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the find value of a+bω+cωSqure/aωsqure+b+cω

Posted by Jitesh Kumar Kumar 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Prove sinπ\6cosπ\6=√3/4

Posted by Rahul Agrawal 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Chapter 3 supplementary material exercise solution

Posted by Praful Tiwari 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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MCQ in all chapter

Posted by Jaspal Raju 6 years, 4 months ago

CBSE > Class 11 > Mathematics

1 answers

Priyanshu Bharadwaj 6 years, 4 months ago

Go through the book of RD Sharma

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Hii Anchal . Solve this- |x+1| +|x|>3

Posted by Manan Aggarwal?? 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Iram Hasan Hasan 6 years, 4 months ago

x<2,x>1

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bc cos A + ca cos B + ab cos C = 1/2(a^2+b^ 2+c^ 2

Posted by Pk Ma 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Solve cos4x

Posted by Kuldeep Chauhan 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Aryan Thakur 6 years, 4 months ago

Cos2(2x) cos 2x

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Convert 15degree 15min 30sec into radian measure

Posted by Gaurav Rathi 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Ayush Vishwakarma?? 6 years, 4 months ago

168/1800 radian

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4 cos theta - 3 sin theta is equal to 4

Posted by Raghwendra Kumar 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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y=x^2+2x-11 / 2(x-3)

Posted by Parul Singh 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Sin(B-C)/Sin(B+C)=b×b-c×c/a×a

Posted by Arti Shah 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Aahwin bought a television for rs.9400.he paid a gst of 18 % for the television. What is the gst and how much is his actual cost price.

Posted by Rekha Jain 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Ankush Upadhyay 6 years, 4 months ago

It's depend on jow much the saler is gaining profit

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किसी समांतर श्रेणी में 60 पद है । इस श्रेणी के प्रथम तथा अंतिम पद क्रमशः 8 और 185 हैं । सार्वांतर ज्ञात कीजिए ।

Posted by Sunder Lal Gota 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Tan3xtan2xtanx=tan3x-tan2x-tanx show that

Posted by Partha Jyoti Dutta 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Sia ? 6 years, 4 months ago

From 3x = 2x + x
Applying tan on both sides..
tan3x = tan (2x+x)
tan3x = (tan2x+tanx)/1-tan2x tanx
(1-tan2x tanx)tan3x = tan2x+tanx
tan3x - tanx tan2x tan3x = tan2x+tanx
tanx tan2x tan3x = tan3x-tan2x-tanx

Hence it is proved.

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x/4<(5x-2)/3-(7x-3)/5

Posted by Lovelykps 1112 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Evaluate : i + 1/i

Posted by Urgain Lhaksam 6 years, 4 months ago

CBSE > Class 11 > Mathematics

7 answers

Kumar Saksham 6 years, 4 months ago

i +1/i = i + i^4/i = i+ i^3 = i -i =0

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Ayush Vishwakarma?? 6 years, 4 months ago

Sahi hai na...

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Urgain Lhaksam 6 years, 4 months ago

Thanks broooo

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Ayush Vishwakarma?? 6 years, 4 months ago

Ni toh 2i/-1 hoga..

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Ayush Vishwakarma?? 6 years, 4 months ago

Are yaar jzb hai tum L.c.m lo

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Urgain Lhaksam 6 years, 4 months ago

How it become 0

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Ayush Vishwakarma?? 6 years, 4 months ago

Zero hoga...

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What is the definition of real number (R)?

Posted by Khushi Hamber 6 years, 4 months ago

CBSE > Class 11 > Mathematics

1 answers

Ayush Vishwakarma?? 6 years, 4 months ago

Those no. Which start from positive natural no..

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If a cone of height 'h' is plced its tip upside and filled with water then there is a space of 8 cm from top to surface of water. Now the cone is inverted and now the space between surface of water and cone's bottom is 2 cm. Find the height of cone .(answer is not 10 cm)

Posted by Manan Aggarwal?? 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Sin 18 find valu

Posted by Sudhir Kumar 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Plss ch3 kae sare important formulae bhej do bhai / bahean...

Posted by Priyanshu Rawat 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Shan? .Va 6 years, 4 months ago

All the formulas in the summary r important. Well mostly v use Tan(x+y) Tan(x-y) Cot(x-y),(x+y) Cos(x+y), (x-y) same with sin And the last 5 formulas r important for the ladt exercise U take a look of cosx+ cosy types also

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Convert the complex numberz=i-1/cosπ/3+isinπ/3 into polar form?

Posted by Yashvi Saxena 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Convert the complex number z=i-1/cosπ/3+isinπ/3

Posted by Yashvi Saxena 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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if the sum of 10 terms of an A.P is 50 and sum of 50 terms is 10. Find the sum of its 60 terms.

Posted by Ankit Prasad 6 years, 4 months ago

CBSE > Class 11 > Mathematics

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Find the value of cos²(2Aplha) where, sin alpha + cos alpha =P

Posted by Tejveer Singh 6 years, 4 months ago

CBSE > Class 11 > Mathematics

1 answers

Kunal J 6 years, 4 months ago

2P + Psin2alpha

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Why cos(180+0) =cos0 is not change in sin0

Posted by Himanshu Parjapati 6 years, 4 months ago

CBSE > Class 11 > Mathematics

3 answers

Kumar Saksham 6 years, 4 months ago

and belongs to no change category

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Kumar Saksham 6 years, 4 months ago

since 180 is an even multiple of 90

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Vasu Aggarwal 6 years, 4 months ago

Because the angle is not changing at y axis for example cos(90+0)=sin0

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Sinx+cosx=1

Posted by Gourav Sah 6 years, 4 months ago

CBSE > Class 11 > Mathematics

1 answers

Kunal J 6 years, 4 months ago

x = 0 or π/2, For general 2nπ + π/2 or 0

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Find the square roots of the following: 1. 1-i 2. -i 3. i 4. 1+i

Posted by Vasu Chaudhary 6 years, 4 months ago

CBSE > Class 11 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

1. Let {tex}x + yi = \sqrt {1 - i} {/tex}
Squaring both sides, we get
x² - y² + 2xyi = 1 - i
Equating the real and imaginary parts
x² - y² = 1 and 2xy=-1... . (i)
{tex}\therefore \;xy = \frac{{ - 1}}{2}{/tex}
Using the identity
(x² + y²)² = (x² - y²)² + 4x²y²
{tex} = {\left( 1 \right)^2} + 44{\left( { - \frac{1}{2}} \right)^2}{/tex}
= 1 + 1
= 2
{tex}\therefore \;{x^2} + {y^2} = \sqrt 2 {/tex} . . . . (ii) [Neglecting (-) sign as x² + y² > 0]
Solving (i) and (ii) we get
{tex}{x^2} = \frac{{\sqrt 2 + 1}}{2}{/tex} and {tex}y = \frac{{\sqrt 2 - 1}}{2}{/tex}
{tex}\therefore x = \pm \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} and {tex}y = \pm \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}
Since the sign of xy is negative.
{tex}\therefore {/tex} if {tex}x = \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} then {tex}y = - \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}
and if {tex}x = - \sqrt {\frac{{\sqrt 2 + 1}}{2}} {/tex} then {tex}y = \sqrt {\frac{{\sqrt 2 - 1}}{2}} {/tex}
{tex}\therefore \sqrt {1 - i} = \pm \left( {\sqrt {\frac{{\sqrt 2 + 1}}{2}} - \sqrt {\frac{{\sqrt 2 - 1}}{2}} i} \right){/tex}