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.
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
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
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
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.
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)
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....
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 ..
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.
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......🎉
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
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.
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
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....
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
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
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😃😃😃😃
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
@@Aditya..433 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😅
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
Numbers from 1 to 20 are written on a board. In every move, You erase any 3 numbers present on the board (say a b and c') .You immediately write 2 new numbers (a+b+c)/3 and 3abc/(a+b+c).Answer the 2 sub-questions based on this information given above. 1.AfterAfter how many steps are you forced to stop. 2. When only 2 numbers are left in this process, it was observed that one number was 19! (19 factorial). Then what must be the other number? Can anyone explain?
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.
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 😃
i solved this question... within 1 minute. I just took cos⁴x common...and then made the whole term in terms of tanx. and then differentiate both sides. Actually i converted all cosx terms to secx terms.. and then to tanx terms. From here I got tan²x = 2/3 directly..and all this took me 30 sec..and 30-35 sec for the rest part..
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.
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.
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) 🥳👍🥳👍🥳👍
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! :)
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
I solved it by the following way : Sin⁴x/2+Cos⁴x/3=1/5 Putting 1=Sin²x+Cos²x in RHS ,we have Sin⁴x/2+Cos⁴x/3=Sin²x+Cos²x/5 =>Sin⁴x/2-Sin²x/5=Cos²x/5-Cos⁴x/3 =>Sin²x(Sin²x/2-1/5)=Cos²x(1/5-Cos²/3) =>Sin²x/Cos²x=(3-5Cos²X/15)/(5Sin²x-2/10) =>Tan²x=(3-5+5Sin²X/3)/(5Sin²x-2/2) (Putting Cos²x=1-Sin²x) =>Tan²x=2(5Sin²x-2)/3(5Sin²x-2) =>Tan²x=2/3 Hence option (a) is correct
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.
I did it also.
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
Ashish sir student?
@@rishavbagri4211 no i was , A4S now i am Pwian
another method: consider x=(sin a) power 2 and y=(cos a) power 2 and solve by ellipse and line's intersection point
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
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
Yess..
Or you can just check the options by making triangles😂
You are right 🙏
@Mathhatter true, I did it similarly by converting sin^4x = (1-cos^2x)^2
Yes and choose the option according to given value
the easisest approach to this problem is by using titus inequality
Sir I solved this by converting all terms to sin and then made a quadratic.
Yes and I also solved the same way but converting it into cos
same
Me too
Same bro
We will get cos²x = 3/5 by this method and on solving further we will get sin²x and then tan
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
Can easily be solved by substituting cos²x with 1-sin²x.
It becomes a quadratic in sin²x.
Yeah it's easy one
Yea mains '22 had harder questions than this adv question
Yeah usse easily ho rha hai, typical approach hai ye
We can write 1/5 as 1/5*(sin²x+cos²x)² youll get a perfect square on putting it to left hand side
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.
Bhai koi badi baat nahi h ye to 10th ka hi question h
@@mvppiro haan bhai. Test ka easiest question hoga probably
gadha hain kya? Ye jee advanced ka question hain. Par iska easier method lagaya tune. Sir ke tarah kar leta to maan jata @@abvd92840
Kaise kiya bro details dena
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)
Yes
In which class you are
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....
I tried this method. It's time consuming but very effective!
Yes bro I also did it in same way😆
@@RajThakkar248 Not very long... Ab itna to karna hi padega bhai....
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 ..
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
Yes good 👍
Simple take lcm and substitute (cos^2x)^2 by (1-sin^2x)^
Further it can be easily done
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.
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......🎉
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
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
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.
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
Put sin²x =a and cos²x=b , then a+b=1. Solve then for a and b, so 3 minutes maximum
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
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....
I did exactly the same
@@boxoftin took like 2 mins.
Divide the whole by cos^4x
sir we can write 1 as sin^2x+cos^2x and then it can be solved very easily
I think the easiest way to solve this question is by using options (if given)
Sir I have done this problem 3rd time and sir jab bhi karta hu maza aata h
Sir , please make video on adv 2022 matrix problem I really excited, how you treat that problem .
❤️🔥🔥🔥
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
Cauchy Schwartz inequality can also be used.
how?
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
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
Best method will be putting 1 =sin²x+cos²x then saare terms ko left me lao and sin²x nd cos²x common leke easily hojayega
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
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😃😃😃😃
One of the best mathematics teacher.. 🙏🙏🤜
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
Almost an oral question for those who knows titu's lemma 💀
Bhai aur aise helpful theorems, concepts, Lammas ke naam batao pls
Dividing the whole equation by cos^4x if we proceed further then we get the value of tan^2x. Easy problem😊😊😊
By quadratic equation this is easy to solve but by your method we learnt another way to solve such question. ...👍
Even I don't know how to solve this problem.. mene dekha toh mene bola ye kya he ye
@@rushikeshpale are u jee aspirant
@@Aditya..433 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😅
@@rushikeshpale don't worry u have still a decent time to crack any exam
Btw where are u from.
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
cs inequality ki baat karra tu
@@Normie4lyf ha cauchy schwarz inequality
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
could you tell me how to solive it using titu's lemma?
Numbers from 1 to 20 are written on a board. In every move, You erase any 3 numbers present on the board (say a b and c') .You immediately write 2 new numbers (a+b+c)/3 and 3abc/(a+b+c).Answer the 2 sub-questions based on this information given above.
1.AfterAfter how many steps are you forced to stop.
2. When only 2 numbers are left in this process, it was observed that one number was 19! (19 factorial). Then what must be the other number?
Can anyone explain?
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.
Full respect
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 😃
Mera 1st optiont aa gaya tan square x =2/3 maine solve kiya Titu lemma / Engel form sedrakyan inequality
Iam in 10nth and yet solved it , it means it is easy question..
Yes it is
Sir I have solved this by options within seconds and after fiunding valve of cos2x and sin2x as 3/5 and 2/5 respectively
Too easy question I solved it in class 10
Ya so easy I solved in kindergarten
It can be done by using cauchy schwaz inequality in 2 lines
Not by cauchy Schwartz inequality ,,,,,,but it can be done by Titu's lemma
@@gauri5926 Titu's lemma is just a special case of Cauchy only
Titus lemma equality holds here
So we can compare 1st and 2nd term
And directly get tan4 x
i solved this question... within 1 minute. I just took cos⁴x common...and then made the whole term in terms of tanx. and then differentiate both sides. Actually i converted all cosx terms to secx terms.. and then to tanx terms. From here I got tan²x = 2/3 directly..and all this took me 30 sec..and 30-35 sec for the rest part..
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
WE CAN ALSO TRY BY PUTTING SIN2 X = 1- COS2 X THEN AFTER SIMPLICATION PUT COS 2 X =T
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.
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
easiest approach is to convert the 1 in RHS to sin^2x + cos^2x then solve in 2 line
Sir sabko cos2x mein convert karke cos2x mein quadratic solve kar sakte hai.. bohot easily ho jata hai
integral of f(x) = f(x) then f(x)=? f(x) is not e power x. please tell sir
I am already solved this question sir
Because doubt question is my favorite
I am a 10th grader and did this que only because of my excellent teacher.
Thanks for uploading such helpful question.
I'm also in 10th but couldn't solve it can you tell me which type of coaching you go
Even I'm 10th
I felt that this was very easy
apne trigno and quadratic gb sir se kiya na?
good job bro
I also did it
Easiest way is to use cauchy swartz angel form or titu's lemma.. just one liner solution
Ye kis chapter mai padhna h?
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.
Sir can u plz bring chapterwise jee main maths solutions of jan attempt??
I think using AM GM inequality would also do the trick...
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.
Sir Apne Dil Jeet Liya ❤️❤️🙏🙏
Kya Shandar Tarike Se solv Kara Apne
Ise quadratic se bhi solve kar sakte h 1st put cos^2x=1- sin^x and 2nd vice versa
SIR I HAD SOLVED THIS QUESTION BY CHECKING YHE OPTIONS .............. 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
U can also try by putting cos^2x= t
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)
Sir I solved by sin2x+cos2x=1 and substituted sin2x as t2
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)
🥳👍🥳👍🥳👍
sir, i have solve it by finding tan x from option , then covert it into sin x and cos x then put in the question , solve in 5 min
Option a,b are correct.Solved in less than 3 mins
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! :)
Differentiate both sides wrt x and ans 3 lines mein nikal jaayega
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
You are my competition
thanks sir aap hamare liye aese hi imp question laya karo and we
love you and your`s maths
Sir I did it using Cauchy Swartz identity
If you use the option for the value of tan^2× it will be much easier
I solved it by the following way :
Sin⁴x/2+Cos⁴x/3=1/5
Putting 1=Sin²x+Cos²x in RHS ,we have
Sin⁴x/2+Cos⁴x/3=Sin²x+Cos²x/5
=>Sin⁴x/2-Sin²x/5=Cos²x/5-Cos⁴x/3
=>Sin²x(Sin²x/2-1/5)=Cos²x(1/5-Cos²/3)
=>Sin²x/Cos²x=(3-5Cos²X/15)/(5Sin²x-2/10)
=>Tan²x=(3-5+5Sin²X/3)/(5Sin²x-2/2)
(Putting Cos²x=1-Sin²x)
=>Tan²x=2(5Sin²x-2)/3(5Sin²x-2)
=>Tan²x=2/3
Hence option (a) is correct
Awesome thinking
put cos^2 x as t and sin^2 x as 1-t.
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
I just started solving by doing L. C. M... And I realised that both option "a" and "c" are right..
Sir apne bohot complex bana diya iss chij ko, isko general approach se bhi banaya jaa saktha hai.
Mja agya sir🙏🏻🙏🏻
5[1.96-sin^2{(cos^-1 (0.02)}° ] find the value of it . Challenge to you sir. 😊😊😊.
I paused video and solved it , converted everything in cos2x and solved the quadratic.
Ig it took 3 mins
Sir I solved it on the knowledge of class 10 trignometery
Sir you are amazing
I think TITU's lemma is a much better approach for such positive real equality problems.
Bhai mene bhi Titus lemma se kiya equality bala
Sir great respect🙏🏻🙏🏻👑
BY CREATING A QUADRATIC IN TERMS OF SINX IS EASIEST WAY. You can get sinx , cosx values in just 3 steps.
Sir again a sinple question bs quadratic bnakr values put krni h
Plz increase the level sir
Convert 1 in RHS into sin²x+cos²x.
Take to LHS and Make groups
sin²x comes out to be 2/5
cos²x comes out to be 3/5
Salute to your explanation sir, even I a 10th class student , understand it without any problem
same bhaii
aapne pcm li bhai?
@@devanshbhardwaj9810 ji ha bhai
@@aditsinghh bhai aapko jee ke liye all the best mai aapka junior 🥰
@@aditsinghh work hard brother
किस्मत सबको मौका देती है
किस्मत सबको मौका देती है
और मेहनत सबको चौका देती है।❤❤
legend solved
Cs lemma se direct ho jayega gurudev
This can be solved very easily by taking derivative on both side
Dude iit is so easyyy 😢😊😊, you guys are lucky , Damnnn those who say JEE is tough. Maybe science is toughhhh