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  • 1 answers

Monika Phalswal 6 years, 3 months ago

This question is same as example 21 of ncert
  • 1 answers

Nishant Bansal 6 years, 3 months ago

1+7i/(4-1-4i) =) 1+7i/3-4i Rationalizing =) (1+7i)*(3+4i)/9+16 =) 3-28+25i/25 =) 25i-25/25 =) i-1 Answer
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  • 1 answers

Pandat Viraj Bhardwaj 6 years, 3 months ago

Sin10.sin30.sin50.sin.70=1/16 (Taking the value of sin30°= 1/2) also sin50°= sin60-10& sin70°= sin60+10(it can be written as this) 1/2 [ sin(60-10)sin(60+10)sin10° 1/2 [sin²60- Sin²10]sin10° 1/2[3/4 - sin²10]sin10°..( sin60=√3/2)&(√3/2)²= 3/4) Now 1/8[ 3sin10°- 4sin³] by using value of sin 3X 1/8 [sin3X] x= 10 1/8[sin3× 10] 1/8[ sin30°] ie. 1/8×1/2= 1/16 Hence proved... #PanDat_Golu❤
2c8
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Nishant Bansal 6 years, 3 months ago

Not possible as a equation of power 2 cannot have value of r as 8
  • 2 answers

Raj Sahu 6 years, 3 months ago

prove ya solve

Raj Sahu 6 years, 3 months ago

Bhai carna kya hai
  • 1 answers

Ritika Goyal 6 years, 3 months ago

It can be written in the form (100-4)whole cube and then it can be solved using binomial theorem
  • 1 answers

Ashish Mishra 6 years, 3 months ago

Its really simple. Let x+2 to rhs and then solve it
  • 2 answers

Mohd Imtiyaz 6 years, 3 months ago

A union B={1,2,3,6,8,9}

Disha Verma 6 years, 3 months ago

A union B = {1,2,3,6,8,9}
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  • 1 answers

Gaurav Seth 6 years, 3 months ago

CosAcos2Acos4Acos8A
=1/2sinA[(2sinAcosA)cos2Acos4Acos8A]
=1/2sinA(sin2Acos2Acos4Acos8A)
=1/4sinA[(2sin2Acos2A)cos4Acos8A]
=1/4sinA(sin4Acos4Acos8A)
=1/8sinA[(2sin4Acos4A)cos8A]
=1/8sinA(sin8Acos8A)
=1/16sinA(2sin8Acos8A)
=1/16sinA(sin16A)
=sin16A/16sinA (Proved)

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Aryan Gupta 6 years, 3 months ago

Because Sin90°= 1 And Cos90°=0 And the relation between Tan a°= sin a°/ cos a° So, Tan90°= sin90°/ cos 90° = 1/0 = infinite
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