This is question is expected to be solved in 20 seconds in SSC CGL exam and worth that only Just reduce all the Tan to Cos then half all the powers of cos and covert to sin It will become 2(sin^2x + sinx)^3 Since sin^2x + sinx = 1 Answer comes out to be 2
Sir all options were integers in exam
Sir the question is wrong . Correct question is {c^12 +t^12 +3(c^10+c^8+t^10+t^8)+c^6+t^6} so the ans is 2 where c=cosx ,t=tanx
3:15 this problem solving idea is also used in finding higheer powers of complex no. a+ib.
Good one
Sinx=Cos^2x (from given equation)
Write all in terms of sinx and you will se a perfect cube.
Simplify and answer will be 2
bro i am stuck in 2 times (sin cube x + sin x)^2... what to do next
Yup did it in the exam
Bhai perfect cube nahi ban raha😂
I am literally on 0 and could do it in the exam with 10th class identities😂😂😂yes the answer was 2
@@statuseditz6401
*sin²x + sinx =1*
sin x = 1 - sin²x
sinx = cos²x 🐶
sinx/cosx = cos x
tan x = cos x *
*2cos¹²x + 6 cos⁸x + 2cos⁶x
=> 2sin⁶x + 6sin⁴x + 2sin³x 🐶
=> 2sin⁶x + 4sin⁴x + *2sin⁴x + 2sin³x*
=> 2sin⁶x + 4sin⁴x + *2sin²x*
=> 2sin⁶x + 2sin⁴x + 2sin⁴x + 2sin²x
=> 2sin²x(sin²x + 1)²
🐣 sin x = (sqrt.5 - 1)/2
🐤 sin²x =
(6- 2sqrt.5)/4
**5 step solution follows for
2sin²x(sin²x + 1)²
= 35 - 15sqrt.5 **
Sir I was in this shift but all option were single digit integers please see to it when u feel right. Rest always helpful for your explanation
Bhai tumse hi gya tha yeh wala question?
@@ICONIC1010Hospital Han matlab mein uss shift mein thaa ye onw of the genius questinos 2 aata hai answer (took 45 seconds in real exm)
@scholar2023 lekin bhai sir ne irrational form me answer kyu Diya hai??.
great piece of maths
How can we divide by sin2x+sinx-1 when it is equal to zero, as division by zero is meaningless
Amazing explanation
Glad you liked it
Ye questions ko convert karna bina root ke ho jayega reduce karte raho bs
Meri shift ka hai ho gya tha 🎉
S=C^2, T=C,
C^12+T^12 +3(C^8+T^8)+c^6 +T^6
# s^6+c^12 +(s^4+c^8)+s^3+c^6
# s^6+s^6+3(s^4+S^4) + 2s^3
# 2(S^2+c^2)^3
= 2 is the correct answer
Tougher than jee advanced(calculation wise)
bhai utna calculation nhi hota genius way mein solve karne se
@scholar2023 kuch question to calculation karwane ke liye hi banaya jata hai ... Usme calculation unavoidable hai chahe kisi bhi way me solve Kiya jai
Sir answer ye nhi tha perfect cube ban rha tha aur integer mein answer tha
@@Blob-q2b ok ji
question bhi alag tha bhai, question me (3cos¹⁰x + 3tan¹⁰x) A.K.A '6cos¹⁰x' absent tha..
Great q
29s1 mai nai tha mai darr gya tha sir mujhe laga mujhse ye miss hoga LOL
same
Fivision by zero?
Division...
This is question is expected to be solved in 20 seconds in SSC CGL exam and worth that only
Just reduce all the Tan to Cos then half all the powers of cos and covert to sin
It will become 2(sin^2x + sinx)^3
Since sin^2x + sinx = 1
Answer comes out to be 2
@@clashott4372 Good 👍
Ye 29 shift 2 me tha .. mere shift me
Hamare shift me cos^10x aur tan^10x bhia tha thrice ke sath
"Nice" ke alaava kya hi bol sakta hu
Achcha ok, thanks for watching
Shift 2 me Ayya Tha
@@jainamchhallani8972 yes
**sin²x + sinx =1**
sin x = 1 - sin²x
sinx = cos²x 🐶
sinx/cosx = cos x
tan x = cos x *
*2cos¹²x + 6 cos⁸x + 2cos⁶x
=> 2sin⁶x + 6sin⁴x + 2sin³x 🐶
=> 2sin⁶x + 4sin⁴x + **2sin⁴x + 2sin³x**
=> 2sin⁶x + 4sin⁴x + **2sin²x**
=> 2sin²x(sin²x + 1)²
🐣 sin x = (sqrt.5 - 1)/2
🐤 sin²x =
(6- 2sqrt.5)/4
**5 step solution follows for ;
2sin²x(sin²x + 1)²
= 35 - 15sqrt.5 **
@@MorphysinceC.E great sir your solution is too good.you are the best.
Question glt h
Cgl question