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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.
| 51. |
ABC is an acute angled triangle with circumcenter O and orthocentre H.If AO=AH, then find the angle A. |
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Answer» OA=R AH=2RcosA 2RcosA=R cosA=`1/2` A=`60^@`. |
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| 52. |
If k be the perimeter of the triangle ABC , then `bCos^2(C/2)` + c`Cos^2(B/2)` is equal toA. `2s`B. `s`C. `0`D. `a+b-c` |
| Answer» Correct Answer - B | |
| 53. |
In `triangleABC, a cosB-b cosA=`A. `(a^(2)-b^(2))`B. `b^(2)-c^(2)`C. `c^(2)-a^(2)`D. `a^(2)+b^(2)+c^(2)` |
| Answer» Correct Answer - A | |
| 54. |
In `triangleABC`, if `a=13, b=14, c=15`, then cosB=A. `(198)/(390)`B. `(394)/(390)`C. `(196)/(390)`D. `(500)/(390)` |
| Answer» Correct Answer - A | |
| 55. |
In any triangle ABC, if `a=" "18 , b=" "24 , c=" "30`, findcosA, cosB, cosCA. `(1)/(5)`B. `(2)/(5)`C. `(3)/(5)`D. `(4)/(5)` |
| Answer» Correct Answer - D | |
| 56. |
If in `DeltaABC,(cosA)/(a)=(cosB)/(b)=(cosC)/(c) and ` side a=2, then the area of the triangle isA. `1` sq. unitsB. `2` sq. unitsC. `sqrt(3)` sqt. UnitsD. `2sqrt(3)` sq. units |
| Answer» Correct Answer - C | |
| 57. |
`tan(cos^(-1)((4)/(5))+tan^(-1)((2)/(3)))=`A. `(17)/(24)`B. `(24)/(17)`C. `(17)/(6)`D. `(6)/(17)` |
| Answer» Correct Answer - C | |
| 58. |
`tan^(-1)((5)/(13))+cos^(-1)((3)/(5))=`A. `tan^(-1)((67)/(19))`B. `tan^(-1)((1)/(59))`C. `tan^(-1)((19)/(67))`D. `tan^(-1)(59)` |
| Answer» Correct Answer - A | |
| 59. |
If `cosA=(3)/(5), cosB=(5)/(13), cosC=(4)/(5)`, then the ratio of sides of triangle isA. `4:5:3`B. `19:13:15`C. `13:60:3`D. `52:60:39` |
| Answer» Correct Answer - D | |
| 60. |
`cos^(-1)((12)/(13))+sin^(-1)((3)/(5))`=A. `sin^(-1)((56)/(65))`B. `sin^(-1)((16)/(65))`C. `sin^(-1)((20)/(65))`D. `sin^(-1)((36)/(65))` |
| Answer» Correct Answer - A | |
| 61. |
निम्नलिखित के मुख्य मानो को ज्ञात कीजिए : `tan^(-1)""(-1)`A. `(3pi)/(4)`B. `(pi)/(4)`C. `(7pi)/(4)`D. `(-pi)/(4)` |
| Answer» Correct Answer - D | |
| 62. |
prove that `cos^3(2x)+3cos2x=4(cos^6x-sin^6x)` |
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Answer» `L.H.S. = cos^3(2x) +3cos2x` `=cos2x(cos^2 (2x) +3)` `=cos2x((2cos^2x-1)^2 +3)` `=cos2x((4cos^4x+1 - 4cos^2x) +3)` `=cos2x(4cos^4x+4 - 4cos^2x)` `=4(2cos^2x -1)(cos^4x+1 - cos^2x)` `=4(2cos^6x + 2cos^2x -2cos^4x - cos^4x- 1 + cos^2x)` `=4(2cos^6x-3cos^4x+3cos^2x-1)` `=4(2cos^6x+3cos^2x(1-cos^2x) - 1)` `=4(2cos^6x+3cos^2xsin^2x - (sin^2x+cos^2x)^3)` `=4(2cos^6x+3cos^2xsin^2x - sin^6x-cos^6x - 3sin^2xcos^2x (sin^2x+cos^2x))` `=4(cos^6x - sin^6x+3cos^2xsin^2x-3cos^2xsin^2x)` `=4(cos^6x - sin^6x) = R.H.S.` |
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| 63. |
Find all the solution of `4cos^2xsinx-2sin^2x=3sinx` |
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Answer» `4cos^2xsinx - 2sin^2x = 3sinx` `=>sinx(4cos^2x - 2sinx -3) = 0` `=>sinx(4(1-sin^2x) - 2sinx -3) = 0` `=>sinx(4-4sin^2x - 2sinx -3) = 0` `=>-sinx(4sin^2x + 2sinx -1) = 0` `=>-sinx = 0 or 4sin^2x + 2sinx -1 = 0` `=>sinx = 0 or sinx = (-2+-sqrt(4-4(4)(-1)))/8` `=>sinx = 0 or sinx = (-1+-sqrt5)/4` `=>sinx = sin 0^@ or sinx = sin18^@ or sinx = sin (-54^@)` `=>x = npi or x = npi+(-1)^n(18^@) or x = npi+(-1)^n(-54^@)` So, these are the three solutions for the given equation. |
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| 64. |
`tan(sin^(-1)((3)/(5))+cos^(-1)((3)/(sqrt(13)))=`A. `(sqrt(13))/(5)`B. `(5)/(sqrt(13))`C. `(17)/(6)`D. `(17)/(9)` |
| Answer» Correct Answer - C | |
| 65. |
`cot x=-sqrt(3)`A. `(pi)/(6), (5pi)/(6)`B. `(7pi)/(6), (11pi)/(6)`C. `(pi)/(6), (7pi)/(6)`D. `(5pi)/(6), (11pi)/(6)` |
| Answer» Correct Answer - D | |
| 66. |
General solution of `tan^(2)x+sec2x=1` isA. `npi, npipm(pi)/(4), ninZ`B. `npi, npipm(2pi)/(3), ninZ`C. `npi, npipm(pi)/(3), ninZ`D. `npi, npipm(pi)/(6), ninZ` |
| Answer» Correct Answer - C | |
| 67. |
Simplify `2sin^2beta+4cos(alpha+beta)sinalphasinbeta+cos2(alpha+beta)` |
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Answer» `2sin^2beta+4cos(alpha+beta)sinalphasinbeta+cos2(alpha+beta)` `2sin^2beta+4cos(alpha+beta)(2sinalphasinbeta+cos(alpha+beta))` `2sin^2beta-1+2cos(alpha+beta)(2sinalphasinbeta+cosalphacosbeta)` `-cos2beta+2cos(alpha+beta)cos(alpha-beta)` `-cos2beta+cos(alpha+beta+alpha-beta)+cos(alpha+beta-alpha+beta)` `-cos2beta+2cosalpha+cos2beta` `cos2alpha`. |
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| 68. |
समीकरण `sin x=(sqrt(3))/(2)` का मुख्य हल ज्ञात कीजिए |A. `(pi)/(3)`B. `-(pi)/(3)`C. `(5pi)/(3)`D. `-(2pi)/(3)` |
| Answer» Correct Answer - A | |
| 69. |
`cot x=-sqrt(3)`A. `npi+(5pi)/(6), ninZ`B. `npi+(7pi)/(6), ninZ`C. `npi+(11pi)/(6), ninZ`D. `npi+(pi)/(6), ninZ` |
| Answer» Correct Answer - A | |
| 70. |
Find the general solution of : `cosx+sinx=1`A. `2npi, 2npi-(pi)/(2)`B. `2npi, 2npi+(pi)/(2)`C. `2npi, 2npi-(pi)/(4)`D. `2npi, 2npi+(pi)/(4)` |
| Answer» Correct Answer - B | |
| 71. |
Find the values of the trigonometric function tan `(19pi)/3` |
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Answer» `tan ((19pi)/3) = tan(6pi+pi/3)` We know, `tan(2npi+x) = tanx` So,`tan(6pi+pi/3) = tan(pi/3) = sqrt3` |
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| 72. |
Simplify `sin(cot^(-1)(cos(tan^(-1)x))),0 ltxlt1`.A. `(x)/(sqrt(x^(2)+1))`B. `(x)/(sqrt(x^(2)+2))`C. `(sqrt(x^(2)+1))/(sqrt(x^(2)+2))`D. `(1)/(sqrt(x^(2)+2))` |
| Answer» Correct Answer - C | |
| 73. |
`cosec^(-1)(-sqrt(2))+cot^(-1)(sqrt(3))=`A. `(-pi)/(4)`B. `(5pi)/(12)`C. `(-pi)/(12)`D. `(pi)/(6)` |
| Answer» Correct Answer - C | |
| 74. |
`cot^(-1)((-1)/(sqrt(3)))` का मुख्य मान ज्ञात कीजिए ।A. `(2pi)/(3)`B. `(-2pi)/(3)`C. `(5pi)/(3)`D. `(4pi)/(3)` |
| Answer» Correct Answer - A | |
| 75. |
Find the general solution of : `cosx+sinx=1`A. `npi+(-1)^(n)(pi)/(4)-(pi)/(4), ninZ`B. `npi+(-1)^(n)(pi)/(4)+(pi)/(4), ninZ`C. `npi+(-1)^(n)(pi)/(3)-(pi)/(4), ninZ`D. `npi+(-1)^(n)(pi)/(6)-(pi)/(4), ninZ` |
| Answer» Correct Answer - A | |
| 76. |
If `tan(A-B)=1a n dsec(A+B)=2/(sqrt(3))`, then the smallest positive values of A and B, respectively, are`(25pi)/(24),(19pi)/(24)`(b) `(19pi)/(24),(25pi)/(24)``(31pi)/(24),(31pi)/(24)`(d) `(13pi)/(24),(31pi)/(24)` |
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Answer» `tan(A-B) = 1 and sec(A+B) = 2/sqrt3` `:. A-B = npi+pi/4 and A+B = 2npi+-pi/6` As, we have to find the smallest positive values of `A` and `B`, `:. A-B = pi/4->(1)` Now, `A+B` can not be `pi/6` as it becmes less than `A-B` as `A` and `b` are positive. So, with smallest positive values for `A` and `B`, `A+B= 2(1)pi-pi/6 = (11pi)/6` `:.A+B = (11pi)/6->(2)` Now, adding (1) and (2), `=>2A = pi/4+(11pi)/6 = (25pi)/12` `=>A = (25pi)/24` `=>B = A-pi/4 = (25pi)/24 - pi/4 = (19pi)/24.` |
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| 77. |
If in ` A B C ,A C`is double of `A B`, then the value of `cot(A/2)cot((B-C)/2)`is equal to`1/3`(b) `-1/3`(c) `3`(d) `1/2` |
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Answer» b=2c-(1) `tan((B-C)/2)=(b-c)/(b+c)cot(A/2)` `cot(A/2)cot((B-C)/2)=(b+c)/(b-c)=(2c+c)/(2c-c)=3`. |
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| 78. |
Prove that:`(sinx-siny)/(cosx+cosy)``=tan``(x-y)/2` |
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Answer» Here, we will use the formulas, `sin A - sin B = 2sin((A-B)/2)cos((A+B)/2)` `cos A + cos B = 2cos((A-B)/2)cos((A+B)/2)` `L.H.S.=(sinx-siny)/(cosx+cosy)` `=(2sin((x-y)/2)cos((x+y)/2))/(2cos((x-y)/2)cos((x+y)/2))` `=sin((x-y)/2)/cos((x-y)/2)=tan((x-y)/2)=R.H.S.` |
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| 79. |
General solution of `cos3x=cos2x` isA. `npi, (npi)/(5), ninZ`B. `2npi, (2npi)/(5), ninZ`C. `npi, (2npi)/(5), ninZ`D. `2npi, (npi)/(5), ninZ` |
| Answer» Correct Answer - B | |
| 80. |
General solution of `cosx-sinx=1` isA. `2npi, 2npi-(pi)/(2), ninZ`B. `2npi, 2npi+(pi)/(2), ninZ`C. `2npi, 2npi-(pi)/(4), ninZ`D. `2npi, 2npi+(pi)/(4), ninZ` |
| Answer» Correct Answer - A | |
| 81. |
If `(sinx)/(siny)=1/2,(cosx)/(cosy)=3/2`, where `x , y , in (0,pi/2),`then the value of `"tan"(x+y)`is equal to(a)`sqrt(13)`(b) `sqrt(14)`(c) `sqrt(17)`(d) `sqrt(15)` |
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Answer» `sinx/siny = 1/2 and cosx/cosy = 3/2` `:. sinx/siny * cosy/cosx = (1/2)/(3/2)` `=>tanx/tany = 1/3` `=>tany = 3tanx` `:. tan(x+y) = (tanx+tany)/(1-tanxtany) = (tanx+3tanx)/(1-tanx(3tanx))` `=>tan(x+y) = (4tanx)/(1-3tan^2x)->(1)` Now, `siny = 2sinx and cosy = 2/3cosx` As, `sin^2y+cos^2y = 1` `:. (2sinx)^2+(2/3cosx)^2 = 1` `=>4sin^2x+4/9cos^2x = 1` `=>4cos^2x(tan^2x + 1/9) = 1` `=>4(tan^2x + 1/9)= sec^2x` `=>4(tan^2x + 1/9)= 1+tan^2x` `=>4tan^2x-tan^2x = 1-4/9` `=>3tan^2x = 5/9` `=>tan^2x = 5/27` `=>tanx = sqrt5/(3sqrt3)` Putting value of `tanx` in (1), `tan(x+y) = (4* sqrt5/(3sqrt3))/(1-3(5/27)) = (4sqrt5sqrt3)/4 = sqrt15` So option `(d)` is the correct option. |
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| 82. |
General solution of `cos2x=sqrt(2)(cosx-sinx)` isA. `npi+(pi)/(2), 2npi+(pi)/(4), ninZ`B. `npi+(pi)/(4), 2npi+(pi)/(4), ninZ`C. `npi+(pi)/(4), 2npi+(pi)/(2), ninZ`D. `npi+(pi)/(2), 2npi+(pi)/(2), ninZ` |
| Answer» Correct Answer - B | |
| 83. |
General solution of `sinx+cosx=sqrt(2)cosalpha` isA. `2npipmalpha+(pi)/(3), ninZ`B. `2npipmalpha-(pi)/(4), ninZ`C. `2npipmalpha+(pi)/(6), ninZ`D. `2npipmalpha+(pi)/(4), ninZ` |
| Answer» Correct Answer - D | |
| 84. |
General solution of `sqrt(3)cosx+sinx=sqrt(2)` isA. `npi+(-1)^(n)(pi)/(4)-(pi)/(6), ninZ`B. `npi+(-1)^(n)(pi)/(4)+(pi)/(6), ninZ`C. `npi+(-1)^(n)(pi)/(4)-(pi)/(3), ninZ`D. `npi+(-1)^(n)(pi)/(4)+(pi)/(3), ninZ` |
| Answer» Correct Answer - C | |
| 85. |
Find the general solution of : `sqrt(3)cosx-sinx=1`A. `2npi+(pi)/(6), 2npi+(pi)/(2), ninZ`B. `2npi-(pi)/(6), 2npi-(pi)/(2), ninZ`C. `2npi+(pi)/(6), 2npi-(pi)/(2), ninZ`D. `2npi-(pi)/(6), 2npi+(pi)/(2), ninZ` |
| Answer» Correct Answer - C | |
| 86. |
If `x=rsinthetacosphi,y=rsinthetasinphi and z=rcostheta, then x^2+y^2+z^2`is independent of`theta,phi`(b) `r ,theta`(c) `r ,phi`(d) r |
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Answer» `x = rsinthetacosphi, y = rsinthetasinphi, z = rcostheta` `:. x^2+y^+z^2 = r^2(sin^2thetacos^2phi+sin^2thetasin^2phi+cos^2theta)` `=r^2(sin^2theta(1-sin^2phi)+sin^2thetasin^2phi+cos^2theta)` `=r^2(sin^2theta-sin^2thetasin^2phi+sin^2thetasin^2phi+cos^2theta)` `=r^2(sin^2theta+cos^2theta)` `=r^2*1 = r^2` `:. x^2+y^+z^2 = = r^2` So, ` x^2+y^2+z^2` is independent of `theta,phi.` |
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| 87. |
Differentiate w.r.t. x: `(i)tan^(-1){sqrt((1+cosx)/(1-cosx))}" "(ii)tan^(-1){sqrt((1+sinx)/(1-sinx))}`A. `(x)/(2)`B. `(-x)/(2)`C. `(pi)/(2)+(x)/(2)`D. `(pi)/(2)-(x)/(2)` |
| Answer» Correct Answer - A | |
| 88. |
If `sin(sinx+cosx)=cos(cosx-sinx),`and largest possible value of `sinxi spi/k ,`then the value of `k`is______. |
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Answer» `sin(sinx+cosx)=cosx(cosx-sinx)` `sin(sinx+cosx)=sin(pi/2-(cosx-sinx))` `sinx+cosx=4pi+(-1)^n[pi/2-cosx+sinx]` `sinx+cosx=pi/2-cosx+sinx` `2cosx=pi/2` `cosx=pi/4` `cos^2x+sin^2x=1` `sin^2x=1-pi^2/16` `sinx=sqrt(16-pi^2)/4` `cosx+sinx=pi-(pi/2-cosx+sinx)` `cosx+sinx=pi/2+cosx-sinx` `2sinx=pi/2` `sinx=pi/4` `k=4`. |
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| 89. |
The solution of the system of equations `sinxsiny=(sqrt(3))/4,cosxcosy=(sqrt(3))/4`are`x_1=pi/3+pi/2(2n+k); n , k in I``y_1=pi/6+pi/2(k-2n); n , k in I``x_2=pi/6+pi/2(2n+k); n , k in I``y_2=pi/3+pi/2(k-2n); n , k in I` |
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Answer» `sinx*siny=sqrt3/4` `cosx*cosy+sinxsiny=sqrt3/2` `cos(x-y)=sqrt3/2` `x-y=2npipmpi/6` `cosxcosy=sqrt3/4` `cosxcosy-sinxsiny=0` `cos(x+y)=0` `x+y=(2n+1)pi/2` `x+y=npi+pi/2` `2x=2npipmpi/6+npi+pi/2` `x=(2n+1)pi/2+pi/4pmpi/2` `2y=4pi+pi/2-2npipmpi/6` `y=(1-2n)pi+pi/4pmpi/2`. |
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| 90. |
General solution of `2tan^(2)x=sec^(2)x` isA. `npi-(pi)/(4), ninZ`B. `npi+(pi)/(4), ninZ`C. `npipm(pi)/(4), ninZ`D. `2npipm(pi)/(4), ninZ` |
| Answer» Correct Answer - C | |
| 91. |
Which of the following expresses the circumference of a circleinscribed in a sector `O A B`with radius `Ra n dA B=2a ?`(a)`2pi(R a)/(R+a)`(b) `(2piR^2)/a``2pi(r-a)^2`(d) `2piR/(R-a)` |
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Answer» We can create a diagram with the given details. Please refer to video for the diagram. From the diagram, `sintheta = r/(OM) = a/(AO)` `=>r/a = (OM)/R` `=>r/a = (R-r)/R` `=>r/a = 1 - r/R` `=>r/a+r/R = 1` `=>r((a+R)/(aR) ) = 1` `=> r = (aR)/(a+R)` `:.` Circumference ` = 2pir = 2pi ((aR)/(a+R))` So, option `(a)` is the correct option. |
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| 92. |
If `3sinbeta=sin(2alpha+beta)`then `tan(alpha+beta)-2tanalpha`is(a)independent of `alpha`(b)independent of `beta`(c)dependent of both `alpha` and `beta`(d)independent of both `alpha` and `beta` |
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Answer» `tan(alpha+beta) - 2tanalpha` `= sin(alpha+beta)/cos(alpha+beta) - sinalpha/cosalpha - tanalpha` `=(sin(alpha+beta)cosalpha - sinalphacos(alpha+beta))/(cosalphacos(alpha+beta)) - tanalpha` `=(sin(alpha+beta-alpha))/(cosalphacos(alpha+beta)) - sinalpha/cosalpha` `=(sinbeta)/(cosalphacos(alpha+beta)) - sinalpha/cosalpha` `=(sinbeta - sinalphacos(alpha+beta))/(cosalphacos(alpha+beta)) ` `=(2(sinbeta - sinalphacos(alpha+beta)))/(2cosalphacos(alpha+beta)) ` `=(2sinbeta - 2sinalphacos(alpha+beta))/(2cosalphacos(alpha+beta)) ` `=(2sinbeta - (sin(alpha+alpha+beta) + sin(alpha-alpha-beta)))/(2cosalphacos(alpha+beta))` `=(2sinbeta - (sin(2alpha+beta) - sinbeta))/(2cosalphacos(alpha+beta))` `=(3sinbeta - sin(2alpha+beta))/(2cosalphacos(alpha+beta))` It is given that, `3sinbeta = sin(2alpha+beta) => 3sinbeta - sin(2alpha+beta) = 0` So, it becomes, `=0/(2cosalphacos(alpha+beta))` `=0` `:. tan(alpha+beta) - 2tanalpha =0` So, given expression is independent of both `alpha` and `beta`. |
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| 93. |
If `secx+sec^2x=1`then the value of `tan^8x-tan^4x-2tan^2x+1`will be equal to0 (b)1 (c) 2(d) 3 |
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Answer» `secx+secc^2x=1` `secx=1-sec^2x=-tan^2x` Squaring both side `sec^2x=tan^4x` `1+tan^2x=tan^4x` Squaring both side `(1+tan^2x)^2=tan^8x` `1+tan^4x+2tan^2x=tan^8x` `tan^8x-tan^4x-2tan^2x+1=2`. |
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| 94. |
If sec `alpha and alpha`are the roots of `x^2-p x+q=0,`then(a) `p^2=q(q-2)`(b) `p^2=q(q+2)`(c)`p^2q^2=2q`(d) none of these |
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Answer» a)`p^2=alpha^2+secc^2alpha+2alphasecalpha` `q(q-2)=alphasecalpha(alphasecalpha-2)` `=alpha^2sec^2alpha-2alphasecalpha` `p^2!=q(q-2)` b)`q(q+2)=alpha^2sec^2alpha+2alphasecalpha!=(alpha+secalpha)^2` c)`p^2q^2=(alpha^2+sec^2alpha+2alphasecalpha)(alpha^2sec^2alpha)` `=alpha^4secc^4alpha+alpha^2secc^4alpha+2alpha^3sec^3alpha` `2q=2alphasecalpha` `p^2q^2!=2q` d)none. |
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| 95. |
Consider a triangle `A B C`and let `a , ba n dc`denote the lengths of the sides opposite to vertices `A , B ,a n dC`, respectively. Suppose `a=6,b=10 ,`and the area of triangle is `15sqrt(3)dot`If `/_A C B`is obtuse and if `r`denotes the radius of the incircle of the triangle, then the value of `r^2`is |
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Answer» We know, `Delta = 1/2ab sinC` `:. 15sqrt3 = 1/2*6*10*sinC` `=>sinC = sqrt3/2` As, `C` is obtuse, so `C = 120^@.` Now, we know, `cosC = (a^2+b^2 - c^2)/(2ab)` `cos120^@ = (36+100 - c^2)/(120)` `=>-1/2(120) = 136-c^2` `=>c^2 = 196` `=>c = 14` `:. s = (a+b+c)/2 = (6+10+14)/2 = 15` Now, `r = Delta/s = (15sqrt3)/15 = sqrt3` `:. r^2 = 3.` |
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| 96. |
If sides of triangle `A B C`are `a , ba n dc`such that `2b=a+c`then`b/c >2/3`(b) `b/c >1/3``b/c |
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Answer» `a+b>c` `2b-c+b>c` `3b-c>c` `3b>2c` `b/c>2/3` Option C is correct. |
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| 97. |
Let `A B C D`be a quadrilateral with area `18`, side `A B`parallel to the side `C D ,a n dA B=2C D`. Let `A D`be perpendicular to `A Ba n dC D`. If a circle is drawn inside the quadrilateral `A B C D`touching all the sides, then its radius is`3`(b) 2(c) `3/2`(d) 1 |
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Answer» With the given details, we can draw the diagram. Please refer to video to see the diagram. From the diagram, In `Delta BEC`, `BC^2 = BE^2+CE^2` `=>(a-r+2a-r)^2 = (2r)^2 +a^2` `=>a = 3/2r` Now, area of the quadrilateral, `(a*2r)+(1/2*a*2r) = 18` `=>6ar = 36` `=>ar = 6` `=>3/2r^2 = 6` `=>r^2 = 4` `=>r = 2` So, option `(b)` is the correct option. |
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| 98. |
Prove that: `((sin7x+sin5x)+(sin9x+sin3x))/((cos7x+cos5x)+(cos9x+cos3x))=tan6x` |
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Answer» Here, we will use, `sinA+sinB = 2sin((A+B)/2)cos((A-B)/2)` `cosA+cosB = 2cos((A+B)/2)cos((A-B)/2)` `L.H.S. = ((sin7x+sin5x)+(sin9x+sin3x))/((cos7x+cos5x)+(cos9x+cos3x))` `=(2sin6xcosx+2sin6xcos3x)/(2cos6xcosx+2cos6xcos3x)` `=(2sin6x(cosx+cos3x))/(2cos6x(cosx+cos3x))` `=(sin6x)/(cos6x) = tan6x = R.H.S.` |
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| 99. |
Find the value of `cot54^@/(tan36^@)+tan20^@/(cot70^@)` |
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Answer» Correct Answer - 2 `cot54^@/(tan36^@)+tan20^@/(cot70^@)=(cot(90^@-36^@))/(tan36^@)+(cot(90^@-70^@))/(cot70^@)` `=tan36^@/(tan36^@)+cot70^@/(cot70^@)=2` |
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| 100. |
If `cos25^0+sin25^0=p ,`then `cos50^0`is(a)`sqrt(2-p^2)`(b) `-sqrt(2-p^2)`(c)`psqrt(2-p^2)`(d) `-psqrt(2-p^2)` |
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Answer» `cos25^@+sin25^@ = p` Squaring both sides, `=>cos^2 25^@+sin^2 25^@+2sin25^@cos25^@ = p^2` `=>1+sin 50^@ = p^2` `=>sin50^@ = p^2-1` `=>cos50^@ = sqrt(1-sin^2 50^@)` `=sqrt(1-(p^4+1-2p^2))` `=psqrt(2-p^2)` `:. cos50^@ = psqrt(2-p^2)` So, option `c` is the correct option. |
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