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  • 1 answers

Shubham Mishra 4 years, 9 months ago

LHS 2tan^-1(cosx) =tan^-1(cosx) + tan^-1(cosx) =tan^-1{(cosx+cosx)/1-cosxcosx)} =tan^-1{(2cosx)/1-cos^2x}
  • 3 answers

Akanksha Kumari 4 years, 9 months ago

Solution of ncert is given in this app

Iqbal Singh 4 years, 9 months ago

Yaa??

Mishti ???? 4 years, 9 months ago

Question no dear please...
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  • 3 answers

Pankaj Kaushik 4 years, 9 months ago

Bro take it as 1/4 {(2sin²x)² and expand square

Iqbal Singh 4 years, 9 months ago

Do u mean this ??

Iqbal Singh 4 years, 9 months ago

Sin4x
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  • 1 answers

Aryan Kumar 4 years, 9 months ago

This can solved by properties of exponent . 9 × 3^x
  • 1 answers

Harsh Kumar 4 years, 9 months ago

What is tan2 x equal to
  • 4 answers

Iqbal Singh 4 years, 9 months ago

Want rubbish is this ??

Tera Baap 4 years, 9 months ago

Jo mujhe report krega wo madherchod rhega issliye jisko madherchod, betichod bnna hai ya fir hai to report kre

Jhalak Mehra 4 years, 9 months ago

square both sides then solve

Khushi Shahi 4 years, 9 months ago

d/dx × (x+y)^1/2 = d/dx × (y). 1/2(x+y)^-1/2(1+dy/dx) = dy/dx. (1+dy/dx)1/2sqrut( x+y )=1. Ans. dy/dx = 2sqrut (x+y) - 1.
  • 4 answers

Chandni Poojara 4 years, 9 months ago

Vedantu math, mathongo..

Mishti ???? 4 years, 9 months ago

U can use vedant u or unacadmey videos on u tube... Ashish sir have charged the premium on so many concept understanding videos dear.... Regards. ??

Jhalak Mehra 4 years, 9 months ago

cbse class videos.

Iqbal Singh 4 years, 9 months ago

Ya ashish sir lecture is much better than others . He explain all the concepts including JEE and all ncert problems and exercise solutions . ?
  • 3 answers

Iqbal Singh 4 years, 9 months ago

What?? Do u mean sin-1x 1/underroot 1+x*2

Raghvendra Singh 4 years, 9 months ago

1/sin

Anshuman Mishra 4 years, 9 months ago

What ?
  • 1 answers

Anshuman Mishra 4 years, 9 months ago

dx/dy = y^y (log y +1)
  • 2 answers

Jhalak Mehra 4 years, 9 months ago

log|secx| + C

Anshuman Mishra 4 years, 9 months ago

x.tan inverse x-1/2 log(1+x^2)
  • 1 answers

Harshita Bhambhani 4 years, 9 months ago

Y=x cosx dy/dx = cosx d/dx(x) +x d/dx(cosx) dy/dx = cosx + x(-sinx) dy/dx = cosx - xsinx d^2y/dx^2 = d/dx (cosx - xsinx) d^2y/dx^2 = d/dx(cosx) - d/dx(xsinx) d^2y/dx^2 = (-sinx) - {(sinx)+x(cosx)} d^2y/dx^2 = -sinx-sinx-xcosx =-(2sinx+xcosx)
  • 1 answers

Tannu Rao 4 years, 9 months ago

|A(adjA)| = |A|^(n-1) = (-2)^2 = 4
  • 1 answers

Anshuman Mishra 4 years, 9 months ago

(e^2x+e^-2x)/2
  • 2 answers

Hardik Kamboj 4 years, 9 months ago

Solve krde bhai

Kshitiz Aggarwal 4 years, 9 months ago

Very easy....
  • 1 answers

Sonam Mishra 4 years, 9 months ago

Log[sec x.cosec x]+C
  • 2 answers

Itzz Aayu 4 years, 9 months ago

1/2arctan (x/2)+c

Suryakant Swain 4 years, 9 months ago

1/2tan^-1(x/2)+c
  • 1 answers

Jhalak Agrawal 4 years, 9 months ago

Since, Sin-¹(1/2) = π/6 And Sin-¹(1/√2) = π/4 So, the required ans is: π/6 + 2(π/4) π\6 + π/2 (π+3π)/6 4π/6 2π/3. Hence 2π/3 is the requured answer.
  • 2 answers

Shraddha ✨ 4 years, 9 months ago

Yes , u r right . The right answer is 13

Sonam Mishra 4 years, 9 months ago

Answer is 13....
  • 1 answers

_Joan_37 ... 4 years, 9 months ago

Suppose a vector AB ⃗ makes an angle θ with a given directed line l (say). ... The vector p ⃗ is called the projection vector and its magnitude I p ⃗ I is called as the projection of the vector AB ⃗ on line l.

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