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  • 1 answers

Sambhav Jain 3 years, 11 months ago

Cos(x+x) = cosxcosx-sinxsinx Cos2x = cos^2 x - sin^2 x = 2cos^2 x - 1 =1 - 2sin^2 x We know that sin(x+y) = sinxcosy+cosxsiny => sin(x+x) = sinxcosx+cosxsinx => sin2x = 2sinxcosx
  • 1 answers

Shivam Sharma 3 years, 11 months ago

dx/dt Y = dx/dt tanx Sec^2x
  • 0 answers
  • 1 answers

Rehmeen Khatoon 3 years, 11 months ago

Infinite set
  • 1 answers

Kalki Kalki 3 years, 11 months ago

First we calculate B-C and then subtract from universal set (U) the answer is come
  • 2 answers

Rehmeen Khatoon 3 years, 11 months ago

tan 60 °=√3

Anurag Singh 3 years, 11 months ago

X=60°
  • 5 answers

Rehmeen Khatoon 3 years, 11 months ago

Yes , because examiners mostly choose the question from this exercise for the exam question paper.

Tanya Bhati 3 years, 11 months ago

Yes u should try as it will make you strong in chapters

Vineeta . 3 years, 11 months ago

Ys of some chapters only but

Kushagra Pandey 3 years, 11 months ago

You should only do mai exercise if you are attempting examination

Seraj ??? 3 years, 11 months ago

Yes
  • 2 answers

Pankaj Singh 3 years, 11 months ago

(x-2)(x^2-1) Solution by u.v method (x-2)(x square+1) u'v+uv' (x-2)'(x^2+1)+(x-2)(x^2+1)' (1+0)(x^2+1)+(x-2)(2x+0) x^2+2(x-2)x+1 3x^2-4x+1

Pankaj Singh 3 years, 11 months ago

First derivative
  • 1 answers

Jagdish 3 years, 11 months ago

Derivative of 2x³ - 3x - 5 is 6x² -3 so at 'x=-5' its derivatives l will be 6(-5)² - 3 = 150-3 = 147
  • 2 answers

Sakshi Aggarwal 3 years, 11 months ago

By first principal F(x) h=0 =tan(x+h)-tan(x)upon h {tanA+B=1+tanAtanB upon tanA-tanB} =1+tanxtanh-tanx upontanx-tanh. h =

Raunak Chauhan 3 years, 11 months ago

Sec²x
  • 3 answers

Anurag Singh 3 years, 11 months ago

CLGQKBC

Sneha Goel 3 years, 11 months ago

CLGQKBC" The pattern followed is - I - skip one letter then the next one is used(K) N - the previous letter is used(M) D -skip one letter then the next one is used(F) I -the previous letter is used(H) A -skip one letter then the next one is used(C) Hope it helps you...

Tanisha Chawla 4 years ago

Hello I think the answer would be - "CLGQKBC" The pattern followed is - I - skip one letter then the next one is used(K) N - the previous letter is used(M) D -skip one letter then the next one is used(F) I -the previous letter is used(H) A -skip one letter then the next one is used(C)
  • 1 answers

Shailendra Dhukia 3 years, 11 months ago

d/dx(sinx)=cosx d/dx(cosx)=-sinx d/dx(tanx)= sec••x ••= square
  • 2 answers

Neelesh Shukla 4 years ago

This answer is wrong

Lucky Sharma 4 years ago

Tan60°
  • 1 answers

Shailendra Dhukia 3 years, 11 months ago

d/dx(secx )= secx.tanx
  • 1 answers
Answer: 5 degree 37 minute 30 second=0.098 Radians.
  • 1 answers

Jagdish 3 years, 11 months ago

Add 20/7 to 3 to get first number then add it again to obtain next number Common difference = (last term - first term)/no. of numbers to be inserted +1 = (23-3)/6+1 = 20/7
  • 3 answers

Nishant Dhakare 4 years ago

Yes not for exams also for you intelligence improvement , in miscellaneous exercises question level os very suprb

Vivek Pandey 4 years ago

Yes v imp

Sameer Tanwar 4 years ago

Yes miscellaneous example and exercise both are important as it covers nice questions which improves your understanding the topic
  • 0 answers
  • 1 answers

Sameer Tanwar 4 years ago

We know that i) sin 2A = 2sinAcosA ii) sin(90-A)= cosA iii) sin30=1/2 LHS = sin 10 sin30 sin50sin 70 =sin 10 × 1/2 × sin (90-40)× sin(90-20) =1/2[sin 10 ×cos 40×cos 20] We multiply with[2 cos 10/2cos 10] =(1/2×1/2cos10)[2sin10cos10×cos20×cos40] =(1/4cos10)[sin20cos20×cos40] =(1/8cos10)[2sin20cos20×cos40] =(1/8cos10)[sin40cos40] =(1/16cos10)[2sin40co40] =(1/16cos10)×sin80 =(1/16cos10)×sin(90-10) =(1/16cos10)×cos10 =1/16 = RHS

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