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InterviewSolution
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.
| 39951. |
Cbse board 10 ka exam kb hoga |
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Answer» Election ki wajha se board paper feb 2019 se start hoga My teacher told 1st week of March But sir hamne to suna h ki up board ka feb me hoga aur cbse ka march me Feb 2019 |
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| 39952. |
THE SUM OF n TERMS OF THE AP IS 4n-n^2, FIND THE COMMON DIFFERENCE. |
| Answer» Common difference = 1 | |
| 39953. |
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that : AB +CD = AD+BC |
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Answer» We know that the tangent segments from an external point to a circle are equal\xa0{tex}\\therefore{/tex} AP = AS ........(1)BP = BQ .......(2)CR = CQ .......(3)DR = DS .......(4)Adding (1), (2), (3) and (4), we get(AP + BP) + (CR + DR) = (AS + BQ + CQ + DS){tex}\\Rightarrow{/tex}\xa0AB + CD = (AS + DS) + (BQ + CQ){tex}\\Rightarrow{/tex}\xa0AB + CD = AD + BC Tq my dear frnd..... It helps me a lot..... ✌️? |
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| 39954. |
Differences between polynomial & algebric expression |
| Answer» An algebraic expression, in which variable(s) does (do) not occur in the denominator, exponents of variable(s) are whole numbers and numerical coefficients of various terms are real numbers, is called a polynomial. (iii) Numerical coefficient of each term is a real number. x x are not polynomials. | |
| 39955. |
How to learning the formula of Trigonometry |
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Answer» By writing 5-6 times in notebook Learn it |
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| 39956. |
Prove that √5 is rational number |
| Answer» It is not rational no. | |
| 39957. |
(cosA-sinA+1)/(cosA+sinA-1)=cosecA+cotA |
| Answer» {tex}L H S=\\frac{\\cos A-\\sin A+1}{\\cos A+\\sin A-1}{/tex}{tex}=\\frac{\\sin A(\\cos A-\\sin A+1)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{\\sin A \\cos A-\\sin ^{2} A+\\sin A}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{\\sin A \\cos A+\\sin A-\\left(1-\\cos ^{2} A\\right)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{\\sin A(\\cos A+1)-(1-\\cos A)(1+\\cos A)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{(1+\\cos A)(\\sin A+\\cos A-1)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{(1+\\cos A)(\\sin A+\\cos A-1)}{\\sin A(\\cos A+\\sin A-1)}{/tex}{tex}=\\frac{1}{\\sin A}+\\frac{\\cos A}{\\sin A}{/tex}= cos A + cot A = RHSProved | |
| 39958. |
Prove 3 root 5 is a rational number |
| Answer» 3√5 is a irrational number not rational number. | |
| 39959. |
Badic proportional theorem |
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Answer» If in triangle, one line is parallel to the third line then ratio of two sides are equal.. It is also known as Thales\' theorem if a line parrallel to third side of a triangle then it will divide other two lines in same ratio/proportion. |
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| 39960. |
What is the shape of square |
| Answer» All sides are equal | |
| 39961. |
Find a & b such that numbers a , 9 , b, 25 form an AP |
| Answer» A2=a+d,A4=a+3d9=a+d25=a+3d-16=-2d8=d,a=1b=a+2db=1+2(8)b=1+16b=17 | |
| 39962. |
If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2±√3,find other zeroes. |
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Answer» -5 and 7 are other zeros of the given polynomial I do not know the answer |
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| 39963. |
Mid point theomer |
| Answer» “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the triangle.” | |
| 39964. |
If a and 1÷a are the zeroes of 4x^2-2x+(k-4)find k |
| Answer» Please check your questions | |
| 39965. |
Two opposite vertices of a square are(-1,2)and(3,2) find the coordinate of other two vertices. |
| Answer» Let ABCD be a square and B (x, y) be the unknown vertex.AB = BC{tex} \\Rightarrow {/tex} AB2 = BC2\xa0{tex} \\Rightarrow {/tex}\xa0(x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2{tex} \\Rightarrow {/tex} x2 + 1 + 2x + y2 + 4 - 4x = x2 - 6x + 9 + y2 + 4 - 4x{tex} \\Rightarrow {/tex}\xa02x + 1 = - 6x + 9{tex} \\Rightarrow {/tex}\xa08x = 8{tex} \\Rightarrow {/tex}\xa0x = 1 ........ (i)In {tex}\\triangle{/tex}ABC, AB2 + BC2 = AC2{tex} \\Rightarrow {/tex}\xa0(x + 1)2 + (y - 2)2 + (x - 3)2 + (y - 2)2 = (3 + 1)2 + (2 - 2)2{tex} \\Rightarrow {/tex}x2 + 1 + 2x + y2 - 4x + 4 + x2 + 9 - 6x + y2 + 4 - 2y = 16 + 0{tex} \\Rightarrow {/tex}\xa02x2 + 2y2 + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16{tex} \\Rightarrow {/tex}\xa02x2 + 2y2 - 4x - 8y + 2 = 0{tex} \\Rightarrow {/tex}\xa0x2 + y2 - 2x - 4y + 1 = 0 ..... (ii)Putting the value of x in eq. (ii),1 + y2 - 2 - 4y + 1 = 0\xa0{tex} \\Rightarrow {/tex}\xa0y2 - 4y = 0\xa0{tex} \\Rightarrow {/tex}\xa0y(y - 4) = 0{tex} \\Rightarrow {/tex}\xa0y = 0 or 4Hence the other vertices are (1, 0) and (1, 4). | |
| 39966. |
Theorem 6.1 prove it please it is very important for me |
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Answer» Hiiii Thank you ... Animesh for being sooo kind Book mei dekho And sorry diagram nhi bna skta Given:-Tri. ABC where DE parallel BCTo proov:- AD/DB AE/ECProof :- area of tri. ADE= 1/2×base×height Area(Ade) =1/2×DB×EN Similarly area(BDE) = 1/2×DB×EN therefore area(ADE) =1/2×AD×EN/1/2×DB×EN=AD/DB - - - 1NOW, In tri. ADE and tri. DECArea(ADE) =1/2×AE×DMArea(DEC) =1/2×EC×DMTherefore, area(ADE)/area(DEC)=1/2×AE×DM/1/2×EC×DM=AE/EC - - - - - 2Tri.BDE and tri.DEC are on same base DE and between the same parallel BC and DE Area(BDE)=area(DEC) - - - 3Therefore, from equation 1,2 and 3 AD/DB=AE/EChence proved Please .....tommorow is my maths exam |
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| 39967. |
If the zeroes of the polinomial (k+4)x²+13x+3k are in A.P. then find the value of k. |
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| 39968. |
If the sum of the zeroes of the polynomial f(x)=2x^3-3kx^2+4x-5 is 6, then find the value of k |
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| 39969. |
If p^ th term of an Ap is q and q^th term is p.find its (p+q)^th term |
| Answer» q = a + (p - 1)d ..... (i)p = a + (q - 1)d ...... (ii)q - p = (p - 1 - q + 1)d{tex}\\frac{{q - p}}{{p - q}} = d{/tex}{tex} \\Rightarrow d = - 1{/tex}Put the value of d in eq (i)q = a + (p - 1) (-1){tex} \\Rightarrow {/tex}\xa0q = a - p + 1{tex} \\Rightarrow {/tex}\xa0a = q + p - 1ap+q = a +(p + q - 1)d= (q + p - 1) + (p + q - 1) (-1)= q + p - 1 - p - q + 1= 0 | |
| 39970. |
3 resistance have 6ohm 12ohm 15ohm how we connected do we have 19ohm |
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Answer» The total resistance is 19/60 ohm not 19ohm 6 andv12 connect in parellel =4. 15 and 4 connect in series |
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| 39971. |
Prove that √1+sin/1-sin+√1-sin/1+sin |
| Answer» LHS = {tex}\\sqrt { \\frac { 1 + \\sin \\theta } { 1 - \\sin \\theta } } + \\sqrt { \\frac { 1 - \\sin \\theta } { 1 + \\sin \\theta } }{/tex}{tex}= \\sqrt { \\frac { ( 1 + \\sin \\theta ) } { ( 1 - \\sin \\theta ) } \\times \\frac { ( 1 + \\sin \\theta ) } { ( 1 + \\sin \\theta ) } }{/tex}+\xa0{tex}\\sqrt { \\frac { ( 1 - \\sin \\theta ) } { ( 1 + \\sin \\theta ) } \\times \\frac { ( 1 - \\sin \\theta ) } { ( 1 - \\sin \\theta ) } }{/tex}{tex}= \\sqrt { \\frac { ( 1 + \\sin \\theta ) ^ { 2 } } { 1 - \\sin ^ { 2 } \\theta } } + \\sqrt { \\frac { ( 1 - \\sin \\theta ) ^ { 2 } } { 1 - \\sin ^ { 2 } \\theta } }{/tex}{tex}= \\sqrt { \\frac { ( 1 + \\sin \\theta ) ^ { 2 } } { \\cos ^ { 2 } \\theta } } + \\sqrt { \\frac { ( 1 - \\sin \\theta ) ^ { 2 } } { \\cos ^ { 2 } \\theta } }{/tex}\xa0{tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}{tex}= \\frac { 1 + \\sin \\theta } { \\cos \\theta } + \\frac { 1 - \\sin \\theta } { \\cos \\theta }{/tex}\xa0{tex}= \\frac { 1 + \\sin \\theta + 1 - \\sin \\theta } { \\cos \\theta }{/tex}{tex}= \\frac { 2 } { \\cos \\theta }{/tex}=\xa0{tex}2sec\\theta{/tex}= RHS | |
| 39972. |
3x+5=11 |
| Answer» 3x + 5 = 113x = 11 - 53x = 6x = 6/3 = 2 | |
| 39973. |
Which term of the AP:3,8,13,18,....is78? |
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Answer» an=a+(n-1)d78=3+(n-1)578-3=(n-1)575/5=(n-1)15=n-115+1=n16=n The given A. P. is 3, 8, 13, 18, .....Here, a = 3 d = 8 - 3 = 5Let the nth term f=of the A. P. be 78Then, an = = a + (n - 1)d78 = 3 + (n - 1) 578 = 3 + 5n - 578= 5n - 25n = 78 + 2 = 80n = 80/5 = 16Hence, 16th term of the A .P. is 78. Hello,\xa0solution:\xa0as we know that nth term of an AP isTn\xa0= a+(n-1)dHere in given APa= 3d= 578 = 3+(n-1)578-3 = (n-1)575/5=n-115 = n-1n = 16\xa0Sixteenth term is 78.\xa0hHop it helps you\xa0 |
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| 39974. |
Is board is now preparing two papers of each subject one for topers and other for average? |
| Answer» I think no Board has no time to prepare two papers of each subject | |
| 39975. |
If p =sec x +tan x then show that sinx=(p)2 -1/(p)2+1 |
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| 39976. |
Find the value of the middle term of the A.P. 7,13,19,........,247. |
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Answer» An = a + (n - 1)d241 = 7 + (n - 1)6241 - 7 = (n - 1)6234 = (n - 1)6234/6 = n - 139 = n - 1n = 40Now middle term =n/2=20An = 7 + (20 - 1)6= 7 + 19 × 6=7 + 114=121 Middle teri |
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| 39977. |
1+sec x-tan x/1+sec x+tan x =sec x-tan x prove it |
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| 39978. |
Is x³+x² is a polynomial ?Is this is quadratic equation ? |
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Answer» But guy\'s focus on the common if we take common then it is a quadratic equation Yes it is a polynomial but not a quadratic equation because highest power of variable is 3. But if we take commen thenX²(x+1) will come which is a quadratic equation Yes x3 + x2 is a polynimail. It is not a quadratic equation because degree is 3. What ??? tumai bau ki |
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| 39979. |
Find the largest no. which divides 615 and 963 leaving remainder 6 in each case. |
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Answer» Since 6 is the remainder in each case Therefore, 615 - 6=609And,963 - 6=957Now required no.=hcf of 609 and 957 957=609×1+348 609=348×1+261 348=261×1+87 261=87×3+0Therefore, required no.=87 To find the largest number which divides 615 and 963 leaving remainder 6 in each case i.e. HCF.Consider HCF be x.In order to make 615 and 963 completely divisible by x, we need to deduct the remainder 6\xa0from both the cases.609 = 3 x 3 x 29957= 3 x 11 x 29⇒ x = 3 x 29 = 87∴\xa0largest number which divides 615 and 963 leaving remainder 6 in each case is 87. |
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| 39980. |
As we all want good percentage in 10th. So if we chose maths B We can\'t opt PCM |
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Answer» What is PCM plz explain Hi?? Yes i am |
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| 39981. |
Solution of lesson 6 |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 39982. |
Solution of lesson 6 and 7 |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 39983. |
Find the middle term of Ap 7,13,19,.247. |
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Answer» And a11 is 76 n is 21. When n is odd,=(n+1/2) th term.21+1/2=11. 11 is the right answer an=247, a=7, d=6, n=?Thenan=a+(n-1)d247=7+(n-1)6(247-7)/6=n-1240/6=n-1n-1=40n=40+1=41Thenn+1/241+1/2=42/2=21Middle term is 21a21=a+20d =7+20(6) =7+120=127Ans:- 127 Middle term of Ap is |
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| 39984. |
If secx=x+1/4x then prove that secx +tanx =2 or 1/2x |
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| 39985. |
Compleating squring. |
| Answer» Ravi aap kha se ho | |
| 39986. |
Completing square how to do |
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| 39987. |
Sin2O+cos2O |
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Answer» Sin20+Cos20 =Sin20+Cos(90-70)=Sin20+Sin70=Sin90=1 = Sin 20+cos (90-70)= sin 20 + sin 70= sin 90= 1 |
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| 39988. |
Show that there is no positive integer n for which √(n+1)+√(n-1) is rational number. |
| Answer» Let us assume that there is a positive integer n for {tex}\\sqrt{n-1}+\\sqrt{n+1}{/tex}which is rational and equal to {tex}\\frac pq{/tex}, where p and q are positive integers and (q\xa0{tex}\\neq{/tex}\xa00).{tex}\\sqrt { n - 1 } + \\sqrt { n + 1 } = \\frac { p } { q }{/tex}......(i)or,\xa0{tex}\\frac { q } { p } = \\frac { 1 } { \\sqrt { n - 1 } + \\sqrt { n + 1 } }{/tex}on multiplication of numerator and denominator by\xa0{tex}\\sqrt{n-1}-\\sqrt{n+1}{/tex}\xa0we get{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( \\sqrt { n - 1 } + \\sqrt { n + 1 } ) ( \\sqrt { n - 1 } - \\sqrt { n + 1 } ) }{/tex}{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( n - 1 ) - ( n + 1 ) } = \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { - 2 }{/tex}or,\xa0{tex}\\sqrt { n + 1 } - \\sqrt { n - 1 } = \\frac { 2 q } { p }{/tex} ........(ii)On adding (i) and (ii), we get{tex}2 \\sqrt { n + 1 } = \\frac { p } { q } + \\frac { 2 q } { p } = \\frac { p ^ { 2 } + 2 q ^ { 2 } } { p q }{/tex}{tex}\\sqrt{n+1}\\;=\\frac{p^2+2q^2}{2pq}{/tex}...............(iii)From (i) and (ii),{tex}\\style{font-family:Arial}{\\sqrt{n-1}\\;=\\frac{p^2-2q^2}{2pq}}{/tex}........(iv)In RHS of (iii) and (iv)\xa0{tex}\\frac{p^2+2q^2}{2pq}\\;and\\;\\frac{\\displaystyle p^2-2q^2}{\\displaystyle2pq}\\;are\\;rational\\;number\\;because\\;p\\;and\\;q\\;are\\;positive\\;integers{/tex}But it is possible only when (n + 1) and (n - 1) both are perfect squares.Now n+1-(n-1)=n+1-n+1=2Hence they differ by 2 and two perfect squares never differ by 2.So both (n + 1) and (n -1 ) cannot be perfect squares. Hence there is no positive integer n for which\xa0{tex}\\style{font-family:Arial}{\\sqrt{n-1\\;}+\\sqrt{n+1}}{/tex} is rational | |
| 39989. |
Sin4x/3 + cos4x/4 = 1/7 find the value of equal to sin8x/27 + cos8x/64 |
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| 39990. |
If A,B,C are interior angle of traingle ABC then prove that sec² B+C\\2 -1=cot² A\\2 |
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| 39991. |
Find the probability of getting a head when a coin is tossed twice? |
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Answer» 1/2 1\\2 |
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| 39992. |
If A,B,C are interior traingle of. ABC. Prove it secw |
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| 39993. |
Find the sum of last ten terms of the A.P.: 8,10,12,14,...,126. |
| Answer» 144 | |
| 39994. |
What is equlid divission lema |
| Answer» Given two positive integers a and b, then there exists two more integers q and r satisfying a=bq+r. | |
| 39995. |
Which form we fill for maths paper options |
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| 39996. |
Formulas of chapter 5 |
| Answer» Common diference formula is a2 - a1 = a3 - a2For any term or any thing formula is an = a + (n-1) dAnd for sum of the AP formula is Sn = n/2 (2a + (n-1) d )To find last term formula is Sn = n/2 (a + l ) | |
| 39997. |
If tan A = 1 and cot B = 1, find the value of tan(A+B) |
| Answer» TanA =1 => tanA = tan45° (tan45°= 1)=> A = 45° ........... (1)Now CotB =1 => cotB= cot45° (cot45°= 1) => B = 45° .............(2)Adding equation 1 and 2 A+B = 45°+45° =90° | |
| 39998. |
Prove that: sec^6A=1+tan^6A + 3 tan^2A × sec^2A. |
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| 39999. |
To prove(1+CotA+TanA)(SinA-CosA)/(Sec³A-Cosec³A)=Sin²A*Cos²APlz answer it it\'s urgent. |
| Answer» L.H.S\xa0= (1 + cotA + tanA) (sinA - cosA)= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA{tex}= \\sin A - \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A - \\cot A\\cos A + \\sin A\\;\\tan A - \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA= sinA tanA - cotA cosA........(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} - \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} - \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} - \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} - \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A - \\cos A \\times \\cot A{/tex}.......(2)From (1) & (2),(1 + cotA + tanA) (sinA - cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } - \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA - cosA.cotA\xa0Hence, Proved. | |
| 40000. |
For board exam if practice NCERT text book completely is it enough |
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Answer» Do it thoroughly. It may help u to score atlest 90% Hello No |
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