 request to all to please give your attendance in the chat box please give your attendance in the chat box also kindly write which all topics do you guys prefer me to take it in the class so that I can give priorities priority to those topics okay circles trigonometry two topics triangular inequalities the guys in two hours class I can only solve three to four topics often are I can get to each and every topic if you want I can increase the class to two and a half hours there is no problem with that okay so I'm not starting I'm do you guys want theorems also or only questions because if I start taking theorems it will take a lot of time for me to take theorems of circles so it's better you refer to my previous classes where where I have already devised theorems of circles and in this class let me only do questions okay so this class and only take problems fine so let me start with first problem I'm first taking circles then I take trigonometry and then I go to triangular inequalities the class may be minimum of two hours it can extend to two and a half hours so this is the question that we need to solve in a moment so suppose this is the circle and this is the diameter AB so this one is diameter AB and this is E and these two points where this line AE and B are intersecting the circle are CD so this is C this is D I am drawing the CB and the center is oh I am drawing this OD so and then I'm drawing the CO so this angle COD angle COD is equal to 40 degrees so what has been given to you it has been given to you that AB is a diameter the second thing which has been given to you that oh is circle sorry circle center AC and BD are produced to meet at E which has which can be seen over here and angle COD is equal to 40 degrees you find angle CD have you solved should I solve the question so what I need to find over here is I need to find this angle if I need to find this angle you should understand that this angle is part of this angle is part of triangle B E C triangle angle CD angle CD can be written as this angle can equivalently be written as angle CEB because this point defaults on line B so there would not be any problem in renaming this particular angle as angle CEB now angle CEB is part of triangle B E C so I need to find out other two angles of triangle other two angles of triangle BC BCE so you look at here this angle angle ACB is equal to 90 degrees why because in a semicircle diameter forms 90 degree angle so the angle ACB is 90 degrees so on other side this angle will also be 90 degrees so why because angle ACB plus angle ECB is equal to 180 degrees this is already 90 degrees so this ECB would also be equal to 90 degrees so I got this angle as 90 degrees now let me try to find out the other angle so I have already got this angle now I need to find out this angle what would be this angle so how to find this angle try to understand this angle has been given as 40 degrees so angle look at this angle angle DBC look at look try to understand let me draw this here what happens is this is called DB here is a line going like this C and this line has not been drawn I am reconstructing this line because this particular chord DB is making what angle this is making angle 40 degrees on the center so this angle is 40 degrees on the center so what would be this angle this angle would be half of this angle 40 degrees so angle DB C is equal to 20 degrees which is equal to angle EBC I have already told that D is equivalent to B so this comes out to be 20 degrees so you know that this is 90 this is 20 so total comes out to be 110 so this angle would be equal to 70 degrees okay let me give you another question okay let me mix and match questions so now I am giving you questions from because there should not be the case that I am doing one topic I am overemphasizing on one topic and leaving another topic so okay I am taking questions from triangular inequalities and the first question which I am giving you is prove that perimeter of a triangle any triangle perimeter of any triangle is greater than some of its three medians solve it I give you two three minutes to solve it okay four people have done it already I'm waiting for others to answer can someone please send me the photocopy of photocopy from your book of trigonometry chapter now look at here this is triangle ABC the perimeter would be AB plus BC plus CA this is medium so this is AD so try to understand here some of the two sides would be greater than the third side so in in triangle ADB AB plus BD is greater than AD okay thank you Dorothy I've got the seeds so please to do the answer that I got the seeds now try to understand now in triangle ADC AC plus DC is greater than maybe you add both of them you get AB plus AC plus what is this BD plus DC is equal to how much that is BC is greater than 2 AD and likewise you will get it for all the medians will get your result or what you can do is okay let me give that question only forget about this right on the question suppose O is any point in the interior of triangle ABC prove that O A plus O B plus O C is greater than half of AB plus BC plus CA okay one of you have done let me give you a question from trigonometry find the value of all the question I've written on the board this question is easy it's not about whether you have it in your syllabus or not when you know that sine 90 minus theta is equal to cos theta cos 90 minus theta is equal to sine theta and tan 90 minus theta is equal to cot theta theta now tan theta multiplied by cot theta is equal to 1 you look at here tan 89 can be written as cot sorry tan 89 can be written as tan 90 minus 1 and 91 as theta is cot theta so this is cot 1 so this is tan 90 tan 89 is nothing but cot 1 and tan 1 and cot 1 months getting multiplied will give me one similarly tan 2 and tan 88 will give me one time 3 and tan 87 will give me one in between I have tan 45 which is already one so what do I get I get the answer as one right another question prove that cos a minus sine a plus one divided by cos a plus sine a minus one is equal to cos a minus cot okay Ritu has done it okay three people have written that they have done it what about others if they are not able to solve let me know I will solve it okay I will solve it so this question one way to attempt this question is to you take from here this is class a take minus from here this is sine a minus one and this is cos a plus sine a minus one so you see here this if this is x and this complete thing is why you get form x minus y and you get x plus y here so now what I do I will multiply this with cos a plus sine a minus one and I will multiply this with cos a plus sine there are two three methods to solve this question so don't think that this is the only method I'm taking the most obvious method by which most of you will start attempting it so this will become nothing but a square minus b square so this will become cos square a minus sine square sorry sine minus a sine minus one whole square let me solve this numerator first and then I'll go to denominator what is this this is nothing but cos square a minus sine square a this will become plus one so this will come out to be minus one and this will be minus two sine a so that will become plus two sine a and denominator would be cos square a plus sine square a plus one minus two sine a so this will be nothing but this is one so one plus one two two minus two sine a so how to solve it so you look at here this is nothing but let me solve here now cos square a can be written as one minus sine square a so numerator is now I'm written one minus sine square a minus sine square a minus one plus two sine a divided by two minus two sine a so this one and one gone you get two sine a minus two sine square a divided by two minus two sine a so what do you get you take sine a common you get two minus two sine a here and divided by two minus two sine a here so I think I'm not getting what I have to get over here because see if I have to get perhaps I've done some calculation mistake I'm just checking for that taken a negative here so this becomes plus sine a minus one then I have multiplied it with oh I have taken wrong multiplication itself so once again let me solve it once again this is cos a minus sine a minus one and this is cos a plus sine a minus one so I'll multiply this with cos a minus sine a minus one and this with cos a minus sine a minus one so what I get here is cos square a this is plus sine a minus one whole square minus two cos a sine a minus one this is my numerator in denominator what I get is I get cos a square minus sine a minus one square so what do I get now I get here just remove this this is nothing but cos square a and this is sine square a plus one minus two sine a and this is nothing but two sine a cos a plus two cos a divided by this is cos square a and this comes out to be minus sine square a minus one and plus two sine a so this is one so I'm writing it one plus one two two minus two sine a minus let me write it as it is two sine a cos a plus two cos a and what I get here is I can convert it into I can convert it into one minus two sine square a and this cost square is and this is minus one plus two sine a so this is nothing but what I have to prove here just let me check cos a plus cos a so look at here now I can prove this so what I do is I'll take this and this gone so I'll take two minus two sine a and here I'll take sine a common so this numerator denominator will give me two minus two sine a here also and what I do here is I write this as two cos a common and one minus two sine a and this is nothing but two sine a now this is not to this is only one because I have taken common two sine a one minus sine a so this and this is gone and this and this is gone this is one by sine a one by sine a can be written as Cossack a and this is two to one cos a by sine a can be written as caught a so this can be done like this now another way to solve this method is solve this question is I took time because I multiplied with wrong rationalizing factor so another way to solve this question is that somehow we utilize the thing cot square a one plus cot square a is equal to Cossack square a now how can you utilize this this particular thing so you need to convert it into cot and Cossack and how do you convert it into cot and Cossack so divide everything by sine a so this gives me cos a by sine a is caught a minus one plus Cossack a and you divide this you get caught a plus one minus Cossack a now if you have this thing so what you get over here is you can multiply this with again I'm not doing the same rationalization again you can multiply this with the rest you can rationalize it by multiplying this with caught a minus one minus Cossack a and then you can utilize this formula one plus cot square a is equal to Cossack square a so by that method also you can solve this question so whichever method you like you can utilize utilize that method now let me give you a question from circle you have to prove that prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic let me know once you are done with the question have you done okay four three people have written that they have done okay Varsha I'll solve it don't worry now try to understand the quadrilateral is found by angular bisector of a cyclic quadrilateral so first of all I have a cyclic quadrilateral so suppose I have a cyclic quadrilateral let me make a better circle for you so I have a cyclic quadrilateral suppose I name the cyclic quadrilateral ABCD this is ABCD and what I do is I have angular bisectors now angular bisectors are making a cyclic quadrilateral so angular bisectors will make cyclic quadrilateral something like this so suppose this is PQRS I have to prove that PQRS is a cyclic quadrilateral now how to proceed it so if I want to prove that PQRS are cyclic quadrilateral I have to prove that this angle which is angle APB what is the property of cyclic quadrilateral that opposite angles are supplementary to each other so opposite angles will add and will give me 180 degree so APB plus CRD is equal to 180 degree if I if I am able to prove this particular equation I shall be able to prove that the cyclic quadrilateral PQRS is cyclic so for that what do I do so I know that some of the triangles is 180 degree so what are the triangles I am taking I am taking triangle APB so in that angle APB plus angle PAB plus angle PBA is equal to 180 degree also if I take triangle CRD angle CRD plus angle RCD plus angle RDC is equal to 180 degree now try to understand that what is this angle PAB angle PAB is this angle now if this line PA is angular bisector then this particular angle PA would be would be half of angle A so I am writing here angle APB plus half of angle A because PAB is half of angle A and PB and this PBA is half of angle B so half of angle B is equal to 180 degree similarly in in this case CRD plus half of angle C plus half of angle D would be equal to 180 degree now what to do in this case so I will add both of them if I add both of them see what happens so this is equation 1 for me and this is equation 2 for me I am adding equation 1 plus equation 2 what do I get I get here angle APB plus angle CRD and half of angles A plus B plus C plus D and that is equal to on right hand side I have 180 plus 180 which is equal to 360 degree now any quadrilateral angle A plus B plus C plus D would be equal to you should understand this would be equal to 360 degree so half of 360 degree is 180 degree so let me write APB plus angle CRD plus 180 degree is equal to 360 degree so this goes on the other side so angle APB plus angle CRD is equal to 360 degree minus 180 degree which is equal to 180 degree that's how you prove this I hope you understood this let me give you another question if you have any question on this you post it on the group I'll try to solve it okay nobody's asking me any doubt so I'm moving ahead let me give you another question the question is from circle itself so look at the question the question tells me that I'm writing this question this is an NCE RT question and it's a very good question it says that ABCD is a parallelogram the circle through ABC intersects CD produced at E prove that AE is equal to ED sorry AD once again this is not ED this is AD the solvents let me know once you are done can solve it let let's other solvents okay Tanna has done it okay anyone else who is solving it otherwise I will solve it one more minute okay I'm solving it now so what needs to be done in this question try to understand ABCD is a parallelogram so let me draw ABCD now CD is produced at E so this is point E now what happens the circle through ABC intersect CD produced at E so it means that there is a circle through ABC and which is meeting this E so these four points ABC are on the circle just let me make this circle there is some circle like this and this E is on the circle so let me join this line this is a so what happens why it is why is on circle already it has been given that ABC are on certain circle and when CD is produced it is meeting intersecting at E so ABC is on circle so hence ABC is a cyclic quadrilateral so this much you could have done without any particular without any difficulty now try to understand if I have to prove a is equal to AD and I know that ABC is a cyclic quadrilateral then to prove that two sides are equal I can only do it with the help of angles and if I prove that this angle is equal to this angle it means that if I am able to prove that angle AD is equal to angle AD E my task would be completed so let me prove this for you now how do I prove this so try to understand angle AD E this angle and on this side angle plus angle AD C is equal to 180 degree now this angle AD C is equal to this angle because in parallelogram opposite angles are equal so angle AD E plus angle ABC is equal to 180 degree now try to understand what can I write after this so I can write that angle one second and there is one more angle I can write which is angle AD E sorry one second I have to write something about I have already written try to understand what I am trying to do I have already taken this angle now you understand that in a cyclic quadrilateral this angle and this angle add to 180 degree so I can write that angle AD plus angle ABC is equal to 180 degree so see here ABC and ABC is common right hand side and right hand side is common so I can say that angle AD is equal to angle AD E and once I say that angle AD is equal to angle AD E it means that these two sides angle A E sorry side A E is equal to side AD what I have done in this question is first I have taken this angle so I have said that angle AD E plus angle AD C is equal to 180 degree and I have replaced this ADC with ABC and I already know that AD plus ABC is 180 degree because they are cyclic quadrilateral so from there I am proving that angle A equal to angle AD this is the case so this is how you had to prove this question this question was good until unless you don't make good diagram in this question you can't solve it let me move to triangular inequality in some what is triangular inequality which is here so let me solve few questions from triangular inequality so I am making a triangle for you just look at here this is P this is Q this is R this is T so PQR is a triangle S is any point in interior and I have already done the construction for you SQ plus SR is lesser than PQ plus RT sorry PR let me know once you are done over most of you have said most of you are saying you have done this so now I am moving to a question on trigonometry let me find which question is good so the question is would that 1 plus tan square E divided by 1 plus cot square E is equal to 1 minus tan square E divided by 1 minus sorry this is not square here 1 minus cot E square and that is equal to tan square E just prove it you've done the question okay three people have told that they have done I'm waiting for others to do it so Kanai has also done it anyone else let me know let me know okay if anyone wants me to do this please type it in the chat box otherwise I will move to the next question because most of you have done this there is no point doing it if anyone wants me to do it please let me know in the chat box Shraddha I will send you the answer don't worry I will send you the answer don't worry okay as no one is saying anything I'm just going to next question so I'm giving one more question of triangular inequality and that is okay take this question only question says that PQRS is a quadrilateral diagonals are PRQS and these diagonals intersect at you have to so show that perimeter of quadrilateral is greater than PRQS solve this question let me know once you have done you are done with the question okay two people have done it or more than two people have done it a lot of people have done it okay so let me move to the question one question this question I'm giving from circles the question is question is I'm writing the question the bisector of angle B of an isosceles triangle ABC with AB is equal to AC meets the circum circle of triangle ABC point P AP and BC I'll make the figure AP and BC are reduced to meet at Q prove that CQ is equal to C so suppose this is the circle and in that circle I draw a line like this then I have another line so this line is going like this and this line is going like this I am joining these two two points this is something like this here this is something like this here now what happens is let me name it for you this is this is A this is B and this is C this is P this is Q I've already told that AB is equal to BC and you have to prove that CA is equal to CQ solve this question let me know once you are done with it once you are done you let me know I'm just waiting okay Ritu I'll solve it I'm waiting for two three more minutes so that others can reply okay I'll solve it for you as most of you are saying that I should solve it I will solve it for you now the first line that you read here is bisector of angle B so bisector of angle B is this line BP so the two angles let me write angle AB B angle ABP is equal to angle P BQ this Q can also be replaced with C because Q and C are on same line PV sorry BQ now try to understand I have to prove that CA is equal to CQ it means that which triangle I have to take I have to take triangle ACQ if I have to take triangle ACQ so in triangle ACQ I have to prove that CQ so I have to prove that this angle I have to prove that angle CAQ and angle CQA is equal if I'm able to prove this I will be able to prove that CQ is equal to CA so what do I need to do in that particular case try to understand now I know that exterior angle of any particular triangle is equal to some of the two angles it means that this angle ACV would be equal to the other two angles and these two angles are the angles for which I'm talking about is equal to summation of these two angles CAQ plus CQA now what is angle now look at here as AB is equal to AC it's an isosceles triangle so this and this is equal it means that angle opposite to this will also be equal so what I'm trying to say is that I'm trying to say you is that I want to take help of this angle to find out this angle if I'm able to find out this angle or this angle then some kind of relationship can be found out so I know that angle ACB if AB is equal to AC so angle ACB is equal to angle ABC and I also know that because line AP is bisector of angle B I know that angle ABC can be written as two times angle PVC PVC pbq whatever you want to write angle PVC I'm here only I'm writing so I'm writing this so this angle ACB would be equal to two times angle PVC so you look at here I'm writing here that two times angle PVC is equal to angle CQA plus try to understand one thing can I take help of this angle can can this angle be written as some other angle tell me guys can this angle I'm giving you a hint can this angle be written as can angle CAQ be written as angle PVC just look just just just try to understand why I'm writing this if you are not understanding this write it in the comment box I'll explain it why I'm writing this because you try to understand here is a chord which is invisible and most of you would have not drawn it this chord will make same angle anywhere on the circle so this chord is making PSE here and PVC here hence angle CAP or CAQ will be equal to angle PVC and that is why I'm replacing this angle CAQ with angle PVC because they come they are formed on the same chord so I can write that angle PVC is equal to angle CQA so this angle PVC is equal to angle CQA so try to understand that angle PVC is equal to angle CQA and angle PVC is also equal to angle CAQ so I can write that angle CAQ is equal to angle CQA and if this is equal then CQ is equal to CA so somebody has posted me I hope you understood this okay so Shantanu I will pass on your message to Akhil which class you are in just just give me your details in the chat I'll pass give me your details your personal details I mean not personal details in the sense which class do you attend and I'll pass on your message to Akhil now let me give another question do you guys want me to continue with the class I'll continue with the class otherwise I'll wrap up okay Shantanu I'll pass on your message others do you want me to give you more questions or you want me to wrap up the class it depends on you you're not saying anything okay let me give you one more question then okay so in this question P is center this is P and this is center now this is X this is Y and this is Z you have to prove that angle XP Z is equal to 2 times angle XZ Y plus angle Y XZ should I try first in two minutes you say sir can you solve this you try for two more minutes this question is so easy I don't just don't want to solve it just look at the arcs properly and utilize the funda that angle formed by any chord on the circle is half of the angle formed by it on the center that's what you have to use it I'm giving you a hint this is very easy okay those who are not getting it try to understand look at chord XY chord XY is forming angle XP Y on the center so I'm writing here line by line chord XY forms angle XP Y on center and same chord XY look at here is forming angle XY Z on circle so angle XP Y would be equal to two times angle XZ Y and and similarly if I take this chord so chord YZ forms angle YPZ on center and angle this angle YXZ on circle so from here I can write YPZ is equal to two times angle YXZ so let me add it angle XP Y plus angle YPZ is equal to two times angle XZ Y plus angle YXZ and this XP Y plus YPZ this is nothing but equal to angle XP Z so this is what you had to prove okay so I have done most of the quest done a lot of questions today so I hope you understood the questions concepts clearly if you feel like me doing more questions you can write it in the chat box otherwise otherwise I wrap up the session okay so as no one is saying anything I'm just wrapping up the session now so thank you so much for attending the class and wish you all the best for your exam