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

11th Term =43 ...,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,............. =› a+10d = 43, ......... ..................................... =› a=43-10d,.......(i),.. ........,,.............................. 17th term = 5+ twice the 8th term, ............... a+16d = 5 + 2(a+7d) ,............,....................... 2d = 5 + a, now putting the value of a form (i) 2d = 5 + 43 -10d, .....,........................................ d = 4, a = 3

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Ak Y 7 years, 3 months ago

Q is right . You have to find an

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Myra Hi 7 years, 3 months ago

I think that it is wrong q . Check the q again . Maybe they v asked to find A OR D

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ANSWER

Prove that sec^A+cosec^A will be never less than 2

Posted by Disha Patidar 7 years, 3 months ago

CBSE > Class 10 > Mathematics

3 answers

Disha Patidar 6 years, 11 months ago

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Nishant Choudhary 7 years, 3 months ago

Please re write the question

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Ishu Sharma 7 years, 3 months ago

Don't know

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(100-100÷100-100)= 2 . This is a true statement . But I'm unable to prove it . Please solve it for me.... ...............

Posted by Anuj Karan 7 years, 3 months ago

CBSE > Class 10 > Mathematics

4 answers

Anuj Karan 7 years, 3 months ago

This is write questions and I have solved it below

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Ishu Sharma 7 years, 3 months ago

It is wrong question

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Gunjan Girdhar 7 years, 3 months ago

Listen......agar hm isko solve kre to isme divide me aa jaega 0 upon 0 Hme pta h 0 ko kissi se multiply ya divide kre to zero hi aaega To agar hm denominator wale 0 se upper wale zero ko 2 pe kaate to 0 aaega But quotient hmara 2 ho jaega So l.h.s become equal to r.h.s

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Anuj Karan 7 years, 3 months ago

(10)^2 -(10)^2 ÷ 10(10-10) =2 A^2 - B^2= (a+b)(a-b) (10+10)(10-10)÷10(10-10) Cut (10-10). by (10-10) = 10+10÷10 20÷10 =2 Hence proved .. AKS

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The product of2no is 70 find the no

Posted by Karanveer Singh 7 years, 3 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Is ,sin(180-30)=sin30

Posted by Anuj Karan 7 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Anuj Karan 7 years, 3 months ago

Friends this statement is completely true.. if we write the table of trigonometry from angle 0 to 360 as we have learnt from 0-90 so we will find its answer . You will learn that table in next standard. Hope i solved it . And also give a correct reason. Ur aks

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Anjali Singh 7 years, 3 months ago

No, if we try it by changing both of them to cos. It will be like this: Sin(180-30)= Sin30 Cos{90-(180-30)}= Cos (90-30) Cos{90-150}= Cos 60 Cos (-60) = Cos 60 Kindly note the angles are different one is negative and other is positive , negative angles don't exist thus the angle is not real .

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9573÷474

Posted by Sneha Choudhary 7 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 7 years, 3 months ago

20.196

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In a class of 48 students , the number of regular students is more than the number of irregular students . Had two irregular students been regular , the product of the number of two types of students would have been 380. (a) Find the number of each type of students in the class. (b) suggest on more value, other than regularity, that a student must possess.

Posted by Sakshi Joshi 7 years, 3 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

CosA-sinA+1/cosA+sinA-1=1/cosecA-cotA

Posted by Reshmi Rao 7 years, 3 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

I have many difficulties in mathematics and i get confused when i tries a question. So, please suggest me the better way to solve any answer

Posted by Guneet Pal Singh 7 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Raunak _ Pandey ?? 7 years, 3 months ago

And try solving previous year question papers , within a given time limit just like u do it in the examination hall Like this you will get some practice before entering the examination hall May be this may look funny but trust me it helps a lot

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Mahanta Bhagat 7 years, 3 months ago

Practice more

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What is the meining of mayank name

Posted by Nitin Mehra 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Sumit Kumar 7 years, 3 months ago

R u gone made

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ANSWER

Prove that tan A /1+ sec A - tan A / 1- sec A = 2cosec A

Posted by Priya Jain 7 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sumit Kumar 7 years, 3 months ago

Ur question is wrong Sec A- k jagha + aayaga miss

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Cos.tita.sin.thita

Posted by Tanu Dhiman 7 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Taiyab Hussain. 7 years, 3 months ago

1/secthita.cosecthita

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ANSWER

0=2f-40÷44+f

Posted by Harmanjot Singh 7 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Urvashi Gupta 7 years, 3 months ago

F=20

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ANSWER

Find the zeros of the quadratic polynomial 4 x^2- 4×-3 and verify the relation between the zeros and its coefficient

Posted by Aman Pundir 7 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 7 years, 3 months ago

p(x) = 4x² - 4x - 3

⇒ 4x² - 6x + 2x - 3

⇒ 2x( 2x -3) + 1(2x + 3)

⇒ (2x + 1) (2x -3)

p(x) = 0

(2x + 1) = 0 or (2x-3) = 0

x = -1/2 or x = 3/2

Hence, -1/2 and 3/2 are the zeroes of p(x).

Sum of zeroes = -1/2 + 3/2 = 2/2 = 1 = Coefficient of x / Coefficient of x²

Product of zeroes = (-1/2)(3/2) = -3/4 = Constant term/ Coefficient of x²

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Prove that line segment between point of contact of two parallel tangents is a dimeter of the circle.

Posted by Deepak Jha 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let PQ and RS be two parallel tangents of a circle where centre is O. Let their points of contact be A and B respectively. Join OA and OB.

Draw OC {tex}\parallel{/tex} PQ or RS
Now PQ {tex}\parallel{/tex} OC and transversal AO intersects them
{tex}\therefore{/tex} {tex}\angle{/tex}PAO + {tex}\angle{/tex}COA = 180^o [{tex}\because{/tex} The sum of consecutive interior angles on the same side of the transverse is 180^o]
⇒ 90°+ {tex}\angle{/tex}COA = 180^o [{tex}\because{/tex} PA is a tangent and OA is the radius through the point of contact]
{tex}\therefore{/tex} {tex}\angle{/tex}PAO = 90^o [ the tangent of any point of a circle is perpendicular to the radius through the property contact]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}COA = 90^o
Similarly,
{tex}\angle{/tex}COB = 90^o
{tex}\therefore{/tex} {tex}\angle{/tex}COA + {tex}\angle{/tex}COB = 180^o
{tex}\Rightarrow{/tex} AB is a straight line through O.