CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

LHS : = CotA-1+cosecA/CotA+1-CosecA. (Divide the both the value by "SinA") = CotA+CosecA-(Cosec^2A-Cot^2A) / CotA-CosecA+1. (1=cosec^2A-cot^2A) = (CotA+CosecA)-(cosec^2A+cot^2A)/CotA-CosecA+1 = (CotA+CosecA)-(CosecA+CotA)(CosecA-CotA) = CotA+CosecA-(CotA-CosecA+1)/CotA-CosecA+1 = CotA+CosecA PROVED!!!! Its a question NCERT EX-8.4 Q5 (v) part...

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At a point on a level ground, the angle of elevation α of a vertical tower is found to be such that 12 5 tan  . On walking 192 m towards the tower, the angle of elevation becomes β such that 4 3 tan  . Find the height of the tower.

Posted by Priyanka Bisht0803 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Yash Preet 6 years, 10 months ago

But solution phone pe nhi paper pe milega

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Yash Preet 6 years, 10 months ago

180 m

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For what positive value of k does the quadratic equation 3 x square minus kx + 3 is equals to zero not have real roots

Posted by Aman Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Aman Sharma 6 years, 10 months ago

Thanks

1Thank You

Ram Kushwah 6 years, 10 months ago

3x^2-kx+3=0

a=3,b=-k,c=3

b²-4ac=(-k)²-4x(3)(3)

=k²-36

For not having real roots

b²-4ac<0

so k2-36<0

k²< 36

k < 6

or k > -6

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1/(cosecA--cotA) --1/sinA =1/sinA -- 1/(cosecA+cotA)

Posted by Deepu Singh 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Shubhi Tyagi 6 years, 10 months ago

For this question you have to change the location of LHS and RHS..... LHS: = 1/(CosecA-CotA) - 1/(cosecA+CotA) =CosecA+CotA+CosecA-CotA/Cosec^2A - Cot^2A =2cosecA RHS: =1/sinA + 1/sinA =2/sinA =2CosecA Proved!!!!

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Yuvraj Verma 5 years, 8 months ago

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What do you mean by incenter

Posted by Avi Tiwari 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Mehak Rathore 6 years, 10 months ago

Point where all angle bisectors meet within a triangle

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The ratio od sum of n terms of two A.P.s is (7n +1):(4n+27).Find the ratio of 'm'th terms.

Posted by Yash Preet 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 10 months ago

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)....(1)
Let’s consider the ratio these two AP’s m^th terms as a_m : a’_m →(2)
The n_th term of AP formula, a_n = a + (n – 1)d
Hence equation (2) becomes,
a_m : a’_m =[ a + (m – 1)d] :[ a’ + (m – 1)d’]
On multiplying by 2, we get
a_m : a’_m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]
= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]
= S_{2m – 1} : S_{’2m – 1}
= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]
= [14m – 7 +1] : [8m – 4 + 27]
= [14m – 6] : [8m + 23]
Thus the ratio of m^_th terms of two AP’s is {tex}\frac{14m-6}{8m+23}{/tex}

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Wetage of all chapter

Posted by Bhupendra Sarode 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

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After how many decimal places will the decimal expansion of 147 upon 120 terminate

Posted by Aanchal Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Anurag Gupta 6 years, 10 months ago

3 places of decimal 120=2*2*2*3*5 147*3/2*2*2*3*5 Now, 147/2*2*2*5 147*5*5/2*2*2*5*5*5 147*25/2³*5³ 3675/(2*5)³ 3675/1000 3.675

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Mayank ... 6 years, 10 months ago

Convert it into the form of 5^m ×2^n.. You will get the answer

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Cbse board the maths exam will very difficult or not

Posted by Priya Raju 6 years, 10 months ago

CBSE > Class 10 > Mathematics

4 answers

Aman Sharma 6 years, 10 months ago

Prepare well, do well and other leave on teachers on teachers, it will be really easy.

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Ayushi Jain 6 years, 10 months ago

Don't know...

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Mayank ... 6 years, 10 months ago

No...

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Aanchal Sharma 6 years, 10 months ago

Yes it will very difficult

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Can we use black pen in board exam please tell me anybody??????

Posted by Sammy'S K 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Ashwini Ramteke 6 years, 10 months ago

by cbse rule you should carry only Blue/Royal blue ball point/Gel and fountain pens only you can use

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Affu 😊 6 years, 10 months ago

No harsh ....only blue pen use in exam ,this is written in admit card

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Sammy'S K 6 years, 10 months ago

Please

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What does D means in trignometry

Posted by Wajahat Wani 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

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Find the mean

Posted by Ashwin Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Mayank ... 6 years, 10 months ago

Where is the question..??

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Abc is an iscoscles triangle right angled at c prove ab2 = 2ac2

Posted by Pranav Attre 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given: ABC is an isosceles triangle right angled at C.

To prove: AB² = 2AC²
Proof : ABC is an isosceles triangle right angle at C.
{tex}\therefore {/tex} AC = BC........(1)
and {tex}\angle{/tex}C = 90^o.......(2)
In view of (2) in {tex}\vartriangle {/tex} ABC,
AB²= AC²+ BC²........By Pythagoras theorem.
AC² + AC² ........Frrom(1)
= 2Ac²

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The number p and Q such that the equation px^2+3x+2q=0 has -6 as the sum of roots and as the product of roots, find the p and q

Posted by Anshit Gangwar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 10 months ago

what is the product of roots

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An army contigent of 16 member is to march behaind an army band of 32 member in a number of coloum in which they can march

Posted by Moon Sidana 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Aman Sharma 6 years, 10 months ago

This is the question of first exercise of 1st chapter in NCERT

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Prove that cotA+cosecA-1

Posted by Jyoti Singh 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If a cosA +b sinA=m and a sinA-b cosA=n prove that a^2+b^2=m2+n2

Posted by Devi Chandana 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 10 months ago

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Tommorow will be math exam

Posted by Data Ram Shankar Lal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Thanks Ki Baarish ........???? 6 years, 10 months ago

Yes

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Find a and d if a-3d , a-2d, a-d ,a+d, a+2d, a+3d are in Ap

Posted by Sonia Dua 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Aliya Qureshi 6 years, 10 months ago

Here, a = a - 3d.............. d = a - 2d - ( a - 3d)........... d = a - 2d - a + 3d........... d = 1d...

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If the circumference of a circle is 22 , then what is the area of a quadrant of a circle ?? Please show me the process of solving this problem.

Posted by Amar Konwar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Aliya Qureshi 6 years, 10 months ago

A quadrant is just one forth part of a circle...... Whenever we need to find area of a quadrant it is the same as that of the sector having theta 90°.......... According to question, Circumference = 22units...... 2 × 22/7 × r = 22........... r = 7/2units....... Area of quadrant = theta/ 360° × 22/7 × r^2........... = 90°/360° × 22/7 × 7/2 × 7/2......... = 77/8.......... = 9.62 units^2

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A solide

Posted by Kiran Mallannavar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gauri ❤ 6 years, 10 months ago

A solide ye kya hai

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aCosA + bSinA =m and aSinA - bCosA =n prove that a2+b2= m2+n2

Posted by Daksh Dubey 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Google User 6 years, 10 months ago

m2+n2=(acos+bcos)^2+(asin-bcos)^2 =a2cos2+a2sin2+2absincos+a2sin2+b2cos2 - 2absincos =a2(cos2+sin2)+b2(cos2+sin2) =a2(1)+b2(1) =a2+b2

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(5+x/5-x)-(5-x/5+x)=15/4

Posted by Nandini Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

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The taxi fare after each km when the fare is ruppee 15 for the first km and ruppee 8 for each additional km. find out it is an a.p or not

Posted by Mohammed Imthiyash 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Priya Darshni 6 years, 10 months ago

Yes it is an ap

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The angle of elevation of the top ofa vertical tower from a point on the ground is 60 degree.from another point 10m vertically above the first its angle of elevation is 30 degree find the height of the tower

Posted by Shubham Bhatt 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given,

Let the height of the tower be h cm.
Now, In {tex}\triangle{/tex}PAB,
{tex}\tan 60^{\circ}=\frac{A P}{A B}{/tex}
{tex}\Rightarrow \sqrt{3}=\frac{h}{A B}{/tex}
{tex}\Rightarrow A B=\frac{h}{\sqrt{3}}{/tex} .....(i)
And, In {tex}\triangle{/tex}PCD,
{tex}\tan 30^{\circ}=\frac{P D}{C D}{/tex}
{tex}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-10}{C D}{/tex}
{tex}\Rightarrow{/tex} CD = {tex}\sqrt{3}(h-10){/tex} ...(ii)
Since, AB = CD, so equation (ii) becomes,
{tex}A B=\sqrt{3}(h-10){/tex} ....(iii)
Equating equation (i) and (iii), we get,
{tex}\frac{h}{\sqrt{3}}=\sqrt{3}(h-10){/tex}
{tex}\Rightarrow h=3(h-10){/tex}
{tex}\Rightarrow{/tex} {tex}2h = 30 {/tex}
{tex}\Rightarrow{/tex} {tex}h = 15 cm{/tex}