Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 26651. |
If a cosA+b sinA=c ‚then prove that asinA-bcosA=±√a^2+b^2-c^2 |
| Answer» We have, {tex}asin\\theta+bcos\\theta=c{/tex}On squaring both sides, we get{tex}(asin\\theta+bcos\\theta)^2=c^2{/tex}(a sin θ)2\xa0+ (b cos θ)2\xa0+ 2(a sin θ) (b cos θ) = c2⇒ a2\xa0sin2\xa0θ + b2\xa0cos2\xa0θ + 2ab sin θ cos θ = c2⇒ a2(1 – cos2\xa0θ) + b2\xa0(1 – sin2\xa0θ) + 2 ab sin θ cos θ = c2 {tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ a2\xa0– a2\xa0cos2\xa0θ + b2\xa0– b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2⇒ –a2\xa0cos2\xa0θ – b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2\xa0– a2\xa0– b2 Taking Negative common,⇒ a2\xa0cos2\xa0θ + b2\xa0sin2\xa0θ – 2ab sin θ cos θ = a2\xa0+ b2\xa0– c2⇒ (a cos θ)2\xa0+ (b sin θ)2\xa0– 2(a cos θ) (b sin θ) = a2\xa0+ b2\xa0– c2⇒ {tex}(acos\\theta-bsin\\theta)^2=a^2+b^2-c^2{/tex}⇒{tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\pm \\sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}\xa0Hence proved, {tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\sqrt{a^2+b^2-c^2}{/tex} | |
| 26652. |
Derivation of CSA and TSA of frustrum of a cone |
| Answer» Refer ti rd sharma | |
| 26653. |
If sinA+cosA |
| Answer» | |
| 26654. |
which term of tha AP :3,15,27....will be 132 more than its 54th term? |
| Answer» Here, first term = a = 3 and common difference = d = 15 – 3 = 12Let nth term of the given AP be 132 more than its 54th term, thenan = a54 + 132⇒ a + (n – 1)d = a + (54 – 1)d + 132⇒ (n – 1) d = (54 – 1)d + 132⇒(n – 1) 12 = (53)12 + 132⇒(n – 1) 12 = 768⇒(n – 1) = 64⇒ n = 65Hence, the 65th term will be 132 more than the 54th term. | |
| 26655. |
When exam date list will be declare |
| Answer» 10 January | |
| 26656. |
Find mode if mean 54 and median 33 |
|
Answer» Abxd 2mean + mode = 3median Mode = -9 |
|
| 26657. |
Find largest number that divide 789, 861, 1069 leaving reminders 7, 11, 15 respectively |
| Answer» | |
| 26658. |
if cosecA + cotA =p then prove that cosA = p×p-1/p×p+1 |
| Answer» Given, cosec θ + cot θ = p...(i)We know that, {tex}cosec^2\\theta-cot^2\\theta=1{/tex}{tex}\\Rightarrow (cosec\\theta+cot\\theta)(cosec\\theta-cot\\theta)=1{/tex}{tex}\\Rightarrow p(cosec\\theta-cot\\theta)=1{/tex}{tex}\\Rightarrow cosec\\theta-cot\\theta=\\frac 1p{/tex}\xa0....(ii)Adding i and ii, we get{tex}2cosec\\theta=p+ \\frac 1p{/tex}{tex}cosec\\theta=\\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow sin\\theta= \\frac{1}{cosec\\theta}=\\frac{2p}{p^2+1}{/tex}We know that,{tex}cos\\theta=\\sqrt{1-sin^2\\theta}=\\sqrt{1- \\frac{4p^2}{(p^2+1)^2}}=\\sqrt{\\frac{p^4+1-2p^2}{(p^2+1)^2}}{/tex}{tex}cos\\theta=\\sqrt{\\frac{(p^2-1)^2}{(p^2+1)^2}}=\\frac{p^2-1}{p^2+1} {/tex} | |
| 26659. |
Hey gyss yaha par 2 Akshita hain |
| Answer» | |
| 26660. |
Vivek kahan gye |
| Answer» | |
| 26661. |
The ratio between the volumes of two sphere is 8:27 ? What is the ratio between their surface area? |
| Answer» 4:9 will be the ratio of their surface area! | |
| 26662. |
Can anyone explain Thales theorem |
| Answer» Without figure ...impossible & here figure ....impossible | |
| 26663. |
Yha abhishek rai kaun hai??? |
| Answer» | |
| 26664. |
The rain water from a roof of 44 m x 20 m drains int |
| Answer» Length of the roof (l) = 44 m,Breadth of the roof (b) = 20 mLet the height of the water on the roof be h m.Volume of water falling on the roof = l{tex}\\times{/tex}b{tex}\\times{/tex}h= 44{tex}\\times{/tex}20{tex}\\times{/tex}h= 880hRadius of the cylindrical vessel\xa0{tex}( R ) = \\frac { 4 } { 2 } = 2 m{/tex}Height of the water in the cylindrical vessel (H) = 3.5 mVolume of the water in the cylindrical vessel =\xa0{tex}\\pi R ^ { 2 } H{/tex}{tex}= \\frac { 22 } { 7 } \\times 2 \\times 2 \\times 3.5{/tex}{tex}= \\frac { 308 } { 7 }{/tex}= 44Volume of water falling on the roof = Volume of the water in the cylindrical vessel{tex}\\Rightarrow{/tex}\xa0880 h = 44{tex}\\Rightarrow h = \\frac { 44 } { 880 }{/tex}{tex}\\Rightarrow h = \\frac { 44 } { 880 } \\times 100{/tex}{tex}\\Rightarrow{/tex}\xa0h = 5 cm{tex}\\Rightarrow{/tex}\xa0Height of the water on the roof is 5 cm. | |
| 26665. |
RIMSHA......,.... |
| Answer» | |
| 26666. |
9586486×546558 |
| Answer» Stupid ??? | |
| 26667. |
Please provide me the slot of previous 10 year question paper of class 10 for board exam |
| Answer» You can check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 26668. |
PQRS is a parallelogram with PQ SQUARE + QR SQUARE=170 and PR=12 then find QS |
| Answer» | |
| 26669. |
(Cosec teta -cot teta)2 =1-cos teta ÷1+cos teta |
| Answer» LHS\xa0{tex}= \\frac{{1 - \\cos A}}{{1 + \\cos A}}{/tex}Multiplying numerator and denominator by 1 - cos A{tex}= \\frac{{(1 - \\cos A)(1 - \\cos A)}}{{(1 + \\cos A)(1 - \\cos A)}}{/tex}{tex}= \\frac{{{{(1 - \\cos A)}^2}}}{{1 - {{\\cos }^2}A}}{/tex}\xa0{tex} \\left[ {\\because (a + b)(a - b) = {a^2} - {b^2}} \\right]{/tex}{tex}= \\frac{{{{(1 - \\cos A)}^2}}}{{{{\\sin }^2}A}}{/tex}\xa0{tex} \\left[ {\\because 1 - {{\\cos }^2}A = {{\\sin }^2}A} \\right]{/tex}{tex}= {\\left( {\\frac{{1 - \\cos A}}{{\\sin A}}} \\right)^2}{/tex}{tex}= {\\left( {\\frac{1}{{\\sin A}} - \\frac{{\\cos A}}{{\\sin A}}} \\right)^2}{/tex}{tex}= {(\\cos ecA - \\cot A)^2}{/tex}\xa0{tex} \\left[ {\\because \\frac{1}{{\\sin A}} = \\cos ec\\;A,\\frac{{\\cos A}}{{\\sin A}} = \\cot A} \\right]{/tex}{tex}= {\\left[ { - 1(\\cot A - \\cos ecA)} \\right]^2}{/tex}= (cot A - cosec A)2Hence proved. | |
| 26670. |
Koi maths ka blue print de do plz |
| Answer» | |
| 26671. |
Prove that tan^2A-tan^2B=cos^2B-cos^2A/cos^2Bcos^2A |
| Answer» | |
| 26672. |
If tanA=tanB than prove that A-B=90° |
| Answer» This question is wrong | |
| 26673. |
If tanA = 2cosA then prove that cos²A +cot²A = 2secA |
| Answer» | |
| 26674. |
Board paper solution |
| Answer» | |
| 26675. |
Board questions paper of class 10 of2013 |
| Answer» | |
| 26676. |
If the HCF of 85 and 153 is expressible in the form of 85m-153,then find the value of m? |
| Answer» M=2 | |
| 26677. |
Find 11th term from the last term of an ap 27,23,19,............-65 |
|
Answer» -25 -25 25 |
|
| 26678. |
If a,b,c are in A.P ,then show that {a(bc)}/bc,{b(ca)}/ca and {c(ab)/ab are also in A.P |
| Answer» abc/bc=a , bca/ca=bc , cab/ab = c | |
| 26679. |
In triangle PQR,right angle at Q PR+QR=25 and PQ=5cm. Prove that sinP+tanA P=5 |
| Answer» {tex}\\frac{{216}}{{65}}{/tex} | |
| 26680. |
Show that one and only one out of n,n+2,n+4 are divisible by 3 ,where n is any positive integer |
| Answer» Esy h bt yha type krna muskil | |
| 26681. |
Show that one and only one out of n,n+2 or n+4 is divisible by 3 , where n is any positive integer |
|
Answer» Plz Write in short ..? Esy h bt yha type krna muskil ????? |
|
| 26682. |
If the ratio of the sum of first n terms of 2 AP\'s is 7n+1: 4n+27 find the ratio of first m terms |
| Answer» Let a, and A be the first terms and d and D be the common difference of two A.PsThen, according to the question,{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \\frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,\xa0{tex}\\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,{tex}\\frac { a + \\left( \\frac { n - 1 } { 2 } \\right) d } { A + \\left( \\frac { n - 1 } { 2 } \\right) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}Putting,\xa0{tex}\\frac { n - 1 } { 2 } = m - 1{/tex}{tex}n-1 = 2m - 2{/tex}{tex}n= 2m - 2 + 1{/tex}or, {tex}n = 2m - 1{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex}Hence,\xa0{tex}\\frac { a _ { m } } { A _ { m } } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex} | |
| 26683. |
Find the value of x in tan3x= sin 45 degree*cos45degree+ sin 30 degree |
| Answer» tan3x=1/√2*1/√2+1/2. ( by using trigonometry table). tan3x=1/2+1/2. tan3x =1. tan3x=tan45. 3x=45. x = 15 degree | |
| 26684. |
1+4+7+...X =207 find value of X |
| Answer» Given, a = 1, d = 4 -1 = 3.Let number of terms in the series be n.{tex}\\therefore \\quad S _ { n } = \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}{tex}\\therefore \\quad \\frac { n } { 2 } [ 2 \\times 1 + ( n - 1 ) 3 ] = 287{/tex}or,\xa0{tex}\\frac { n } { 2 } [ 2 + ( 3 n - 3 ) ] = 287{/tex}or, {tex}n [3n - 1] = 574{/tex}or, {tex}3n^2- n - 574 = 0{/tex}{tex}3n^2 - n - 574 = 0{/tex}{tex}3n(n - 14) - 41(n - 14) = 0{/tex}{tex}(n - 14)(3n - 41) = 0{/tex}n - 14 = 0 or 3n - 41 = 0n = 14 or 3n = 41The 14th term is x.{tex}\\therefore{/tex}\xa0{tex}a + (n - 1)d = x{/tex}1 + 13\xa0{tex}\\times{/tex}\xa03 = x{tex}1 + 39 = x{/tex}x = 40 | |
| 26685. |
If cosecø-sinø=a3Set-cosø=b3Then,,prove that:a2b2(a2+b2)=1 |
| Answer» | |
| 26686. |
find the largest number that will divide 398,436 & 542 leaving remainder 7,11 &15 respectively |
| Answer» The largest positive integer that will divide 398, 436 and 542 leaving reminders 7, 11 and 15 respectively is the HCF of the numbers (398 – 7), (436 – 11) and (542 – 15) i.e. 391, 425 and 527.HCF of 391,425 and 527:HCF of 425 and 391:425 = 391 × 1 + 34391 = 34 × 11 + 1734 = 17 ×2 + 0HCF of 425 and 391 = 17527 = 17 × 31Similarly, HCF of 17 and 527 = 17So, HCF of (391,425 ,527) = 17∴ Required number is 17. | |
| 26687. |
GivenTanA=nTanBSinA=mSinBProveCos(square)A=m(square)-1/n(square)-1 |
| Answer» | |
| 26688. |
49 ka whole square kya ha |
|
Answer» 2401 So stupid ??? |
|
| 26689. |
2 marks |
| Answer» | |
| 26690. |
How many we gets when we multiple 2188 by10096? |
| Answer» 22090048 | |
| 26691. |
What is euclid division lemmma |
|
Answer» Abe chup be a=bq+r EDL |
|
| 26692. |
Who is single likee in this group |
|
Answer» Aur hai its first time so I leave u...... It is not Teri maa Sajan it is your mother Teri maa |
|
| 26693. |
Is there Anyone who know the value of sin 18° |
|
Answer» Kya hai.. Yes |
|
| 26694. |
X²+6x-(a²+2a-8)=0 |
| Answer» The given equation is\xa0{tex}x^2\xa0+\xa06x -\xa0(a^2 +\xa02a - 8) = 0{/tex}Comparing it with {tex}Ax^2\xa0+ Bx +C = 0,{/tex} we get{tex}A = 1,\\ B = 6\\\xa0and\\ C = -(a^2 +\xa02a - 8){/tex}{tex}\\therefore{/tex}\xa0{tex}D =B^2\xa0- 4AC ={/tex} (6)2\xa0- 4(1)(-(a2 +\xa02a - 8))= 36 +4a2\xa0+ 8a - 32= 4a2\xa0+ 8a\xa0+\xa04 = 4(a2\xa0+ 2a + 1) = 4(a + 1)2\xa0> 0So, the given equation has real roots, given by{tex}\\alpha = \\frac { - B + \\sqrt { D } } { 2 A } = \\frac { - 6 + \\sqrt { 4 ( a + 1 ) ^ { 2 } } } { 2 \\times 1 } = \\frac { - 6 + 2 ( a + 1 ) } { 2 }{/tex}\xa0{tex}= -3 + (a + 1) = a - 2{/tex}{tex}\\beta = \\frac { - B - \\sqrt { D } } { 2 A } = \\frac { - 6 - \\sqrt { 4 ( a + 1 ) ^ { 2 } } } { 2 \\times 1 } = \\frac { - 6 - 2 ( a + 1 ) } { 2 }{/tex}\xa0{tex}= -3 - (a + 1) = -a - 4 = -(a + 4){/tex}Hence,\xa0(a - 2)\xa0and -(a + 4)\xa0are the roots of the given equation. | |
| 26695. |
What is mensuration cycle ? |
| Answer» When egg is not fertilized then it is shed off through endometrial wall every month. | |
| 26696. |
Find the number of terms (i)- 7,13,19....,205 |
|
Answer» 34 So easy ?? |
|
| 26697. |
If 1/x+2,1/x+3 and 1/x+5 are in A.P. find the value of x |
|
Answer» Equation will be come in quadratic form So easy ?? |
|
| 26698. |
find quardratic equation whose zeroes-4÷5&1÷3 |
|
Answer» 15xka square -17x +4 So easy ?? |
|
| 26699. |
If tan A=√2-1,prove that tan A/1+tan A×tan A=√2/4. |
| Answer» So easy ?? | |
| 26700. |
Date sheet for cbse class 10th examination |
| Answer» It will be release most probably this week | |