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Tan A + cot A = 7 => (tan A + cotA)^2 = 49 (squaring on both sides) => tan^2A + cot^2A + 2(tanA × cotA) = 49........(1) . Now, as we know that, tanA× cotA= 1 . So, put the value of tanA × cotA in (1), we get, tan^2A + cot^2A + 2 = 49 => tan^2A + cot^2A = 47.

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Sahil ???? 6 years ago

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Ramkali required rupees 2500 after 12 weeks to send her daughter to school she saved rupees hundred in the first week and increase of reclusive in rupees 20 every week find whether she will be able to send her daughter to school after 12 weeks or not

Posted by Yudhveer Soni 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years ago

First saving = 100
second saving =120
third saving = 140
clearly this is an A.P. series with a=100,
d=20
we are to finf the sum of 12 terms of this series
so,
sum =( n/2)*(2*a+(n-1)*d)
sum=(12/2)*(2*100+(12-1)*20)
sum=6*(200+11*20)
sum=6*(200+220)
sum=6*420
sum=2520
so she would be able to save Rs. 2520 in 12 months and therefore will be able to send her daughter to school

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Find the sum of 1-6+2-7+3-8+.......100term

Posted by Nidhi Gupta 6 years ago

CBSE > Class 10 > Mathematics

3 answers

Abhishek Kumar 6 years ago

-500

2Thank You

Aadya Singh ? 6 years ago

-250

2Thank You

Sahil ???? 6 years ago

-250

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For which value of a and b does not following pair of linear equations have an infinite many number of solutions? 2x+3y=7. (a-b)x+(a+b)y=3a+b-2

Posted by Narendra Chouhan 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Narendra Chouhan 6 years ago

Please give me solution

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Seema can embroider 1/5 of a kurta in 4 days .How long will it take her to embroider it completely

Posted by Yash Raj 6 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If the sum of the squares of zeroes of the quadratic polynomial fx=x^2-8x+k is 40. Find the value of k

Posted by Ganga Sharma 6 years ago

CBSE > Class 10 > Mathematics

1 answers

A.T'S Gaming 6 years ago

Tu ek baar apne aap try karo

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Lesson 5 are not understanding

Posted by Jigar Panda 6 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

In a triangle ABC right angled at B . Side BC is trisected at a point D and E Prove that: 8AE^2 = 3AC^2 +5AD^2

Posted by Bedika Punshi 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Aditi .D 6 years ago

ABC is a triangle right angled at B, and D and E are points of trisection of BC. Let BD = DE = EC = x Then BE = 2x and BC = 3x In Δ ABD, AD² = AB² + BD² AD² = AB² + x² In Δ ABE, AE² = AB² + BE² AE² = AB² + (2x)² AE² = AB² + 4x² In Δ ABC, AC² = AB² + BC² AC² = AB + (3x)² AC² = AB² + 9x² Now, 3AC² + 5AD² = 3(AB² + 9x²) + 5(AB² + x²) 8AB² + 32x² 8(AB² + 4x²) = 8AE² ⇒ 8AE² = 3AC² + 5AD² Hence proved.

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Math ka 4chapter ma discrimint kaise yad kra

Posted by Akanksha Rani 6 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Find two consecutive positive integers sum of whose squares is 365

Posted by Rajbir Singh 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Aditi .D 6 years ago

let the no. be x and x+1 then we have x²+(x+1)²=365 2x²+2x-364=0 x²+x-182=0 x²+14x-13x-182=0 (x+14)(x-13)=0 That gives x=13 Avoiding negative value one number is 13 and other one 14.

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How i can get full marks in math exam also in board exam

Posted by Vaibhav Tiwari 6 years ago

CBSE > Class 10 > Mathematics

2 answers

Aditi .D 6 years ago

First solve all the ncert qurstion and when u think that particular chapter' s all the textual question are done the go only go for gides and value based question after that u can set timer and solve question so it will be easy to complete the paper and perfectly

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Payal.. ??Singh 6 years ago

Only ek hi step hhh practice pratice pratice.... Mera bi yhi haal tha but maths ko apni life ka part banao iski practice krna skip n kro.. Hmesha 2hours do uss... Your maths will be perfect...

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Ncert ch 1 example 2

Posted by Suneha Tanwar 6 years ago

CBSE > Class 10 > Mathematics

0 answers

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The largest number which divides 70 and 125 leaving remainder 5 and 8 respectively is

Posted by Jayveer Rangi 6 years ago

CBSE > Class 10 > Mathematics

4 answers

Alex M 6 years ago

The required number divideds 70 - 5 =65 and 125 - 8 = 117. Hence, Required number is the H.C.F of (65 and 117 ). 65 = 5 × 13 117= 3 × 3 × 13 H.C.F = 13 Therefore, The largest number which divideds 70 and 125 is 13

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Jayveer Rangi 6 years ago

1750

1Thank You

Hardeep Singh Bhatti 6 years ago

1Thank You

⛩Singh Empire ???? 6 years ago

2Thank You

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Ch 12 vilum

Posted by Mansi Singh Roy 6 years ago

CBSE > Class 10 > Mathematics

0 answers

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Probability of event lie b/w

Posted by Nitin Tomar 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Shubham Kumar 6 years ago

Number of events/total number of outcomes

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If one zero of the polynomial is 2-root5 and product is 1 form a quadratic polynomial

Posted by Disha Garg 6 years ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If sina+sin^2a=1 so prove that cos^2a+cos^4=1

Posted by Radheyshyam Dahal 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Navya ????‍?❤ 6 years ago

SinA + sin (sq) A =1 SinA =1- sin(sq)A _(1) By identity Sin(sq)A+cos(sq) A =1 1-sin(sq) A =cos(sq)A On putting this value in _(1) Sin A=cos(sq)A On squaring Sin(sq) A= cos (raised to4) A _(2) We have to find Cos(sq) A + sin (sq) A from _(2) =1 by identity

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Hcf of 168 and 126

Posted by Rajat Singh 6 years ago

CBSE > Class 10 > Mathematics