by sin a to bx, correct? And we know that alpha beta are the roots of this equation so everything has been derived from this equation only. So finally this equation will have these 2 as your roots, correct? Now this is like finding the product of the roots. Product of the roots we all know it's sin, it's going to be c by a in this case that will be sin square a. So this 2 product is going to become minus sin square a for you, okay? And in a similar way which I am not going to do right now in the interest of time and space these 2 terms will actually be the product of the roots of such a equation where you will get the product as cos square a. Now how do we get this? I will just give you a hint for this. You just write sin theta by sin a as 1 minus cos theta by cos a that means write it like this and square both the sides and if you square both the sides in a similar way you will get a quadratic in cos theta by cos a just like I got a quadratic in sin theta by sin a, okay? And from here you will get the expression as minus of cos square a. So when you add these 2 terms your answer will be minus sin square a minus cos square a which is going to be minus 1 so option number 2 becomes correct. So yes, Atmesh is the first one to answer this correctly and he is the only one to answer this correctly. So yes it is slightly higher than J e main level but of course you can expect such questions to come in the April session given that it is happening little late. So don't expect easy questions unlike what you had this time in J e main to happen in April. Is that fine guys? Any question with respect to this? So the moral of this story was to make a quadratic whose roots are these 2 terms okay and make another quadratic whose roots are these 2 terms and play with the concept of product of the roots. So moving on to the next question now, question number 2 of the day. If x, y, z are all positive acute angles then the least value of this expression is. Now guys I am expecting you to answer this pretty quickly because it is a super simple question. Hello, good afternoon to all of you who have joined the session, Sanjana, Vishis, Shweta. Alright, so Rohan is the one who is giving the answer first, Rohan says option 3. Okay, Vaishnavi, Vishis also go with this. Guys this becomes a very easy question. There are some keywords which you need to be watchful about positive acute angle. Positive acute angle means all these expressions which we have over here tan, cot, etc. They will all be positive and whenever it comes about the least or the greatest values AMGM inequality proves to be very, very handy, right? So I will be using the same concept over here as well. So if you break this term by opening the brackets you will realize you will get terms like tan x, cot y, then you get tan x, cot z, you get tan y, cot z, you get tan y, cot x, then you get tan z, cot x, and then you will get tan z, cot y, okay? So these are the terms and all of them are positive terms. So each of the terms, each of the tan x i's and cot x, x i's will be positive. So I can say the arithmetic mean of it 1, 2, 3, 4, 5, 6 terms are there will always be greater than equal to the geometric mean. Geometric mean would be the product of all of them. So let's keep multiplying them tan x, cot z, tan y, cot z, tan y, cot x, tan y, cot x, tan z, cot x, tan z, cot y, whole to the power of 1 by 6, okay? So this tan x will get cancelled with some cot x, okay? Again tan x gets cancelled with cot x, cot z gets cancelled with tan z, tan y gets cancelled with cot y, again cot z, this gets cancelled, cot y, tan y gets cancelled leaving you with 1. So this desired expression, this desired expression will always be greater than 6. So let me call this expression as e for the time being, so let's call this as e. So e would be greater than equal to 6. So the least value of this expression would be 6, which is option number 3 is correct. So here guys, more of the story is, please keep the AMGM inequality in mind while solving such kind of a problem. So don't think that it's only applicable to series-based questions can be used anywhere as per the given problem. Is that fine? Any questions, any concerns with this? Please type in in your chat box if you feel you have not understood any step of this solution. Great. So let me now pose the next question at you. So theta 1, theta 2, theta 3 are angles of first quadrant. If tan theta 1 is equal to cos theta 1, so this is the first set of information given to us, tan theta 1 equal to cos theta 1, tan theta 2 is cos c theta 2 and cos theta 3 is equal to theta 3, which of the following is not true? Now read this kind of problem very, very carefully. Which of the following option is not true? So atmesh again gives the first response, he says option 1, anyone else? So atmesh wants to withdraw his answer, he's saying it may not be option 1. Great. Guys, these type of problems are better solved using graphs. So don't hesitate from using graphs. You realize from graphical view, you will get the answer in no time. Swetha says option 3, atmesh gives option 2. What about others? Purvekh, Visist, Rohan, Sanjana, Sai, Shreyas, Lalitha, Mukhabh guys, I want it to be a responsive class, it's not about just me solving the problem. Okay, so mostly people, yeah, I'm sure it's back now. Okay, so we talk about these three graphs. One is tan x graph, we all know tan x graph will be increasing like this. So we're just talking in the first quadrant, okay, y equal to x will be below this line because we know y equal to x is always below the graph of tan x. And we have the graph of cos x like this, okay. And we also have the graph of cosecics like this. So I'm drawing it beyond 90 degrees, this is 90 degrees, okay. Now guys, if you look at this expression very, very carefully, you would realize that the point where the cos x graph and tan x graph meet is at theta, so they meet at theta 1, that means both the equations are satisfied, that means these two graphs y equal to tan x and y equal to cos x graph, they meet at x equal to theta 1, correct. That means this angle over here, I'm just showing it on the x axis, this is going to be theta 1, okay. And if you see this one, so this is addressed, if you see this one theta 3, theta 3 is where y equal to x, y equal to cos x meet, right, these two meet at x equal to theta 3. So as you can see this part will be theta 3, slightly higher than, slightly higher than or greater than theta 1, okay. And tan x, tan x will meet cosec of x graph, that means these two graphs, these two meet at x equal to theta 2. So definitely from this graph I can figure out that theta 1 is lesser than theta 3 and theta 3 is lesser than theta 2, correct. Now 1, 1 is not the right option because 1 is correct, because theta 1 is less than theta 2, I have to identify which is the wrong option, remember that. And here theta 1 is less than theta 3, which is also correct. And 2 is our right answer because it's not theta 2 which is lesser than theta 3, it's actually theta 3 which is lesser than theta 2, okay. So option number 3 is correct and I feel Shweta is the only one who gave the right answer, well done Shweta, very good. So guys never underestimate the power of graphs while solving such questions, correct. Okay, so this should have been solved by most of you. Any question with respect to this, let's move on to the next question then. So we have around 25 questions for this session, so we need to be a little faster. So we would appreciate a faster response from all of you. Next is a super, super easy question, question number 4, find the maximum value of the expression 1 plus sin of pi by 6 plus theta plus 2 cos of pi by 3 minus theta for all real values of theta, for all real values of theta, okay. Atmesh says option number 3, Lalitha also backs him up. All right, so almost everybody is going with option number 3 to be correct, so let's simplify this. Guys, when I see this kind of questions, I'm basically trying to address the maximum or minimum value of such expression. So try to bring this entire expression in this form, that's the trick for solving such questions. So what I'll do is I'll expand sin of pi by 6 plus theta using the compound angle identities which is going to be sin of 30 degree which is half cos theta plus cos 30 degree which is root 3 by 2 times theta and this becomes 2 times cos pi by 3 which is half cos theta plus root 3 by 2 sin theta, okay. Now try to group up terms having cos theta in it, so when you group up terms having cos theta in it, we get 3 by 2 cos theta, okay. And here I get root 3 by 2 plus root 3 sin theta which actually you can write it as root 3, if you take root 3, you get 3 by 2, so root 3 times 3 by 2 sin theta. Now it becomes a very easy question now because we know that the max value of this is going to be under root of a square plus b square plus c and minimum value is going to be minus under root a square plus b square plus c, however in this question we only require the max value so that will become under root of 3 by 2 whole square plus 3 root 3 by 2 whole square which is going to be 1 plus half under root of 9 plus 27 which is going to be under root of 36 which is going to be 6, so your answer is going to be 4 that's option number 3 is correct. Well and those who answered they answered it correctly so nobody was wrong here, so guys any question with respect to this problem I am assuming this problem was super simple for you to solve, so let's move on to the next question that's your question number 5 of the day, if the median of triangle abc through a is perpendicular to ab then which of the following option is correct, guys since a request to all of you make honest attempt to solve these problems please do not try to copy each other's answer which I am sure you must not be doing it but again this is the last lap of your journey please be honest with you, with yourself. Yeah, I have already done the diagram for you, so you can refer to this diagram, so ad is the median over here and it's perpendicular to ab as well alright so purbeck is saying option 2, shweta things option 4 none of these okay sure sure take your time there's no hurry please be accurate that's my only request to all of you it's worth spending an extra one minute and getting it right rather than saving a one minute and getting it wrong alright Rowan goes with option 3 so does Saim here alright let's try to solve this problem to solve this problem I'll do one small construction okay I'll just make a small construction I'll extend this line forward okay okay and from see I will drop a perpendicular here as well so this is also 90 degree okay let me call this point as n okay now let me call this angle here as angle alpha okay so from the diagram I can say that tan of a tan of angle a please remember this is your angle a this entire thing is your angle a okay tan a is tan pi minus alpha right and we know that tan pi minus alpha is negative of tan alpha correct now tan of alpha over here from this diagram if you see that's going to be minus cn by an cn by an right now if you see at these two triangles triangle b cn and triangle b da these two triangles are clearly similar triangles these two triangles are clearly similar triangles because they have a common angle and from this construction this is parallel to this correct now since b is equal to DC I can say bd by bc is 1 is to 2 and so would be da by cn right so I'm just applying proportionality thing over here bd by bc so let me show it from the diagram bd by bc will be equal to da by cn so I can write minus 2 cn as minus cn as minus 2 da or ad okay divided by an now an so an let's figure out what is an over here now an is also equal to a b because the triangles are similar to each other so an is also equal to a b so I can write this an as a b so what am I getting I'm getting minus 2 ad by a b right what is ad by a b ad by a b is basically nothing but tan of this angle b right so I can write this as minus 2 tan of angle b so I started with tan a and ended at minus of 2 tan b which implies which implies tan of a plus 2 tan of b is equal to 0 correct that means option number 3 is the right answer over here so the only person to answer this correctly was Rohan so Rohan was the first one and the only one to answer this question correctly well done Rohan so guys here the trick was this construction this construction was the trick over here is that fine any question any concerns with respect to this question if no then I can move on to the sixth question of the day all right so let's move on to the next question next question is we have a triangle abc and it has been given that I think by mistake tan b by 2 has been written twice so let me just erase this up I just ignore one of the tan b's so tan a by 2 tan b by 2 tan c by 2 are in arithmetic progression then cos a cos b cos c are in options are a p g p h p none of these now this should be a very simple basic standard problem that we get in properties of triangles the guys request you all to solve it very carefully without making any mistake remember accuracy always dominates speed yes done anyone keep the time limit in your mind guys okay so Sai says option one how about others okay so I don't think so this was a difficult question actually when you start with the fact that it's a it's a geometric progression sorry it's an arithmetic progression we can say tan a by 2 plus tan c by 2 okay you can do one thing you can just okay let me just complete the statement so we can write tan b by 2 minus tan a by 2 is equal to tan c by 2 minus tan b by 2 okay let me write this as sine b by 2 by cos b by 2 minus sine a by 2 by cos a by 2 and this also I can write it as sine c by 2 cos c by 2 minus sine b by 2 cos b by 2 okay so this will become sine b by 2 cos a by 2 minus sine a by 2 cos b by 2 divided by cos b by 2 cos a by 2 and on this side also I'll have sine c by 2 cos b by 2 minus sine b by 2 cos c by 2 divided by cos c by 2 cos b by 2 okay so let me cancel this off now this expression over here is as good as saying sine of b minus a by 2 okay and this expression over here is as good as saying sine of um c minus b by 2 correct now let us cross multiply so from this step I'll cross multiply this with this and this with this so what I will get I'll get sine b minus a by 2 into cos c by 2 and on this side I will get sine c minus b by 2 into cos of a by 2 okay and we all know that in a triangle in a triangle abc right I can write c as a plus b pi minus a plus b let me write it properly pi minus a plus b so c by 2 will be pi by 2 minus a plus b by 2 so cos c by 2 will actually become sine b plus a by 2 okay similarly here it will become sine c minus b by 2 into sine c plus b by 2 correct let me just clear up this step now this becomes sine square b by 2 minus sine square a by 2 is equal to sine square c by 2 minus sine square b by 2 correct which is nothing but 2 sine square b by 2 is equal to sine square a by 2 plus sine square c by 2 which is nothing but saying that 1 minus cos b is 1 minus cos a by 2 and 1 minus cos c by 2 so cancelling the factor of 1 and taking the 2 on the other side it becomes 2 cos b is equal to cos a plus cos c which clearly implies that the terms cos a cos b cos c will also be in arithmetic progression that means option number a will become or option number one will become correct so the first one to answer this correctly was sine here okay so guys it's just the use of your arithmetic property arithmetic progression properties and of course the conditional identities related to properties of triangles and it was just a matter of using your trigonometric identities and simplifying it again treat this as uh slightly lesser than day many levels so it was a simple question to be very frank is it fine any question with respect to this can you move on to the seventh question okay so let's move on to the seventh question here hope you would have read this question alpha beta gamma are acute angles cos theta is sine beta by sine alpha cos phi is sine gamma by sine alpha and cos theta minus phi is equal to sine beta sine gamma then the value of tan square alpha minus tan square beta minus tan square gamma any idea anyone how to proceed so let me slowly start giving you a hint let me start with the third conditions sine of theta minus phi is equal to sine beta sine gamma okay so when you expand this you actually get cos theta cos phi plus sine theta sine phi is equal to sine beta sine gamma correct now what I'll do next is I'll send this to the other side that means sine theta sine phi I would write it as cos theta cos phi minus sine beta sine phi let's say square both the sides so I don't care about the sine change okay let's square okay so this becomes a sine square theta into sine square phi sine square phi and this is a let's say you keep it as cos theta cos phi minus sine beta sine gamma whole square okay now cos theta is known to me so sine square theta will be nothing but 1 minus cos square so I can write it as 1 minus sine square beta by sine square alpha okay similarly sine square phi could be written from here that is cos phi expression which is again 1 minus sine square gamma by sine square alpha cos theta cos phi is just the product of these two so I can write it as sine beta sine phi sorry sine beta sine gamma by sine square alpha and let the other term be as such that is minus sine beta sine gamma whole square so let's guys simplify this I can write this as sine square alpha minus sine square beta into sine square alpha minus sine square gamma and on this side I can write it as correct me if I'm wrong sine square beta sine square gamma I can take common and I'll be left with 1 minus sine square alpha whole square sine square alpha whole square all right so now let us start multiplying the terms let's multiply the terms which I have on the left hand side okay and let's multiply the terms and let's open the terms which I have on the right hand side okay so while expanding it I will get sine to the power four from here also sine to the power four is when these two terms multiply and also get sine to the power four alpha term when this square is opened up okay so I can directly club sine to the power four alpha and I can take common so this will have the coefficient as sine square beta sine square gamma so I can take this as common okay terms containing just science sine alpha would be minus sine alpha from here and here I would get sine square beta plus sine square gamma and from this side I will get minus two sine square beta sine square gamma okay I think other terms will get cancelled off other terms will get cancelled off like sine square beta into sine square gamma from both the sides will get cancelled off fine now sine square alpha I can drop from both the sides I can drop a sine square alpha from both the sides so I'll make this as two and I'll cancel this off which implies sine square alpha I can write it as sine square beta plus sine square gamma minus two sine square beta sine square gamma okay whole divided by one minus sine square beta sine square gamma now why I'm doing all these steps I'm doing all these steps to get to the expression of tan square alpha I want to see whether I can express tan square alpha as some constant plus plus these these terms okay so now what will be cos square alpha cos square alpha will be one minus sine square alpha so let's find cos square alpha by subtracting one minus this so if I do one minus this correct me if I'm wrong I will get something like this whole divided by one minus sine square beta sine square gamma right hope I've not missed out on any term I think this will become plus yeah small mistake over here this will become plus over here this term will be not there so this is the final term that you see let me rephrase it in a proper way so it will be one minus sine square beta minus sine square gamma plus sine square beta sine square gamma by one minus sine square beta sine square gamma now this is our first expression this is our second expression let's divide them because by dividing them only we can get tan square alpha so dividing one by two I will get tan square alpha and that would be sine square beta plus sine square gamma minus two sine square beta sine square gamma divided by one minus sine square beta minus sine square gamma plus sine square beta sine square gamma okay so here if you see if I try to take one minus sine square gamma if you see this term this term is actually made up of one minus sine square beta into one minus sine square gamma right which is actually cos square beta minus by cos square beta cos square gamma so I'm taking it over here because of lack of space so this can be written as sine square beta plus sine square gamma minus two sine square betas sine square gamma by cos square beta cos square gamma correct now in the numerator what I will do I will split this two as two parts I will write this as let me erase some something let me just erase the lower part of this because I need some space to write on so I'm just erasing this part so I can say tan square alpha I can write it as this I will break it up as sine square beta minus sine square beta sine square gamma plus sine square gamma minus again sine square beta sine square gamma whole divided by cos square beta cos square gamma now take sine square beta common from here if you take sine square beta common it will give you one minus sine square gamma take sine square gamma from here it will become one minus sine square beta okay whole divided by let me again make some space whole divided by cos square beta cos square gamma which is nothing but sine square beta cos square gamma plus sine square gamma cos square beta divided by divided by cos square beta cos square gamma okay now individually divide these two so it will give you tan square beta so when these two terms divide each other you will get tan square beta when these two terms divide each other you will get tan square gamma so what am I getting I'm getting tan square alpha is equal to tan square beta plus tan square gamma which clearly implies that tan square alpha minus tan square beta minus tan square gamma is going to be zero which means option number two is correct now guys this is a super super lengthy question it's one of those questions which you should never attempt which you should never attempt typically it will not be asked even if in a GE advanced question because there are so many simplification involved and you have to keep switching between sine square and cos square terms and of course the roadmap was simple I needed to reach somehow to the expression of tan square alpha in terms of tan square beta and tan square gamma so moving on to the next question which is the eighth question for the day so we have x and y as real numbers satisfying the condition sine x sine y plus 3 cos y plus 4 sine y cos x is equal to root of 26 find the value of tan square x plus cot square y is it 9 into 17 is it 205 is it 1 by 7 16 plus 9 by 17 and or none of these yes anybody any idea anybody who's trying please let me know if nobody's trying I can start solving it but I'd expect somebody to try it out all right sure sure sure take your time okay let's see the left hand side of this expression if you see left hand side this actually is c cos y and from these two terms if I take a sine y common I get sine x plus 4 cos x if you treat this to be like of the form a cos y plus b sine y we all know that this expression will always be less than equal to under root of a square plus b square right we all know that correct so this implies I can treat this expression that is 3 cos y plus sine y sine x plus 4 cos x to be less than equal to under root of a square in this case a square will be 9 plus b square b square will be sine x plus 4 cos x the whole square okay right in turn we also know that we also know that sine x plus 4 cos x will always be less than equal to under root of 1 plus 16 correct right 1 square plus 4 square which is 1 plus 16 correct so sine x plus 4 cos x whole square will always be less than less than 1 plus 16 which is 17 correct yes or no that means this expression will always be less than equal to under root of 9 plus 17 which means which means it will always be less than equal to 26 correct correct now here it is written that it is equal to root 26 if it is equal to root 26 means the three terms which are involved over here okay should be in such a ratio that sine x by sine y should be equal to cos y by 3 should be equal to sine y cos x by 4 then only this equality will exist this equality will exist when these three terms are in this ratio is that fine is it clear the equality will exist right the equality will exist only when sine x sine y e is equal to cos y by 3 is equal to sine y cos x by 4 so this term by 4 now can anybody give me the reason why it happens anybody can give me the reason why this ratio has to be equal just think over it think think in the line of when should am be equal to even the equality exists just think in lines of that yeah shweta to get the maximum value the terms must be equal to each other that's fine right but here this ratio is also coming right so why is this ratio coming correct when the terms are equal when these terms are equal why is this particular expression equal to its maximum value okay so yes most of you have given the answer so for the equality to exist the terms must be equal now in this case if it is so can I say tan of x tan of x I can actually obtain from comparing these two terms so tan of x will be nothing but sine y sine y gets cancelled so it will be 1 by 4 right the moment I realize this square of it will be 1 by 16 so this actually qualifies to be my answer but let me still complete it what will be caught why caught why I can obtain from here so caught of y cos of 5 by sine of y will be equal to 3 cos x by 4 and if tan x is equal to 1 by 4 it implies cos of x will be 4 divided by root of 17 so this will be 3 into 3 by 4 into 4 divided by root of 17 so 4 4 will get cancelled off so caught of 5 will become 3 by root 17 right so square of this will give you 9 by 17 and therefore this ratio therefore this term is equal to the third option which is 1 by 16 plus 9 by 17 so option number 3 becomes correct in this case so now let us move on to the next problem which is problem number 9 so u is given to you as 1 plus cos theta times 1 plus cos 2 theta minus sine theta sine 2 theta v is given to you sine theta times 1 plus cos 2 theta plus sine 2 theta times 1 plus cos theta then u square plus v square is which of the following okay this problem can be solved easily by using some special substitution so try that out but be careful not to use any substitution which may result into multiple options giving you the same answer option one is according to Purvik that is the option for him what about others Sanjana also thinks the same all right so if you open the u term you see cos theta plus cos of 2 theta plus cos theta cos 2 theta minus sine theta sine 2 theta correct which is actually cos theta plus cos 2 theta and this becomes cos of 3 theta because this follows the compound angle identity isn't it now I can always this is 1 plus also I forgot add 1 plus yeah now I can combine 1 plus cos 3 theta together and this will become 2 cos 3 theta by 2 into cos theta by 2 and this also becomes 2 cos square 3 theta by 2 plus 2 cos 3 theta by 2 cos theta by 2 that becomes 2 cos 3 theta by 2 and the brackets will have cos 3 theta by 2 plus cos theta by 2 okay similarly for expression we also I can write this as sine theta plus sine 2 theta plus sine 3 theta and this I can write it as 2 sine 3 theta by 2 cos theta by 2 and this I can break it up as sine 3 theta by 2 cos 3 theta by 2 so take sine 3 theta by 2 common you will get cos theta by 2 plus cos 3 theta by 2 okay now this is your view okay so if you see this term is same in both of these expressions correct so if you square these two terms and add them you will get 2 sine square 3 theta and this you can actually take common this will become cos theta by 2 plus cos 3 theta by 2 that will come as common and these two terms will combine to give you a fourth correct okay and inside here also we can actually combine these two terms and write four times 2 cos theta cos theta by 2 square okay so this becomes 16 cos square theta into cos square theta by 2 that actually becomes 4 times 1 plus cos 2 theta into 1 plus cos theta if you divide this 16 as 2 2 in this and keep the 4 aloof so let me write it properly so this 16 I will break it up as 4 times 2 into 2 then this term becomes this and this term becomes this which is clearly option number a which is correct so option number a in this case is correct so I think poor week was the first one to answer this followed by sanjana and saim here well done guys however I would suggest use some special values of theta like you may go for 5 by 6 would not be a good option because then these two will give me the same option I think 5 by 4 would be a better option to try with okay so what what substitution did you use guys poor wake sanjana you use 0 okay if you take 0 doesn't these two give the same options in fact no these two these two will give you the same option 0 will give me yes this is this will be different yeah there's chances that these two would have been same then there would be a problem okay so you're lucky enough to get the first option correct else they could have been a clash between second second and fourth options okay anyways so be a safe if you are substituting us with some special value make sure none of the options become the same with that particular substitution so we'll move on to the next question or question number 10 for the day so this question goes like this theta 1 theta 2 theta 3 are three values lying between 0 to 3 pi and tan theta is lambda then the value of mod of tan theta 1 by 3 tan theta 2 by 3 etc yeah there's just some lambda which they have equated tan theta 2 that means if you solve these two equation if you solve this equation these are the roots of this equation lambda is some real number it could be any value it could be any value and your answer doesn't depend on that value actually guys try to think or try to formulate a equation where theta 1 by 3 theta 2 by 3 and theta 3 by 3 will be the roots so which equation will have these two as the roots or for that matter tan theta 1 by 2 tan theta 2 by 2 and tan theta 3 by 3 which equation will have these three as your roots try to think in that terms okay so purvig has given a response option 3 because it's pretty simple I don't see a you know trouble in this problem just express tan of theta as tan of 3 theta by 3 okay which is actually tan of theta by 3 minus tan cube theta by 3 by 1 minus 3 tan square theta by 3 okay and this is equated to lambda actually correct so if you simplify this that means if you cross multiply this I see something like this so it gives me tan cube theta by 3 minus 3 lambda tan square theta by 3 minus 3 tan theta by 3 plus lambda equal to 0 now if you treat your tan theta by 3 as your x so it becomes a cubic equation like this correct and this cubic equation will have the roots as this equation will have roots as tan theta 1 by 3 tan theta 2 by 3 tan theta 3 by 3 okay so here we are dealing with product of roots 2 at a time so I'm dealing with the products of roots 3 at a time we all know that it's going to be sorry 2 at a time it's going to be c by a right so c by a in this case will be minus 3 right so that would be let's say I call this as product of alpha beta okay but I have to find a mod of this mod of this will be this which is actually going to be 3 right so option number 3 is going to be correct so by any chance is the second term supposed to be theta 2 by 3 plus theta 3 by 3 second term which term oh yes yes yes yeah there's a typing error yeah sorry about that is that fine that's a small typo anyways is this fine so basically this is based on uh transformation of the equation according to the given requirement so apart from this changed no other question no other concern so we can move on to the next question so let's move on to question number 11 for the day find the value of summation of tan square r pi by 24 from r equal to 1 till r equal to 11 yes guys any response see let's start with the value of r as one so when you put one you get tan square pi by 24 okay and the last term will be tan square 11 by 24 11 pi by 24 so let's group these two terms okay so I'm grouping these two terms okay let me write it on this side so tan square pi by 24 and tan square 11 pi by 24 I'm writing together why I'm writing together is because this 11 pi by 24 could actually be written as pi by 2 minus pi by 24 right because pi by 2 could be written as 12 pi by 24 so 12 pi by 24 minus pi by 24 will be 11 pi by 24 so they they could be club I mean they are related to each other similarly tan square 2 pi by 24 I can club up with tan square 10 pi by 24 okay similarly tan square 3 pi by 24 can be clubbed with tan square 9 pi by 24 and again tan square 4 pi by 24 can be clubbed with tan square 8 pi by 24 and tan square 5 pi by 24 could be clubbed with tan square 7 pi by 24 and that last term that is tan square 6 pi by 24 will be left without a pair but anyway this is going to be tan square pi by 4 that's 1 okay now we all know that pi by 12 is 2 minus root 3 tan pi by 12 is 15 degree tan 15 degree is 2 minus root 3 correct and tan tan pi by 12 which is actually cot pi by 12 is reciprocal of this reciprocal of this is 2 plus root 3 okay so these two values are known to us okay similarly tan 3 pi by 24 which is actually pi by 8 that's going to be pi by 8 is going to be root 2 minus 1 so tan 9 pi by 24 will be the reciprocal of this reciprocal of this will be root 2 plus 1 correct so these two are also known okay and this is pretty simple this is tan square pi by 6 which is 1 by root 3 whole square and this is reciprocal of 1 by root 3 which is root 3 whole square that's fine so we know all these underlined values the only thing that we don't know are the values of the terms which are left off that means I need to know the value of this term this term this term this term and this term correct now let us look into a identity we all see that there are terms like tan square theta plus cot square theta now can I can I convert this in terms of two theta angle can I make this in terms of two theta angle so I can write this as sine square theta by cos square theta plus cos square theta by sine square theta that gives me sine to the power four theta plus cos to the power four theta by sine square theta cos square theta which actually can be written as sine square theta plus cos square theta whole square minus minus two sine square theta cos square theta and if you divide throughout with sine square theta cos square theta you get expression like one by sine square theta cos square theta minus two put a factor of four over here so it becomes I need to clear up some space because I need some space let me just try to excuse things over here so it becomes four cosecant square two theta minus two okay which if you want you can write it as two plus four cot square two theta now why I'm interested in this is because I can now write the value of certain terms for example for this case I can write the answer as two plus four cot square pi by 12 correct so this case I can write the answer as two plus four cot square five pi by 12 and these values could be easily obtained isn't it so let me just fill in the blanks now whatever values we have already found out we'll be filling in the values there over there so let me just clear up the space over here so the first term is going to give me four cot square pi by 12 cot square pi by 12 is going to be two plus root three the whole square okay and I have the second term as again two minus root three the whole square two plus root three the whole square plus root two minus one the whole square root two plus one the whole square plus one by three plus three plus this term over here will be two plus four cot five pi by 12 cot five pi by 12 is tan of 15 degree is sorry cot of 75 degree which is cot of 15 degree which is two plus root three again so it's two plus two minus root three the whole square plus one plus one okay now if you expand this if you expand this we'll see a lot of things happening for example if you take care of all the constant terms only like two from here one from here three from here again two from here and one third from here it will become two plus three a five seven eight eight plus one by three so these terms I'm just taking care of so I'll have three plus three nine eleven eleven plus three plus three six eight eight plus one by three correct and we'll have four times two square plus three and here also I'll have two square plus three two square plus three from here I'll have two plus one two plus one okay and last term would be four times two square plus three okay so let's sum this up if you sum this up you'll get the term total as 20 to 53 by three so option number three becomes correct in this case again for such kind of a problems you have to be ready with the values of the non-aligned angles or the sub-sub-multiple angles of those allied angles that we know that is 30 degree you know 22 and a half degree 15 degree 18 degree all those things have to be known to you prior to solving this problem of course you will be knowing it through a lot of practice anyways is that fine all right so moving on to the 12th problem now to find the value of this particular expression please don't use calculator etc not going to be helpful okay wishes says option three anyone else who thinks the same see guys it's simple if you take the lcm lcm is going to be sine 25 degree sine 70 degree and sine 85 degree correct here you will get a sine 25 cos 25 sine 70 cos 70 he'll get sine 85 cos 85 okay just provide a factor of two two two so this will become two signs sorry this will become sine 50 degree sine 140 degree sine 170 degree bye so far so good now guys how many of you recall that just try to recall this conditional identity that if a plus b plus c is 180 degree then sine 2a plus sine 2b plus sine 2c will actually become 4 sine a sine b sine c right this is one of the most widely used conditional identity related to properties of triangles okay so keep such identities in mind because it makes your life really very simple if you see these angles belong to a triangle 25 plus 75 is 95 95 plus 85 is 180 degree so this criteria is met by 25 75 and 85 correct right so on top you see sine 2a this is your 2a this is your 2b this is your 2c correct and here you see 2 sine a sine b sine c so if you know this identity you can directly write the answer as 2 sine 25 sine 70 sine 85 divided by 2 sine 25 sine 70 sine 85 okay the model of the story is you should be well aware of certain angles certain identities conditional identities which which will save a lot of time for you in the exam so option number three becomes correct here is that fine is it okay any questions all right so moving on to the next question so this is question number 13 for you consider the inequality consider quantities x1 x2 x3 till x10 such that all these quantities lie between minus one to one and x1 cube plus x2 cube plus x3 cube all the way till x10 cube is equal to zero then find the maximum value of this expression find the maximum value of this expression okay shares says option one how about others so there's no sign of any tignometric ratio over here but still I'm dealing with this problem in the discussion on tignometry chapter so this itself is a big hint all right so in the interest of time I'll be solving this problem so let's say I assume all the x i's to be sine theta i's okay so this condition is automatically satisfied because of this assumption correct now what is given to me summation of sine cube theta i is equal to zero correct right so can I say four summation sine cube theta i will also be zero so can I write this as three summation sine theta i minus summation sine three theta i this will also be zero now how can I say this I can say this on the fact that if sine theta i is equal to if summation sine theta i is equal to zero or you can just say sine theta i is equal to zero then it also implies sine three theta i will also be zero that means three times summation sine theta i will be equal to summation of sine three theta i correct now this is nothing but three times summation x i because your sine theta i is actually your x i okay and summation three three theta i I can say that this term will always be less than equal to 10 correct so this expression will also be less than equal to 10 correct that means summation of x i will always be less than equal to 10 by 3 which means the minimum value which means the minimum value of summation x i from i equal to 1 to 10 will be 10 by 3 that means option number one yeah why does this this because any sine angle is always between minus 1 to 1 right so sine of three theta i will always die between minus 1 to 1 so its maximum value is going to be 1 so what i'm saying let's say the maximum value of each sine three theta i is 1 so you are summing it from 1 to 10 so 1 plus 1 plus 1 till 10 times so it will always be less than equal to 10 okay so option number a or option number one becomes correct in this case is that fine any question with respect to this please do ask if not let's move on to the next problem which is problem number 14 for the day so in a triangle abc cot a by 2 cot b by 2 cot c by 2 is equal to 4 then the triangle is acute angle obtuse angle right angle none of these okay so rohan has given the response as option number one okay wishes also says option number one anybody else so on the rest is option b okay it's an obtuse angle triangle okay guys i have a very important identity that i wanted to ask you how many of you are aware of this identity cot a by 2 plus cot b by 2 plus cot c by 2 is actually cot a by 2 cot b by 2 cot c by 2 okay in any triangle abc are you aware of this identity if i take the arithmetic mean of this term if i take the arithmetic mean of this term this will always be greater than equal to if you're not aware then you need to revise this angle since this is true can i substitute in place of the this term i can substitute this term over here so i can say cot a by 2 cot b by 2 cot c by 2 by 3 would always be greater than equal to cot a by 2 cot b by 2 cot c by 2 to the power of one third correct yeah that means cot a by 2 cot b by 2 cot c by 2 to the power of 2 by 3 will always be greater than 3 now raising both sides to the power of 3 by 2 raising both sides to the power of 3 by 2 this will go off that means this expression will always be greater than equal to 3 root 3 correct 3 root 3 is approximately 3 into 1.732 right which will be greater than 5 actually right here it is given equal to 4 is this possible is it possible the answer is no because if this expression is always greater than 3 root 3 it can never be equal to 4 so there is no such triangle possible so none of these would be the answer guys this was a googly question and you all were ticked so again note down in your to-do list that you have to revise your conditional identities i'm i'm finding that none of you are aware of your conditional identities so these are basic things that j is expecting you to come prepared with is that fine so none of you got this right never mind we'll take up the next problem which is the 15th problem now the number of solutions satisfying that equations tan 4 theta equal to cot 5 theta and sin 2 theta equal to cos theta in the interval 0 to 2 pi sorry guys today there will be no break because we are already running behind schedule because we had to complete 25 problems today so it's just 25 minutes left and you have to complete 10 more problems so anyways i'll be sharing this word that problem pdf with you so please do solve it as homework and i've only sent a worksheet of main and advanced level questions on the group so please just solve them no need to no go for any other resource material it has got solutions also attached with it option 4 okay so let's check this so sin 2 theta let's start with this equation first sin 2 theta is 2 sin theta cos theta equal to cos theta so take cos theta common so it becomes 2 sin theta minus 1 equal to 0 so there are two possibilities cos theta could be 0 or sin theta could be half this gives you theta value as pi by 2 and 3 pi by 2 in this interval and this gives theta value as pi by 6 and 5 pi by 6 okay now guys all i need to do is i need to see which of these values satisfies this condition as well because both have to be simultaneously true see the connector and so if i put pi by 2 i get 0 and cot 5 by 2 is also 0 so this satisfies it okay when i put 3 pi by 2 i get tan 6 pi which is 0 and cot cot 15 pi by 2 that's again 0 so both 0 0 again satisfies it so this is also acceptable pi by 6 will give me tan 2 pi by 3 and this will give me cot 5 pi by 6 tan 2 pi by 3 will be equal to minus of root 3 and cot 5 pi by 6 is also negative of root 3 okay so this also satisfies it now tan 5 pi by 6 will give me tan i'll give me 10 pi by 3 and here i get cot 25 pi by 6 this is going to be tan 3 pi plus pi by 3 3 pi plus pi by 3 will be positive so it will be 1 by root 3 and this will be cot 4 pi plus pi by 6 that's again going to be 1 by root 3 sorry this will be a root 3 my bad this will be root 3 and this will also be root 3 okay so option number 3 is correct all the four options are going to satisfy this okay let's move on to the next problem which is problem number 16 the fruits of the equation sin 2x plus pi by 18 cos 2x minus pi by 9 equal to minus of 1 by 4 lying between 0 to 2 pi okay shear says okay this is the answer for the previous one yes any idea so let me multiply both sides with 2 first and i can write this as cos 2x minus 2 pi by 18 actually pi by 9 that's going to be minus of half so this is going to be sin of 2 sin a cos b so 2 sin a cos b we know that it's sin a plus b plus sin a minus b so to sin a plus b would be nothing but 4x by the way this is 10 degree you can write this as 10 degree over here and this is 20 degrees easy to deal with the figures of angles so this will give you minus 10 degree and here i will get plus sin of a minus b that is sin 30 degree that's going to be minus half okay so sin 4x minus 10 degree is giving you a minus half minus half which is minus 1 so the possible values that i can get this is this can be sin 270 degrees okay so sin 4x minus 10 degree is this so 4x minus 10 degree is 270 degree so 4x is 280 degrees so x is 70 degrees and we know this is going to be a periodic function this is a periodic function with a period of 2 pi by 4 2 pi by 4 is pi by 2 which is 90 degrees so the next angle will be 90 degree to it so which is 160 again 90 degree to it which is 250 again 90 degree to it which is 340 degrees so 1 2 3 4 so 4 possible solutions are there in the interval 0 to 2 pi so option number 2 becomes correct in this case i is back up easy question shouldn't have left this is that fine so let's move on to the next one that's question number 17 for today so cos square pi by 3 cos x minus 8 pi by 3 is 1 find the number of solutions in the interval 0 to 10 pi okay third option how about others guys cos square some angle is equal to 1 what does it mean can you say that angle has to be of the form n pi correct yes or no that means pi by 3 cos x would be equal to n pi plus 8 pi by 3 and if you divide by pi by 3 throughout you get 3n plus 8 correct right n being integer now this is already 8 which means 3n has to be such that cos x is contained between minus 1 to 1 right so only possible value of n that happens over here is minus of 3 correct so if cos x is minus if n is minus of 3 it implies that cos x can only be minus of 1 and cos x is minus of 1 only at odd multiples of pi so x could be pi 2 pi 3 sorry pi 3 pi 3 pi 5 pi 7 pi 9 pi so 1 2 3 4 5 solutions are possible which is option number 3 is correct so yes shares Bhaktram was the only person to get this right well that's yes moving on to the next problem now question number 18 this should be an easy question guys i'm sure you must be able to do this all of you yes guys first of all you can drop sin x and say sin x could be zero right sin x could be zero okay if sin x is zero x is all multiples of pi which is none none of these options okay so we just side you know sin x equal to zero solution so just sideline it so you get 2 cos 4 cos square x minus 2 sin x is equal to 3 okay which is 4 minus 4 sin square x minus 2 sin x is equal to 3 which is 4 sin square x plus 2 sin x minus 1 equal to 0 correct so i can use the formula minus b plus minus and the root b square minus 4 ac plus 16 by 2 a that's actually minus 1 plus minus root 5 by 4 correct so sin x could be these two values now this gives me two values one sin x is root 5 minus 1 by 4 and other is sin x is minus 1 minus root 5 by 4 but this is not possible i guess no both no both are possible yeah both are possible now this comes from 18 degree right sin 18 degree value is this which is actually pi by 10 so your solution could be x is n pi plus minus 1 to the power n pi by 10 correct so here it gives you n pi uh what about this guys okay now if you see if you just take this minus sign out it gives you 1 plus root 5 by 4 okay now how many of you remember that sin 36 degree actually gives you that sorry sin uh cos 36 degree gives you that cos 36 degree gives you root 5 plus 1 by 4 right yes i know it can be true a perfect uh 1 plus root 5 by 4 can be less than 1 if i'm not wrong if you check uh root 5 plus 1 by 4 yeah it can be less than now so cos 36 is this so which angle will give you that means sin 54 is this sin 54 is root 5 plus 1 by 4 correct correct so sin of minus 54 will be minus of root 5 plus 1 by 4 okay which actually gives you the general value as n pi plus minus 1 to the power n minus 3 pi by 10 3 pi by 10 is uh 54 degrees correct so your possible answer could be option number 2 which is n pi plus minus 1 to the power n plus 1 this additional n plus 1 comes because of this negative sign over here okay 3 pi by 10 so option number 2 is correct in this case all right so we'll move on to the last question of the day the sum of the solutions of sin pi x plus cos pi x equal to 0 in the interval 0 to 100 yes guys anyone all right okay wishes says option 2 four weeks says option 4 okay let's discuss this this is clearly root 2 sin pi x plus pi by 4 right and this is 0 which means sin pi x plus pi by 4 is 0 which implies a pi x plus pi by 4 is actually a multiple of pi okay so i can say pi x is n pi minus pi by 4 drop a factor of pi so x is equal to this where n is an integer okay now your range of value of x possible over here is from 0 to 100 so if i say x is from 0 to 100 that means this is between 0 to 100 so n should be between 1 by 4 to 104.25 correct in other words i have to find the sum of x from n n could be only 1 to less than 100.25 is 100 that means i have to sum up n minus 1 by 4 from n equal to 1 till 100 that's going to be n into n plus 1 by 2 minus 1 by 4 n that's going to be 100 into 101 by 2 minus 100 by 4 so 50 this becomes 5050 minus 25 that's 5025 which is option number 4 is correct option number 4 is correct in this case is that fine guys so alright we'll end the session at this problem this was your 19th problem how there's six more problems to go which you can actually do it as a homework and submit it on the group personally to me okay so hope this session was useful to you in understanding what are the things that you need to again revise and revisit especially please focus to properties of triangles the inequalities the conditional identities those are important part and of course trigonometric equation also you guys have messed up somewhere so please revise that as well okay so thank you so much all of you for coming online today for the live class over and out from my side and I've already sent you the assignment on trigonometry so please complete it and we can discuss if at all you are not able to understand any solution okay thank you so much guys bye bye take care