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If sec theta + tan theta then find value of cossec theta

Posted by Hardik Panwar 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Bhavya Vardhan Jain 6 years, 9 months ago

Question is incomplete bro

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A bag contains 8 red balls and x blue balls. the odd against drawing a blue ball are 2:5.what is the value of x

Posted by Sahana Shree 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

#Miss_Raagini ???? 6 years, 9 months ago

Hey dude..... ur answer is here...!!! total no. of balls = 8 + x odd against = no. of red/ no. of blu balls 8/x 8/x = 2/5 2x = 40 x = 20 thus no. of blue balls is 20=x thanks....!

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Page no. 283 in ncert maths . Questions no. 16 Plz ans ..

Posted by Inayat Ali Ali 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 9 months ago

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If x+a is a factor of the polynomial x²+px+q and x²+mx+n prove that a=n-q/m-p

Posted by Tenzin Deki 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Since x + a is a factor of x²+ px + q
then (-a)²- pa + q = 0
{tex}\Rightarrow{/tex}a² = pa -q ..........(i)
also (x + a) is a factor of x²+ mx + n then we get
(-a )²- am + n = 0
{tex}\Rightarrow{/tex} a² - am + n= 0
{tex}\Rightarrow{/tex}a²= am - n........(ii)
From eq(i) and (ii), we get
am - n = ap - q
{tex}\Rightarrow{/tex}am - ap = n - q
Hence, a = {tex}\left[ \frac { n - q } { m - p } \right]{/tex}

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One zero of the polynomial 3xcube + 16xsquare + 15x - 18 is 2/3 find the other zero of the polynomial

Posted by Reena Goyal 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 9 months ago

Q. One zero of the polynomial 3x³ + 16x² + 15x - 18 is 2/3 find the other zero of the polynomial

Answer:

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a+b ka whole cube

Posted by Raaj Meena 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

#Miss_Raagini ???? 6 years, 9 months ago

a³+b³+3ab(a+b)

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Raaj Meena 5 years, 8 months ago

a cube plus b cube plus 3ab (a+b)

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Prove that cos2theta =sin2tgeta

Posted by Kushal Gandhi 6 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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Express sin 70°+85°in terms of trigonometric ratios of angles between 0°and 45°

Posted by Harsha Theja 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 9 months ago

sin(70+85)

=sin 155

=sin(180-25)

=sin (2x90-25)

=sin 25

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prove pythagorous

Posted by Jatin Yadav 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Gaurav Seth 6 years, 9 months ago

Pythagoras Theorem

Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

Given: A right triangle ABC right angled at B.

To prove: AC² = AB² + BC²

Construction: Draw BDAC

Proof :

We know that: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

ADBABC

So,(Sides are proportional)

Or, AD.AC = AB²... (1)

Also,BDCABC

So,

Or, CD. AC = BC²... (2)

Adding (1) and (2),

AD. AC + CD. AC = AB² + BC²

AC (AD + CD) = AB² + BC²

AC.AC = AB² + BC²

AC² = AB² + BC²

Hence Proved.

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Mayank .?? 6 years, 9 months ago

See textbook

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Two horses are considered fir a race. The probability of selection of first horse is 1/4 and that is second is 2/3. Find the probability a. Both will be selected b. One of them will be selected c. None will be selected

Posted by Sahana Shree 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Natwar Kalwani 6 years, 9 months ago

Check the question correctly

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In an AP the first term is 2 and the sum of the first five terms is one fourth of the sum of the next five terms . Show that the 20th term is -112

Posted by Raj Goswami 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 9 months ago

a=2,sum of first five terms =1/4th sum of next five terms

this gives,, 4 times of sum of first five terms =sum of next five terms

sum of first five terms + sum of next five terms = to sum of first ten terms

sum of first five terms + 4 times of sum of first five terms=sum of first ten terms

5 times of sum of first five terms =sum of first ten terms

5*n/2[2a+4d]=n/2[2a+9d] :(a=2}

5*5/2[4+4d]=10/2[4+9d]

25/2[4+4d]=5[4+9d]

5[4+4d]=2[4+9d]

20+20d=8+18d

so,d= -6

a₂₀= a+ 19d

= 2 + 19(-6)

= 2 - 114

= -112

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Prove that the line segment joining the mid point of adjacent sides of aquadrilateral form a parallegram.

Posted by Rajesh Kumar Didwadiya 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 9 months ago

Prove that the line segments joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram

GIVEN A quadrilateral ABCD in which P. Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. TO PROVE PQRS is a parallelogram.

CONSTRUCTION Join AC.
PROOF In A ABC, P and Q are the mid-points of AB and BC respectively.

PQ || AC

In Δ ACD, R and S are the mid-points of CD and DA
respectively.

SR || AC

From equations (i) and (ii), we have

PQ II AC and SR || AC

PQ || SR.

Similarly, by considering triangles ABD and BCD, we can prove that PS || QR.

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Why we find inner surface area of combine solids like cylender is mounted on hemisphere

Posted by Vivek Patidar 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

#Miss_Raagini ???? 6 years, 9 months ago

If it is asked in question then v hv 2 find yrr.. ?

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Calculate the HCF of 3³×5 and 3²×5²

Posted by Jain Ramesh 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Affu 😊 6 years, 9 months ago

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Shreya Shalini 6 years, 9 months ago

45 is the HCF

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The length of the hypotenunes of a right tringle exceed the length of its base by 2 cm and exceed twice the length of its altitude by 1 cm find each side

Posted by Dharmendra Kumar 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 9 months ago

Let hypotenuse be a

Given, base = a-2 ---------------- (1)

Given, altitude = (a-1)/2. ---------------- (2)

By Pythagoras theorem

a^2 = (a-2) + ((a-1)/2)^2

a^2 = (a^2-2a+1)/4 + a^2-4a+4

4a^2 = a^2-2a+1+4(a^2-4a+4)

4a^2 = a^2-2a+1+4a^2-16a+16

a^2 - 18a + 17 =0

(a-17)(a-1) = 0

a = 17

Substitute a = 17 in (2) we get

and lenght of each side = 17 - 1/2

= 16/2

= 8

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Sin theta _ cos theta + 1/sin theta + cos theta _1 =1/sec theta_ tan theta

Posted by Rajmohan Singh 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Rajmohan Singh 6 years, 9 months ago

I can't understand that

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Affu 😊 6 years, 9 months ago

See in ncert chapter trignometry example 15

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how prove the ex 8.4

Posted by Ranjan Bhai 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Ranjan Bhai 6 years, 9 months ago

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Sin square thitha plus cos four thitha is equal to sin four thitha plus cos square thitha prove that

Posted by Prashant Kumar 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 9 months ago

{tex}\begin{array}{l}\sin^4\theta-\cos^4\theta\\=(\sin^2\theta+\cos^2\theta)((\sin^2\theta-\cos^2\theta)\\\sin^4\theta-\cos^4\theta=\sin^2\theta-\cos^2\theta\\or\;\;\sin^4\theta+\cos^2\theta=\cos^4\theta+\sin^2\theta\\or\;\sin^2\theta+\cos^4\theta=\;\sin^4\theta+\cos^2\theta\end{array}{/tex}

Hence proved

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A box contain 12 which some are red in colour if 6 more red ball are put in the box and a ball us drawn at random , the probability of drawing a red ball doubles then what is was before . Find the no. Of red ball

Posted by Rohit Singh 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Simmiii ?? 6 years, 9 months ago

Let no. Of red balls =x Total balls =12 Probability=x/12 If 6 more red balls are put in a box then Total balls 12+6=18 No. Of red balls=x+6 Probabily =x+6/18 X+6/18=2*x/12 By cross multiply x=3ans

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Water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10kmper hours. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Posted by Yash Goswami 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Divya Yadav 6 years, 9 months ago

Volume of water in 1 min=6×1.5×10000/60(speed) =1500m cube Volume of water in 30 min= 45000 Volume of water flowing= volume of water In 30 min from canal = for irrigation 45000= a×8/100 a= 562500 m square Area irrigated in 30 min = 562500 m square i Hope this will help you??

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What is the formula for area of segment of a circle

Posted by Pr!Nce Pr@J@P@T! 6 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Pr!Nce Pr@J@P@T! 6 years, 9 months ago

Thanx usha and simmili both of you

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Simmiii ?? 6 years, 9 months ago

Area of required sector-areaof triangle

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Usha Yadav 6 years, 9 months ago

If theta is equal to or less than 90,then area of segment of circle is πr^2theta/360-r^2/2×sin theta............ And if theta is greater than 90 then,area of segment of circle is πr^2theta/360-r^2×sin theta/2 ×cos theta/2.........Hope it will help you ????

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Prove that 1-tan square theta divided by cot square theta -1 is equal to tan square theta..

Posted by Zindagi Aayat 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Usha Yadav 6 years, 9 months ago

Sec^2theta/cosec^2 theta =1/cos^2×sin^2 theta/1=sin^2 theta /cos^2 theta =tan^2 theta.......Hope it will help you ????

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If -5 is the root of the quadratic equation 2x^2 + px - 15 = 0 amd the quadratic equation p(x^2 + x )k = 0 has equal roots then find the value of k

Posted by Shubham Jangra 6 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Simmiii ?? 6 years, 9 months ago

Firstly -5 is put in eq 1 then the value of p is put in 2 eq

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Shikhar Singh 6 years, 9 months ago

p=7 and k=0

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Ship.. ??? 6 years, 9 months ago

-5 pheheli eq. Meh x place par dalke p ki value nikalo. Then use the value of p for the 2nd eq. And find th value of k

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If Sec+Tan = p Find the value of Cosec.

Posted by Vibhanshi Jain 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Shikhar Singh 6 years, 9 months ago

Is this was helpful for you? Please tell me.

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Shikhar Singh 6 years, 9 months ago

1/cosA + sinA/cosA=p 1+sinA/cosA=p On squaring on both sides. (1+sinA)^2/(cos^2)=p^2 (1+sinA)^2/(1-sinA^2)=p^2 (1+sinA)/(1-sinA)=p^2 Now by changing sin into cosec and simplify the eq. and then take cosec common and you will get the answer.

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If the two two digit numbers are placed one after the other to form a four digit number such that the larger number is placed first. The difference between the four digit number and the sum of two two digit mumber is 5742. Then find the numbers.

Posted by Shikhar Singh 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 9 months ago

Let the numbers are xy=10x+y and ab=10a+b

then 4 digit no=xyab=1000x+100y+10a+b

Now 1000x+100y+10a+b-(10x+y+10a+b)=5742

990x+99y=5782

or 10x+y=58

so first No=58

For 2nd No . other condition is required so the question is incomplete

By trial and error method:

take a=1 and b=1

Then 5811-(58+11)=5811-69=5742

again if a-=1 b=2

Then 5812-(58+12)=5812-70=5742

again if a-=2 b=3

Then 5823-(58+23)=

=5800+23-58-23=5800-58=5742

So So many solutions are possible but fist number 58 is sure

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Show that in a right triangle the square of the hypotenuse is equal to this sum the square of the other two sides. Thus if ABC is an isosceles triangle right angled at c prove that a b square is equal to 2 AC square

Posted by Sahil Ansari 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given : {tex}\triangle \mathrm{ABC}{/tex} is right angle at B
To prove: {tex}A C^{2}=A B^{2}+B C^{2}{/tex}

Construction: Draw {tex}B D \perp A C{/tex}

Proof:In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}ABD

{tex}\angle{/tex}ABC={tex}\angle{/tex}ADB=90

{tex}\angle{/tex}A is common

{tex}\angle{/tex}ABC{tex}\sim{/tex}{tex}\angle{/tex}ADB
{tex}\Rightarrow \frac{A D}{A B}=\frac{A B}{A C}{/tex}
{tex}\Rightarrow{/tex} AD. AC = AB² ...(i)
Similarly {tex}\Delta \mathrm{BDC} \sim \Delta \mathrm{ABC}{/tex}
{tex}\Rightarrow \frac{C D}{B C}=\frac{B C}{A C}{/tex}
{tex}\Rightarrow{/tex} CD.AC = BC² ...(ii)
Adding (i) and (ii)
AD.AC + CD. AC= AB² + BC²
AC(AD + CD) = AB² + BC²
AC {tex}\times{/tex} AC = AB² + BC²
AC² = AB² + BC²
Hence Proved