You will learn a lot from this JEE Advanced 2009 Problem | Aman Sir Maths

[@anirudhathakur7350]

Super easy... Converted cos^4x in (1-sin^2x)^2 then sin^2x = t Solve the quadratic, and then put sin^2x =2/5 value in all options Got option A and B in 1 min.

[@quintonjazz7204]

I did it also.

[@kushagramishra1729]

Sir I solved this by converting all terms to sin and then made a quadratic.

[@ghostunstoppable6297]

Yes and I also solved the same way but converting it into cos

[@aradhyasingh4043]

same

[@Zx-cl5dr]

Me too

[@Vikash-mb9zi]

Same bro

[@Vikash-mb9zi]

We will get cos²x = 3/5 by this method and on solving further we will get sin²x and then tan

[@udaymaheshwari1702]

another method: consider x=(sin a) power 2 and y=(cos a) power 2 and solve by ellipse and line's intersection point

[@qbish_]

yes ,it works but why does it make sense , we have only one variable,converting it would look like X^2/(2/5) +Y^2/(3/5) = 1 ,giving X=+root(2/5),Y=+root(3/5) ,but why does it make sense

[@AimzIIT]

Sir it can be easily solved by Titu's Lemma we know (x1)²/A+(x2)²/B >=[x1+x2]²/A+B And equality exists iff (X1)/A = =X2/B So this equation has equality and satisfies Titu lemma

[@rishavbagri4211]

Ashish sir student?

[@AimzIIT]

@@rishavbagri4211 no i was , A4S now i am Pwian

[@JeeNeetMemes]

The easiest approach in this question is to divide the whole equation by cos4x and converting all terms to tanx and then solving the equation

[@aashishkumar9658]

Yess.. Or you can just check the options by making triangles😂

[@studywithsourodeep]

You are right 🙏

[@krishgarg2806]

@Mathhatter true, I did it similarly by converting sin^4x = (1-cos^2x)^2

[@devcoachingclasses1]

Yes and choose the option according to given value

[@mr.singhbist]

the easisest approach to this problem is by using titus inequality

[@parthmadaan3399]

Sir I have done this problem 3rd time and sir jab bhi karta hu maza aata h

[@abvd92840]

14:08 I'm from 10th class and I solved this question, just simply divide both sides by cos⁴X, convert all terms into forms of tan²X and tan⁴x you'll get a quadratic equation with equal roots.

[@mvppiro]

Bhai koi badi baat nahi h ye to 10th ka hi question h

[@abvd92840]

@@mvppiro haan bhai. Test ka easiest question hoga probably

[@abhirupkundu2778]

gadha hain kya? Ye jee advanced ka question hain. Par iska easier method lagaya tune. Sir ke tarah kar leta to maan jata @@abvd92840

[@khuranapranshu]

Kaise kiya bro details dena

[@div_07]

Can easily be solved by substituting cos²x with 1-sin²x. It becomes a quadratic in sin²x.

[@MdWasim-mj9wj]

Yeah it's easy one

[@shiwoshiwoismyactualname]

Yea mains '22 had harder questions than this adv question

[@akshatagarwal6308]

Yeah usse easily ho rha hai, typical approach hai ye

[@parthdanve6511]

Easiest method is to take cos^4x common to make tan^4x on LHS and send to other side as sec^4x. Then write sec^4x as (tan^2x+1)^2. We get biquadratic in tanx or quadratic in tan^2x using this method. After that just substitute values and check for tan^2x. Do the same method for the equation options and substitute correct value of tan^2x to check whether its true

[@mriduldoesminecraft6882]

simple approach write sin^4x as (sin^2x)^2=(1-cos^2x) ^2 then obtain a quadratic in cos^2x and cos^2x=3/5 then basic trigo needed for giving answer a) and b)

[@befactfull525]

Yes

[@befactfull525]

In which class you are

[@priyankachhajer1702]

One of the best mathematics teacher.. 🙏🙏🤜

[@Alexander_1212-j1v]

By quadratic equation this is easy to solve but by your method we learnt another way to solve such question. ...👍

[@rushikeshpale]

Even I don't know how to solve this problem.. mene dekha toh mene bola ye kya he ye

[@Alexander_1212-j1v]

@@rushikeshpale are u jee aspirant

[@rushikeshpale]

@@Alexander_1212-j1v you will amaze but I really don't know which entrance exam should I give. And just 5 months remain for boards it feels like I am dumbest person now😅

[@Alexander_1212-j1v]

@@rushikeshpale don't worry u have still a decent time to crack any exam

[@Alexander_1212-j1v]

Btw where are u from.

[@anonymous-jg3ec]

I think the easiest way to solve this question is by using options (if given)

[@ranistanly5340]

We can write 1/5 as 1/5*(sin²x+cos²x)² youll get a perfect square on putting it to left hand side

[@ritamroy5758]

Sir u upload good vdos. I appreciate u. But this vdo is already uploaded in Math Booster channel. But here putting sin^2x = a and cos^2x = 1 - a, we get the solution very fast.

[@Lalan0_0]

Almost an oral question for those who knows titu's lemma 💀

[@AyushRajpurkar]

Bhai aur aise helpful theorems, concepts, Lammas ke naam batao pls

[@debadritoduttaedits]

Exactly bruh😂😂 this would be the easiest problem if it appeared in ioqm

[@deeppandey5654]

Iam in 10nth and yet solved it , it means it is easy question..

[@notayush2064]

Yes it is

[@vinayak9828]

Divide the entire equation by cos and then write sec ^2 as 1 + tan ^2 and then make quadratic in tan^2 put tan^2 = u and solve

[@JeeNeetMemes]

Yes good 👍

[@arpitagarwal82]

sin square x = y and cos square x = z substitute karke original eq becomes 15y^2 + 10z^2 = 6 and another identity eq y + z =1 ..... solve both equations. Eleminating z will give an easy eq (5y-2)^2 = 0. hence sin^2x = 2/5....

[@RajThakkar248]

I tried this method. It's time consuming but very effective!

[@udaygoyal5569]

Yes bro I also did it in same way😆

[@arpitagarwal82]

@@RajThakkar248 Not very long... Ab itna to karna hi padega bhai....

[@anandchaurasia5992]

Sir kuchh jyada dimag laga liye hai 1-take l.c.m and express in a simple equation by cross multiply 2-break cos^4x into sin 3-then let sin^2x=a 4-solve the quadratic equation by spliting method 5- then we get value of sin^2x 6-then find p,b,h. Then we can find all value 7-bhannat sir ka koi aur que try karo😃😃😃😃

[@shreyasidubey7042]

Sir ,I have done it by using simple quadratic equation For it I have converted one of them in such way that it become a function in Terms of sin or cos then after by using quadratic I have done it ........ Thanku for such interesting problems......🎉

[@9anishantgarg184]

divide both sides by cos^4 x and create a quadratic by taking tan^2 x = y and make a quadratic..options will be managed in two steps

[@pepethe123]

I am a 10th grader and did this que only because of my excellent teacher. Thanks for uploading such helpful question.

[@Tryha4d]

I'm also in 10th but couldn't solve it can you tell me which type of coaching you go

[@arjunphaneesh6051]

Even I'm 10th I felt that this was very easy

[@poonammittal4603]

apne trigno and quadratic gb sir se kiya na?

[@pushkardev3090]

good job bro

[@mamtaggupta]

I also did it

[@harshitghotiya]

Easiest way is to use cauchy swartz angel form or titu's lemma.. just one liner solution

[@Maths_fun_by_himanshu]

Ye kis chapter mai padhna h?

[@cuteff8811]

Full respect

[@NIRAjKUMARबिहार]

I am already solved this question sir Because doubt question is my favorite

[@meetkathrecha1448]

Sir , please make video on adv 2022 matrix problem I really excited, how you treat that problem . ❤️🔥🔥🔥

[@yuvrajsaini3369]

it would get solved in seconds if we would have used titu's lemma , but thank you aman for giving us a fundamental approach to solve this problem . It helped in clearing some basic concepts

[@thetenniszone123]

could you tell me how to solive it using titu's lemma?

[@rajanbal5316]

We can do this by writing above equation as sin^4x/(5-3) + cos^4x/(5-2) = 1/5 taking 5 common from the denominator of both sides sin^4x/(1-3/5) + cos^4x/(1-2/5) = 1 now compare each part of the above equation with ( sin^2x + cos^2x = 1 ) we know that sin^4x/(1-3/5) = sin^2x and cos^4x/(1-2/5) = cos^2x therefore, sin^2x = 1-3/5 = 2/5 and cos^2x = 1-2/5 = 3/5 so tan^2x = 2/5

[@rajvishwakarma758]

Cauchy Schwartz inequality can also be used.

[@ahornypotato5397]

how?

[@srijansinha2218]

Simple take lcm and substitute (cos^2x)^2 by (1-sin^2x)^ Further it can be easily done

[@wantedgamer2235]

Sir Apne Dil Jeet Liya ❤️❤️🙏🙏 Kya Shandar Tarike Se solv Kara Apne

[@deveshjoshi5415]

I have easy soln Let sin²x= t It form quadratic and at end we get Sin²x=2/5 and cos²x=3/5 and find whatever given so option get a and d

[@i.not_iman]

im in 10th grade...i solved it by turning it into a quadratic equation in one variable i.e. by supposing Sin²x = y!! i don't find why the question is challenging!? i think most of my friends in 10th can solve it as well.. 😅 The way you solved it is critical tho! :)

[@nibaranghosh9480]

Sir, I solved this by converting all terms to cos and then made a quadratic. I found only option (a) tan²x = 2/3. And in second answer I saw of yours solution.

[@ansafmustafa5863]

Bhai second answer ke liye jo tumhari quad se sinx^2 aur cos square ki value aa rahi hai uska power 4 karke divide option wale no. be se karo

[@milianxhighlights]

Agar ham jyada na soche aur isko quadratic banake solve kare jaha sin^2x = t put kare toh easily saare jawab mil jayenge

[@_dhiru_4507]

thanks sir aap hamare liye aese hi imp question laya karo and we love you and your`s maths

[@karthiktatikonda8583]

take lcm and then divide by cos^4x and use the idenitiy sec^2x = 1+ tan^2x , you will get tan^2 x as 2/3 then take tanx = +-underoot2/3 draw the triangle then youll get sin x = root2/root5 and cosx = root3/root5 after simplifying we get option a and b as correct ..

[@neonftw9658]

Dude iit is so easyyy 😢😊😊, you guys are lucky , Damnnn those who say JEE is tough. Maybe science is toughhhh

[@dineshnahar8557]

Too easy question I solved it in class 10

[@atharvajoshi6338]

Ya so easy I solved in kindergarten

[@sohammohanty5875]

Sir u're heavily underrated 🥲

[@shrawon3605]

But the amount of our time he wastes is huge So he's rightly rated

[@IamHINDU.JaiSiyaRam]

still you watch him

[@mcplayervineet4461]

Araay bhai Jo loog srif paiso kaay pichaay bhaag taay hai unka yahi haal hota hai

[@aimbotarmy6342]

@@mcplayervineet4461 matlab??

[@abhinavshyam9564]

Sir sabko cos2x mein convert karke cos2x mein quadratic solve kar sakte hai.. bohot easily ho jata hai

[@twinopedia7652]

sir we can write 1 as sin^2x+cos^2x and then it can be solved very easily

[@thezerothandtheinfinite]

First check the options: (tanx)^2= 2/3 (tanx)^2 +1= 1/(cosx)^2= 5/3 (cosx)^2=3/5 and (sinx)^2= 2/5 (cosx)^4= 9/25 and (sinx)^4= 4/25 substitute in equation, you'll get 1/5=1/5, which is true, so option a is correct and you now have (sinx)^4 and (cosx)^4, so just square and substitute in the remaining two options

[@Sd4ss]

Sir great respect🙏🏻🙏🏻👑

[@DeepeshSonkar-b1q]

write sin^4xas sin^2x(1-cos^x) and cos^4x as cos^2x(1-sin^2x) take them on one side since sin^2x and cos^2x can only be zreo if they are multipled by zero , hence we directly get sin^2x as 2/5

[@moses7860]

Put sin²x =a and cos²x=b , then a+b=1. Solve then for a and b, so 3 minutes maximum

[@mr.technicalverma7506]

Kya dialogue maarte ho yaar aap 😂👏👏maja aa jaata hai

[@nirmalupadhyay4548]

SIR I HAD SOLVED THIS QUESTION BY CHECKING YHE OPTIONS .............. 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@cataminz933]

write b as a+b - a and then take lcm such that a is common in both sinusoidal terms and convert the given equation into quadratic by using a² - b² identity

[@rajneeshyadav4001]

Mja agya sir🙏🏻🙏🏻

[@Unknowngaming07]

Best method will be putting 1 =sin²x+cos²x then saare terms ko left me lao and sin²x nd cos²x common leke easily hojayega

[@lokendrasingh2810]

Let sin²x=a,cos²x=b,so a+b=1 and a=1-b a²/2+b²/3=1/5 3a²+2b²=6/5 3(1-b)²+2b²=6/5 3(1+b²-2b)+2b²=6/5 3+3b²-6b+2b²=6/5 5b²-6b+9/5=0 25b²-30b+9=0 25b²-15b-15b+9=0 5b(5b-3)-3(5b-3)=0 (5b-3)²=0 b=3/5 a=2/5

[@adityashukla3742]

The Cauchy - Schwartz Identity seized the game in seconds!!

[@cataminz933]

how?

[@AdityaKumar-gv4dj]

Sir apne bohot complex bana diya iss chij ko, isko general approach se bhi banaya jaa saktha hai.

[@mathomagics2712]

easiest approach is to convert the 1 in RHS to sin^2x + cos^2x then solve in 2 line

[@satviktripathii]

easiest method is to convert cos x into sin x and solve the equation to get sinx..convert to tanx .......for second part just put the value of sin x & cos x obtained to get 1/125

[@eihridtxf-hl7wx]

It seemed easy sir. Sab ko cos me lake karne se bhut jaldi bn gaya question

[@anjalidwivedi2057]

I am in class 10 I'll give boards after 9 days and I solved this one by using quadratic equation and basic trigonometry.Took me just 1 min.... This was easy....cant believe these type of questions are ed in JEE

[@Arya-o7j]

You are my competition

[@ayushawasthi07]

It can be done by using cauchy schwaz inequality in 2 lines

[@gauri5926]

Not by cauchy Schwartz inequality ,,,,,,but it can be done by Titu's lemma

[@sirak_s_nt]

​@@gauri5926 Titu's lemma is just a special case of Cauchy only

[@harshitdodani6180]

Sir you are amazing

[@Manish_babu-ft4up]

Sir , on solving tan^x=a/b now if we see tan^2x is equi🤔valent to sin^4/2+cos^4/3 so if we write (tan^2)^3 which is directly equal to 1/125

[@AbheeshnaDey]

Dividing the whole equation by cos^4x if we proceed further then we get the value of tan^2x. Easy problem😊😊😊

[@kavyanshtyagi2563]

sir aise hi mast mast maths ke saval lekar ayiea

[@aashutoshpathak8973]

I think TITU's lemma is a much better approach for such positive real equality problems.

[@TOOFAANFACTS]

Bhai mene bhi Titus lemma se kiya equality bala

[@adityarawat3968]

Sir i am in class 10th and solved this question by converting cos⁴x in termas of sinx....and made all thing quadratic...and easily solved

[@rajaspirant544]

legend solved

[@aritrabarman8763]

Sir can u plz bring chapterwise jee main maths solutions of jan attempt??

[@Allinonegaming-r2y]

Mera 1st optiont aa gaya tan square x =2/3 maine solve kiya Titu lemma / Engel form sedrakyan inequality

[@himesh6707]

Sir mai 10th mai hu aur mujhe ek question mai doubt hai Q- sinθ + 2cosθ=1 then we have to prove 2sinθ-cosθ=±2 My steps to solve the question : Given, --> Sinθ+2cosθ=1 or 2cosθ=1-sinθ Dividing cosθ both sides 2 = 1/cosθ - sinθ/cosθ 2 = secθ - tanθ .... (1) As, 2 = secθ - tanθ Therefore, by the identity ( sec²θ - tan²θ=1) secθ + tanθ = 1/2 .... (2) Adding (1) and (2) We get, secθ = 5/4 = HYPOTENUSE/ BASE Let, H=5x and B = 4x By Pythagorean triplet, Perpendicular will be 3x Therefore, Sinθ = P/H = 3x/5x = 3/5 Cosθ = B/H = 4x/5x = 4/5 We have to prove - 2sinθ - cosθ= ±2 Taking LHS, = 2(3/5) - (4/5) { sinθ = 3/5 and cosθ = 4/5) = 6/5 - 4/5 = 2/5 But, 2/5 ≠ 2 Sir please tell me what mistake I did here.

[@theprakhar733]

sin θ+2cos θ=1 (given) Now sq both sides - (sin θ+ 2cos θ)² =1 sin² θ+ 4 cos² θ + 4sinθcosθ=1 sin² θ + 4(1-sin² θ) +4sinθcosθ=1 sin² θ + 4 - 4sin² θ +4sinθcosθ=1 Now - Taking sin² θ to RHS 4- 4sin² θ+4sinθcosθ=1- sin² θ So: 4-4sin² θ+4sinθcosθ=cos² θ Now taking all terms to LHS except 4 - 4=4sin² θ + cos² θ- 4sinθcosθ So we get - 4= (2sinθ - cos θ) ² Now sq root on both sides - 2sinθ - cos θ=+-2 Hence Proved Hope you understood

[@janu1742]

I am in 10th class .I loved this question .and I already solved some these type of question.

[@vitthalrastogi2829]

किस्मत सबको मौका देती है किस्मत सबको मौका देती है और मेहनत सबको चौका देती है।❤❤

[@highermathematicsbya.p.t2293]

Sabse easy approach by titu,s inequality

[@AdityaKumar-xo3ot]

a²/x + b²/y >= (a+b)²/x+y ; Equality holds when a/x=b/y Use this inequality and get your answer within 30 seconds

[@Normie4lyf]

cs inequality ki baat karra tu

[@AdityaKumar-xo3ot]

@@Normie4lyf ha cauchy schwarz inequality

[@befactfull525]

Sir I have solved this by options within seconds and after fiunding valve of cos2x and sin2x as 3/5 and 2/5 respectively

[@anmoltyagi7107]

This can easily be solved by using titu's lemma

[@Bittu_Kumar333]

I am in 10th; going to give boards after 10 days. I have solved this question just by using quadratic equation and simple trigo. Thank you sir!

[@maneeshsk4375]

@@tanmaydhule7706 it's not about solving this particular problem. This is actually a very easy one can be solved using multiple methods. The thing he wanted to convey in the video was the identity used while framing the question.

[@ashaydwivedi420]

@@maneeshsk4375its okay yaar 10th me to mujhe sec2 x = 1 + tan2 x bhi yaad nahi rehta rha

[@ajal3025]

Sir again a sinple question bs quadratic bnakr values put krni h Plz increase the level sir

[@abhirajsrivastava6000]

Sir jee aapka idea great

[@challenge8346]

Sir we want you daily upload min one question of Jee Advance.

[@shahidpervaiz5216]

Simple We can write as 👇 15*((Sinx)^2)^2 + 10* ((Cosx)^2)^2 = 6 Substitute (Cosx)^2 = 1 - (Sinx)^2 Once simplified we get 25*(Sinx)^4 - 20*(Sinx)^2 + 4 = 0 Let p = (Sinx)^2 Then equation becomes 25*p^2 - 20*p + 4 = 0 Once simplified it gives Sinx = SQR (2/5) So x = arc Sin (2/5) 🥳👍🥳👍🥳👍

[@tapasbhakta9944]

I just started solving by doing L. C. M... And I realised that both option "a" and "c" are right..

[@itxx__vivek]

Sir u are best in maths

[@maddenom]

I think using AM GM inequality would also do the trick...

[@Fsinthechat10]

I actually managed to solve this in 5 min by converting the equation into a quadratic equation

[@DivyanshSingh-mp5uh]

Sir you can reduce the steps it will more easy to do this by taking cos4x and sin4x as 1 and then by dividing it by a/b cos 4x and will neglect the negative sin as the tan may lie in any of quadrant? Is this method right pls Tell 😃

[@rameshn282-u7q]

integral of f(x) = f(x) then f(x)=? f(x) is not e power x. please tell sir

[@S2FunReacts22]

Sir I solve this using Titu's Lemna And I got Correct option within 4 min ..

[@anushkamittal6805]

You can easily solve it by forming a quadratic equation (by the way I am in 10th and any good student of class 10th can solve this)

[@pepethe123]

What I did is- first of all I took LCM Than converted whole LHS into sin terms but putting on identity than I got 5sin⁴x-4sin²x=-4/5 than I assumed sin²x=t and make a quadratic and found the value of t by applying quadratic/shree dhar Acharya formula and found sin²x and so on....

[@boxoftin]

I did exactly the same

[@atharvsharma1866]

@@boxoftin took like 2 mins.

[@utkarshgangwar2120]

Divide the whole by cos^4x

[@reettwikpati8654]

Nishant Vora sir is the God of Mathematics ❤

[@kiran6178]

Yes

[@pranavpathak1459]

We know sin^2x+cos^2x=1 So in this question comparing coefficients is the easiest way to solve.No need of applying formula this much.Simplest way here is coefficients comparing.

[@dragonslayer7364]

Sir aapne Jo ye question ka example karwaya na waisa hi same question ek 10 ki maths ki book 15spq arihant mein hai pg no 17 question no 19 for 2020

[@purnaroy460]

Very good explanation sir ❤

[@Aditya_Mishra-ASKM]

Sir I solved it on the knowledge of class 10 trignometery

[@enthusiastic123]

Sir good explanation 😊

[@sairevanth2616]

Another method: write [sin^4x]/2 + [cos^4x]/3 = 1/5 into 5[sin^4x]/2 + 5[cos^4x]/3 = 1. compare this equation with sin^2x + cos^2x = 1. rest is easy.

[@mathology568]

Sir I solved by sin2x+cos2x=1 and substituted sin2x as t2

[@Mathalay]

5 +4-3 = 5+1 = 6, here operation was not done from left to right, but result is the same. I opine, the answer should be 1. Operation 2(3) should be performed before division.

[@kavyaarora1348]

WE CAN ALSO TRY BY PUTTING SIN2 X = 1- COS2 X THEN AFTER SIMPLICATION PUT COS 2 X =T