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  • 2 answers

Sadhu Meher 5 years, 9 months ago

ii)tan13A-tan9A-tan4A=tan13A tan9A tan4A L.h.s Tan13A=tan(9A+4A) Tan13A=tan9A+tan4A/1-tan9A .tan4A Tan13A(1-tan9A tan4A)=tan9A+ tan4A Tan13A-tan13Atan9 Atan4A=tan9A+ tan4A Tan13A- tan9 A-tan4A=tan13 Atan9A tan4A L.h .s= R.h.s (Prove)

Techno Burst 5 years, 9 months ago

Hhhhhhj
  • 0 answers
  • 3 answers

Piyush Giri Goswami 5 years, 9 months ago

You have to be with the politics until you grown up and become capable of making ur own decisions

Piyush Giri Goswami 5 years, 9 months ago

Wrong.but my friend this is politics

Ilu Rathore 5 years, 9 months ago

Yes
  • 1 answers

Kartik Gandhi 5 years, 9 months ago

X should be strictly less than 1 .....this would give u domain in interval form
  • 0 answers
Jj
  • 1 answers

K Nayati 5 years, 9 months ago

Gfdgfdsf
  • 3 answers

Pratiksha Pandey 5 years, 9 months ago

Sin x =cos x

Khushreet Aulakh 5 years, 9 months ago

f(x) = sinx Then f(x)'= cosx

Roshan Singh 5 years, 9 months ago

Answer
  • 2 answers

Bhushan Verma 5 years, 9 months ago

Me to check kar raha tha ? answer aate hai ya nahi

Ekta E 5 years, 9 months ago

Yeh 11 ki maths mein kha hai idiot
  • 1 answers

Gaurav Seth 5 years, 9 months ago

Step-by-step explanation:

Let the set of students who offered mathematics, statistics and physics  be M, S and P respectively. Let a be number of students who offered all 3 subjects.  So we get

40 – a  +  a + 20 – a  + 8 = 65

So 68 – a = 65

 A = 3

Now students who offered Mathematics  

 15 + 10 – a + a + 40 – a

 65 – a

65 – 3 = 62

Now students who offered Statistics

12 + 10 – a + a + 20 – a

42 – a  

42 – 3 = 39

Now students who offered any of three subjects are

15 – 12 + 8 + 10 – a + 40 – a + 20 – a + a

105 – 2 a  

105 – 2 x 3

105 – 6

= 99

So number of students who did not offer any subjects were 100 – 99 = 1

  • 1 answers

Sid Siddharth 5 years, 9 months ago

If a=2+√3 Then a-1/a=(2+√3)-1/(2+√3) =(2+√3)^2-1/2+√3 =6+4√3/2+√3 Now By rationalysation methode = -12+2√3
  • 2 answers

Sumit Maurya 5 years, 9 months ago

1/2

Omanshi Singh 5 years, 9 months ago

0.5
  • 3 answers

Anshika Agrawal 5 years, 10 months ago

cos^2x - sin2x

Sid Siddharth 5 years, 10 months ago

2cos^2x-1

Nirmalya Madhu 5 years, 10 months ago

1-2sin^2x
  • 1 answers

Sid Siddharth 5 years, 10 months ago

X=2nπ+-π/2 X=nπ+(-1)^n.π
  • 6 answers

Prince Raj 5 years, 9 months ago

For this answer you have to go in class LKG and read .

Ekta E 5 years, 10 months ago

10

Ramanamma Ch 5 years, 10 months ago

10

Anant Dongre 5 years, 10 months ago

100

Robinpreet Kaur 5 years, 10 months ago

10

Shivam Sharma 5 years, 10 months ago

20
  • 0 answers
  • 0 answers
  • 2 answers

Ekta E 5 years, 9 months ago

Union or intersection?

Ekta E 5 years, 10 months ago

Toh question toh batao kya chahiye
  • 2 answers

Sid Siddharth 5 years, 10 months ago

Let A=[a,b,c] B=[d,e,f] Then A-B=0 And also A'=[d,e,f] B'=[a,b,c] Then now A'-B'=0 Hence proves

Md Sheikh 5 years, 10 months ago

Yes
  • 1 answers

Ekta E 5 years, 10 months ago

By applying formula f'(x) =f(x+h)-f(x) /h
  • 1 answers

Sid Siddharth 5 years, 10 months ago

nπ-π/6,nbelongs to z
  • 1 answers

Sid Siddharth 5 years, 10 months ago

Putting the calues os cosA,cosBAns cosC also the u can find ans
  • 1 answers

Sid Siddharth 5 years, 10 months ago

Cosx.e^sinx
  • 4 answers

Ekta E 5 years, 10 months ago

-cosec square

Dhruv Yadav 5 years, 10 months ago

-cosec2^x

Vivek Lather 5 years, 10 months ago

-cosec^2x

Sandeep Singh 5 years, 10 months ago

may be -cosec2 square
  • 1 answers

Divanshu Rawat 5 years, 10 months ago

(A union B)'= A' intersection B' (A intersection B )'= A' union B'

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