Given, sec {tex}\theta{/tex} = x + {tex}\frac{1}{4 x}{/tex}
We know that, tan {tex}\theta{/tex} = {tex}\pm \sqrt{\sec ^{2} \theta-1}{/tex}
{tex}=\pm \sqrt{\left(x+\frac{1}{4 x}\right)^{2}-1}{/tex}
{tex}=\pm \sqrt{x^2 + \frac{1}{(4x)^2}+2.x. \frac{1}{(4x)}-1}{/tex}
{tex}=\pm \sqrt{x^2 + \frac{1}{(4x)^2}+ \frac{1}{(2)}-1}{/tex}
{tex}=\pm \sqrt{x^2 + \frac{1}{(4x)^2}- \frac{1}{(2)}}{/tex}
{tex}=\pm \sqrt {\left(x-\frac{1}{4 x}\right)^2}{/tex}
{tex}=\pm\left(x-\frac{1}{4 x}\right){/tex}
Now, sec {tex}\theta{/tex} + tan {tex}\theta{/tex} = {tex}\left(x+\frac{1}{4 x}\right) \pm\left(x-\frac{1}{4 x}\right){/tex}
i) sec {tex}\theta{/tex} + tan {tex}\theta{/tex} = {tex}\left(x+\frac{1}{4 x}\right) +\left(x-\frac{1}{4 x}\right){/tex}
{tex}=2x{/tex}
ii) sec {tex}\theta{/tex} + tan {tex}\theta{/tex} = {tex}\left(x+\frac{1}{4 x}\right) -\left(x-\frac{1}{4 x}\right){/tex}
{tex}=\frac{1}{4x}-(-\frac{1}{4x}) \\=\frac{1}{4x}+\frac{1}{4x}\\=\frac{2}{4x}{/tex}
{tex}=\frac{1}{2 x}{/tex}

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For some a and b ,HCF of 55 and 210 is 210a + 55b ,then find the value of a and b

Posted by Vini Negi 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Sudhanshu Yadav 6 years, 10 months ago

Here hcf of 55 and 210 is 5 then by euclid division process _210=55×3+45_a 55=45×1+10_b 45=10×4+5_c 10=5×2+0_d Taking (c) steps as 5=45-10×4 5=45-(55-45×1)×4 5=45×5-55×4 5=(210-55×3)×5-55×4 5=(210×5)+{55×(-19)} Thus,a=5and b=-19

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Yogita Ingle 6 years, 10 months ago

HCF of 210 & 55
210 = 55× 3 + 45 ….....(i)
55 = 45 × 1 +10 ….........(ii)
45 = 10 ×4 +5 …...........(iii)
10 = 5 ×2 + 0
hence HCF of 210 & 55 = 5
now from (iii), we get
45 = 10 ×4 + 5
so 5 = 45 – 10×4
5 = 45 – (55 – 45)×4
5 = 45 – 55×4 + 45×4
5 = 45 ×5 – 55×4
5 = (210 – 55×3) ×5 – 55×4
5 = 210×5 – 55×15 – 55×4
5 = 210×5 – 55×19
5 = 210 x + 55 y
where x = 5, y = –19

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ANSWER

what is phythagoras theorem? how can it proved? plz answer this question

Posted by Jatin Arora 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Nehadris Bose 6 years, 10 months ago

P^2=H^2+B^2or H^2=P^2+B^2 OR B^2=P^2+H^2

1Thank You

Shivam Sharma 6 years, 10 months ago

Take a triangle ABC right angeled at B . Draw BD perpendicular to AC. 1.prove triangle ABC congruent to triangle BDC. 2.By CPCT take BC/AC=BD/BC to get BC^2=BD.AC and mark it as 1. 3.similarly prove triangle ABC congruent to triangle BDA to get AB/AD=AC/AB to get AB^2=AD.AC mark it as 2. 4. Now add 1 and 2 to get AB^2+BC^2=BD.AC + AD.AC =AC (BD+AD) =AC(AC) =AC^2 Hence proved

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Shivam Sharma 6 years, 10 months ago

According to pythagoras theorem ,In a right angled triangle the square of hypotenuse is equal to sum of square of its prependicular and base . i.e. h^2=p^2+b^2

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ANSWER

2cosA-2secA

Posted by Aryan Rawat 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Nikita Sharma 6 years, 10 months ago

2 cosA-2/cosA. (2cos²A -2)/cosA. [2(cos²A-1)]/cosA. [2sin²A]/cosA. 2sinA*tanA.

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ANSWER

A man employed to count 10710rupees. He counts at the rate of 180 rupees per minute for half an hour . After this he counts at the rate of 3 rupees less every minute than the preceding minute . Find the time taken by him to count the entire amount

Posted by Riya Raj 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Sec+ tan=p then the value of cosec

Posted by Vishwash Jain 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

Secθ+tanθ=p ----------------------(1)
∵, sec²θ-tan²θ=1
or, (secθ+tanθ)(secθ-tanθ)=1
or, secθ-tanθ=1/p ----------------(2)
Adding (1) and (2) we get,
2secθ=p+1/p
or, secθ=(p²+1)/2p
∴, cosθ=1/secθ=2p/(p²+1)
∴, sinθ=√(1-cos²θ)
=√[1-{2p/(p²+1)}²]
=√[1-4p²/(p²+1)²]
=√[{(p²+1)²-4p²}/(p²+1)²]
=√[(p⁴+2p²+1-4p²)/(p²+1)²]
=√(p⁴-2p²+1)/(p²+1)
=√(p²-1)²/(p²+1)
=(p²-1)/(p²+1)
∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1)

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ANSWER

To prove sin2A+cos2A=1

Posted by Vitthal Gajjal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 10 months ago

How to prove here its too long, plz check it on good, hope yu can understand ?

3Thank You

ANSWER

Is there any chapter wise blueprint

Posted by Robin Rawat 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 10 months ago

Ya, yu can check it in Google...........

2Thank You

ANSWER

Prove that √2 id irrational

Posted by Ayush Singh 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Mansa Gaur 6 years, 10 months ago

Let √2 be a rational number So it is written in a form of p/q and a co prime number So: √2 =p/q √2/q=p And we see that √2 is a irrational number so √2/q is also a irrational number and it is equal to p which is an rational so our assumption is wrong it is a irrational number. I hope this Answer will help you ?

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An equilT

Posted by Yuvraj Bhati 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

A train takes 2 hour less for journey of 300 km if its speed is increased by 5 km/h from its usual speed . Find th usual speed of the train

Posted by Shaikh Ahmad 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Mohit Dhillon 6 years, 10 months ago

Let the usual speed of train be x km per hour So usual time is 300/x hours Spees changed Is x+5 kmph Time taken =300/x+5 hours ATQ 300/x - 2=300/x+5 After solving this eq we will get X=25

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ANSWER

Find the value of (-1)^n+(-1)^2n+(-1)^2n+1+(-1)4n+2 , where n its a positive odd integer

Posted by Last Bencher 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Last Bencher 6 years, 10 months ago

Pls give the steps

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Deepanshi Tayal 6 years, 10 months ago

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ANSWER

No. Not divisible by 8, at random between 1 and 100 . Find the probability

Posted by Sahil Kumar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Dr Devesh Parchwani 6 years, 10 months ago

Here n=98 Number divisible by 8 are(8,16....96) M=12 p(A)=m/n=12/98

3Thank You

ANSWER

1+1+1=3-1-1-1-1

Posted by Md Danish 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Aditya Khandelwal 6 years, 10 months ago

Answer is -1

1Thank You

Aditya Khandelwal 6 years, 10 months ago

Rakhi is right

2Thank You

Rakhi Pal 6 years, 10 months ago

What type of Question is this?

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ANSWER

1add1add1is equal to (3-1-1-1-1)under vinculum

Posted by Md Danish 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Summative Assessment_2_2014_2015 set 1 answersheet

Posted by Birjesh Goyal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Aditya Khandelwal 6 years, 10 months ago

?????

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ANSWER

Full mid point theorem

Posted by Tript Preet???? 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Priyanshi Mishra 6 years, 10 months ago

See ncerr

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ANSWER

A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm redpectively.find the capacity and surface area of the bucket . Find the cost of the milk required to completely fill the bucket , at the rate of Rs.25 per litre ( use π =3.14)

Posted by Riya Rao 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

capacity or volume of a frustum of a cone is given by

here we have R= 20cm r = 10 cm , h = 30 cm

putting all these values in the volume formula

Now the for the surface area calculate first the slant height l

using Pythagorean theorem as

now surface area is given as

and total surface area is

Now the capacity of the bucket is 22 ltr

so cost of 1 ltr milk is Rs 25 so

cost of 22 Ltr milk will be = 25×22 Rs.

= Rs. 550

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ANSWER

Find the area of minor segment of a circle of radius 14 cm, when its central angle is 60 degree. Also find the area of the corresponding major segment

Posted by Deepanshu Nawab 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Tript Preet???? 6 years, 10 months ago

84.87-102.66 =17.84cm2 (approximately)

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Rajneesh Kaurav 6 years, 10 months ago

Major 1540/3

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Gaurav Seth 6 years, 10 months ago

Radius of the circle, r = 14 cm

The chord AB subtends an angle 60^o at the centre of the circle.
Now,

Area of the minor segment = Area of the sector AOB - Area of ΔAOB

= 102.67 - 84.87
= 17.85 cm²

Now,
Area of the major segment = Area of the circle - Area of the minor segment

=616 - 17.8

= 598.2 cm²

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ANSWER

A thief runs with a uniform speed of 100 m/min. After 1 min. A policeman runs after the thief to catch him. He goes with a speed of 100 m/min. in the first minute and inreases speed by 10 m/min. every scceeding minute. After how many minutes the policeman will catch the thief.

Posted by Vipul Amrania 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Rajneesh Kaurav 6 years, 10 months ago

5 minutes

1Thank You

Gaurav Seth 6 years, 10 months ago

Let the police catch the thief in ‘n’ minutes
Since the thief ran 1 min before the police start running.
Time taken by the thief before he was caught = (n + 1) min
Distance travelled by the thief in (n+1) min = 100(n+1)m
Given speed of policeman increased by 10m per minute.
Speed of police in the 1st min = 100m/min
Speed of police in the 2nd min = 110m/min
Speed of police in the 3rd min = 120m/min
Hence 100, 110, 120… are in AP
Total distance travelled by the police in n minutes =(n/2)[2a+(n – 1)d]
= (n/2)[2 x 100 +(n – 1)10]
After the thief was caught by the police,
Distance traveled by the thief = distance travelled by the police
100(n+1)= (n/2)[2 x 100 +(n-1)10]
200(n + 1) = n[200 + 10n – 10]
200n + 200= 200n + n(10n – 10 )
200 = n(n – 1)10
n(n – 1) – 20 = 0
n^{2 }– n– 20 = 0
n^{2 }– 5n + 4n – 20 = 0
n(n – 5) + 4(n – 5) = 0
(n – 5) (n+4) = 0
(n – 5) = 0 or (n + 4) = 0
n= 5 or n= -4
Hence n= 5 since n cannot be negative
Therefore the time taken by the policeman to catch the thief = 5minutes

3Thank You

ANSWER

Prove that n raise to power 3 is divisible by 6 if n is any +ve integer.

Posted by Prabh Singh 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

n³- n = n (n²- 1) = n (n - 1) (n + 1)

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.

Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n (n-1) (n+1) = n³- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)

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Prove the volume of frustrum and its surface area too??

Posted by Abhishek Tyagi 6 years, 10 months ago

CBSE > Class 10 > Mathematics