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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. |
If `-1lexle0` then `sin^(-1)x` equalsA. `pi-sin^(-1)sqrt(1-x^(2))`B. `tan^(-1)(x)/sqrt(1-x^(2))`C. `-cot^(-1)sqrt(1-x^(2))/(x)`D. none of these |
Answer» Let `sin^(-1)x=theta` Then `x =sin theta` Now `-1 lt x lt 0 rarr-(pi)/(2)ltthetalt0` `pi-sin^(-1)sqrt(1-x^(2))` `=pi-sin^(-1)(cos theta)` `As-(pi)/(2)ltthetalt0` `therefore 0ltpi-thetalt(pi)/(2)rarr 0ltpi-sin^(-1)sqrt(1-x^(2))lt(pi)/(2)` `therefore sin^(-1)x ne pi-sin^(-1)sqrt(1-x6(2))` So option (a) is not correct we have `tan^(-1)(x)/sqrt(1-x^(2))` `=tan^(-1)(sin htheta)/(sqrt(1-sin^(2)theta)=tan^(-1)(tan theta)=theta =sin^(-1)x` Thus option (b) is correct We have `-cot^(-1)sqrt(1-x^(2))/(x)=cos^(-1)(cos theta)//(sin theta)=-cos^(-1)(cot theta)` `=cosT^(-1)(cos(-theta))` `-theta` `-sin^(-1)x` Thus option (c ) is also not correct |
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52. |
Find the value of `sin^(-1)(cos(sin^(-1)x))+cos^(-1)(sin(cos^(-1)x))`A. `(pi)/(4)`B. `(pi)/(2)`C. `(3pi)/(4)`D. 0 |
Answer» We have `sin^(-1){cos(sin^(-1)x)}+cos^(-1){sin(cos^(-1)x)}` `=sin6(-1){cos((pi)/(2)cos^(-1)x)}+cos^(-1){sin(cos^(-1)x)}` `sin^(-1){sin(cos^(-1)x)}+cos^(-1){sin(cos^(-1)x)}=(pi)/(2)` |
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53. |
If `sin^(-1)sqrt(x^2+2x + 1) + sec^(-1)sqrt(x^2 + 2x + 1) = pi/2; x!= 0,` then the value of `2sec^(-1)(x/2) + sin^(-1)(x/2)` is equal toA. `-(pi)/(2)` onlyB. `{-(3pi)/(2),(pi)/(2)}`C. `(3pi)/(2)` onlyD. `-(3pi)/(2)` only |
Answer» We have `sin^(-1)sqrt(x^(2)+2x+1)+sec^(-1)sqrt(x^(2)+2x+1)=(pi)/(2)` `sin^(-1)sqrt(x^(2)+2x+1)+cos^(-1)sqrt(1)/(x^(2)+2x+1)` `rarr x^(2)+2x+1=1 rarr x=0 -2` for x =0 we find that `sec^(-1)(-1)+sin^(-1)(-1)` `=2xxpi-(pi)/(2)=(3pi)/(2)` |
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54. |
If `-1lexle0` then `sin{tan^(-1)((1-x^(2))/(2x))-cos^(-1)((1-x^(2))/(1+x^(2)))}` is equal toA. 1B. `-1`C. 0D. none of these |
Answer» We know that `tan^(-1)=-pi+cot^(-1)(1/x)if xlt0` `therefore {tan^(-1)(1-x^(2))/(2x) -cosf^(-1)(1-x^(2))/(1+x^(2))}` `=sin{-pi+co^(-1)(2x)/(1-x^(2))-cos^(-1)(1-x^(2))/(1+x^(2))}` `=sin{-(pi)/(2)-2tn^(-2)tan^(-1)x+2tan^(-1)x+2tan^(-1)x}` `=sin(-(pi)/(2))=-1` |
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55. |
The value of `cos [tan^-1 {sin (cot^-1 x)}]` isA. `sqrt(x^(2)+2)/(x^(2)+3)`B. `sqrt(x^(2)+2)/(x^(2)+1)`C. `sqrt(x^(2)+1)/(x^(2)+2)`D. none of these |
Answer» We have `cos[tan^(-1) sin(cot(-1)x)]` `cos{tan^(-1)(1)/sqrt(1+x^(2))}=sqrt(1+x^(2))/sqrt(2+x^(2))=sqrt(1+x^(2))=sqrt(1+x^(2))/(2+x^(2))` |
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56. |
If `|cos^-1((1-x^2)/(1+x^2))| < pi/3,` then x belongs to the intervalA. `[-1//sqrt(3),1//sqrt(3)]`B. `(-1//sqrt(3),1//sqrt(3))`C. `(0,1//sqrt(3))`D. none of these |
Answer» We have `|cos^(-1)((1-x^(2))/(1+x^(2)))|lt(pi)/(3)` ltrbgt `rarr 1/2 lt (1-x^(2))/(1+x^(2)) lt (pi)/(3)` `rarr 1+x^(2)lt2-2x^(2)lt2+2x^(2)` `rarr 1+x^(2)lt2-2x^(2) and 2-2x^(2)lt2+2x^(2)` `rarr 3x^(2)-1lt0and 4x^(2)gt0 rarr -(1)sqrt(3)ltxlt(1)/sqrt(3)` |
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57. |
Which of the following pairs of function/functions has same graph?`y=tan(cos^(-1)x); y=(sqrt(1-x^2))/x``y=t a n(cot^(-1)x);y=1/x``y="sin"(tan^(-1)x); y=x/(sqrt(1-x^2))``y="cos"(tan^(-1)x); y=s in(cot^(-1)x)`A. `y = tan (cos^(-1) x), y = (sqrt(1 - x^(2)))/(x)`B. `y = tan (cot^(-1) x), y = (1)/(x)`C. `y = sin (tan^(-1) x), y = (x)/(sqrt(1 + x^(2)))`D. `y = cos (tan^(-1) x), y = sin (cot^(-1) x)` |
Answer» Correct Answer - A::B::C::D (1) `y = tan (cos^(-1)x) and y = (sqrt(1 -x^(2)))/(x)` or `y = tan (tan^(-1).(sqrt(1 -x^(2)))/(x)) = (sqrt(1 -x^(2)))/(x) and y = (sqrt(1 -x^(2)))/(x)` `:. D_(1) = [-1, 1] - {0} and D_(2) = [-1, 1] {0}` So functions are identical and hence they have same graph (2) `y = tan (cot^(-1) x) and y = (1)/(x)` `:. y = tan (tan^(-1).(1)/(x)) adn y = (1)/(x)` or `y = (1)/(x) adn y = (1)/(x)` `:. D_(1) = R - {0} and D_(2) = R - {0}` So, functions are identical and hence they have same graph (3) `y = sin (tan^(-1) x) and y = (x)/(sqrt(1 + x^(2)))` or `y = sin (sin^(-1).(x)/(sqrt(1 + x^(2)))) = (x)/(sqrt(1 + x^(2))) and y = (x)/(sqrt(1 + x^(2)))` or `D_(1) = R, D_(2) = R` So, functions are identical and hence they have same graph. (4) `y = cos (tan^(-1) x) and y = sin (cot^(-1) x)` or `y = cos (tan^(-1) x) = cos [cos^(-1).(1)/(sqrt(1 + x^(2)))] = (1)/(sqrt(1 + x^(2)))` and `y = sin (cot^(-1) x) = sin (sin^(-1).(1)/(sqrt(1 + x^(2)))) = (1)/(sqrt(1 + x^(2)))` `:. D_(1) = R and D_(2) = R` So, functions are identical and hence they have same graph. |
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58. |
If z = `sec^(-1) (x + 1/x) + sec^(-1) (y + 1/y)`, where xy< 0, then the possible value of z is (are)A. `(8pi)/(10)`B. `(7pi)/(10)`C. `(9pi)/(10)`D. `(21pi)/(20)` |
Answer» Correct Answer - C::D `xy lt 0` `rArr x + (1)/(x) ge 2, y + (1)/(y) le -2` or `x + (1)/(x) le -2, y + (1)/(y) ge 2` `x + (1)/(x) ge 2` `rArr sec^(-1) (x + (1)/(x)) in [(pi)/(3), (pi)/(2))` `y + (1)/(y) le -2` `rArr sec^(-1) (y + (1)/(y)) in ((pi)/(2), (2pi)/(3)]` `rArr z in ((5pi)/(6), (7pi)/(6))` |
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59. |
The value of `k(k >0)`such that the length of the longest interval in which the function `f(x)=sin^(-1)|sink x|+cos^(-1)(cosk x)`is constant is `pi/4`is/ are8 (b) 4(c) 12 (d)16A. 8B. 4C. 12D. 16 |
Answer» Correct Answer - B `f(x) = sin^(-1) |sin kx| + cos^(-1) (cos kx)` Let `g(x) = sin^(-1) |sin x| + cos^(-1) (cos x)` `g(x){(2x,0 le x le(pi)/(2)),(pi,(pi)/(2) lt x le (3pi)/(2)),(4pi - 2x,(3pi)/(2) lt x le 2pi):}` `g(x)` is periodic with period `2pi` and is constant in the continuous interval `[2n pi + (pi)/(2), 2n pi + (3pi)/(2)] ("where " n in I) and f(x) = g(kx)`. So, `f(x)` is constant in the interval `[(2npi)/(k) + (pi)/(2k), (2n pi)/(k) + (3pi)/(2k)]` Thus, `(pi)/(4) = (3pi)/(2k) -(pi)/(2k)` or `(pi)/(k) = (pi)/(4)` or `k = 4` |
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60. |
If `cos^(-1)(x/2)+cos^(-1)(y/3) = theta`, prove that `9x^2- 12xycostheta+ 4y^2= 36 sin^(2)theta`A. 36B. `-36 sin^(2) theta`C. `36 sin^(2) theta`D. `36 cos^(2) theta` |
Answer» We have `cos^(-1)(x)/(2)+cos^(-1)(y)/(3)=theta` `rarr cos^(-1)(xy)/(6)sqrt(1-(x^(2))/(4)sqrt(1-(y^(2))/(9))}=theta` `rarr xy-sqrt(4-x^(2))sqrt(9-y^(2))=6costheta` `rarr xy-6cos theta^(2)=(4-x^(2))(9-y^(2))` `rarr-12xy cos theta + 36 cos^(2) theta =36 -4y^(2)-9x^(2)` `9x^(2)+4y^(2)-12xy cos theta =36 sin^(2) theta` |
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61. |
The sum of the series `cot^-1 2 + cot^-1 8 +cot^-1 18+cot^-1 32................`isA. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(6)`D. none of these |
Answer» Correct Answer - B | |
62. |
Prove that `sin^(-1) (3/5) +cos^(-1) (15/17)+ sin^(-1) (36/85)=pi/2` |
Answer» `cos^(-1)15/17+sin^(-1)(3/5sqrt(1-(36/85)^2)+sqrt(1-(3/5)^2)*36/85)` `cos^(-1)15/17+sin^(-1)(3/5*77/85+4/5*36/85)` `cos^(-1)(15/17)+sin^(-1)(375/425)` `cos^(-1)(15/17)+sin^(-1)(15/17)=pi/2` |
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63. |
Solve for x, `tan^(-1)2x+tan^(-1)3x=pi/4` |
Answer» `tan^(-1)2x+tan^(-1)3x=pi/4` `rArrtan^(-1)((2x+3x)/(1-6x^2))=pi/4 " only if "(2x)(3x) lt 1` `rArr tan^(-1)((5x)/(1-6x^2))=pi/4 "i.e.,"6x^2 lt 1` `(5x)/(1-6x^2)=pi/4` `:.x in((-1)/sqrt6),1/sqrt6)` =1 `rArr 5x=1-6x^2` `rArr 6x^2-5x-1=0` `rArr(6x-1)(x+1)=0` `x=-1 or x=1/6` But x = -1 does not lie in domain of x therefore `x=1/6` |
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64. |
Let `alpha=som^(-1)((36)/(85)),beta=cos^(-1)(4/5)a n dgamma=tan^(-1)(8/(15))`then`cotalpha+cotbeta+cotgamma=cotalphacotbetacotgamma``tanalphatanbeta+tanbetatangamma+tanalphatangamma=1``tanalpha+tanbeta+tangamma=tanalphatanbetatangamma``cotalphacotbeta+cotbetacotgamma+cotalphacotgamma=1` |
Answer» `alpha=sin^(-1)(36/85),sinalpha=36/85,tanalpha=36/77` `beta=cos^(-1)(4/5)=sinbeta=3/5,tanbeta=3/4` `gamma=tan^(-1)(8/15),tangamma=8/15` `tan(alpha+beta+gamma)=(sumtanalpha-pitanalpha)/(1-sumtanalphatanbeta)` `=(36/77+3/9+8/15-36/77*3/4*8/15)/(1-(36/77*3/4+8/15*3/4+8/15*36/77))` `tan(alpha+beta+gamma)=oo` `alphaa+beta+gamma=pi/2` `1=tanalphatanbeta+tanbetatangamma+tanalphatangamma` `cotalpha+cotbeta+cotgamma=cotalphacotbetacotgamma`. |
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65. |
If `cot^(-1)(sqrt(cosalpha))-tan^(-1)(sqrt(cosalpha))=x ,`then `sinx`is`tan^2alpha/2`(b) `cot^2alpha/2`(c) `tan^2alpha`(d) `cotalpha/2` |
Answer» `cot^-1(sqrtcosalpha) - tan^-1(sqrtcosalpha) = x` `=>tan^-1(1/sqrtcosalpha) - tan^-1(sqrtcosalpha) = x` `=>tan^-1((1/sqrtcosalpha -sqrtcosalpha)/(1+1/sqrtcosalpha*sqrtcosalpha)) = x` `=>tan^-1((1-cosalpha)/(2sqrtcosalpha)) = x` `=>tan x = (1-cosalpha)/(2sqrtcosalpha)` `=>cotx = 1/tanx = (2sqrtcosalpha)/(1-cosalpha)` `=>cosecx = sqrt(1+cot^2x) = (1+cosalpha)/(1-cosalpha)` `=>sinx = 1/(cosecx) = (1-cosalpha)/(1+cosalpha)` `=>sinx = sin^2(alpha/2)/cos^2(alpha/2) = tan^2(alpha/2)` So, option `(a)` is the correct option. |
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66. |
If `sec^(-1)x=cosec^(-1) y` then `cos^(-1)(1/x)+cos^(-1)(1/y)`=A. `pi`B. `(pi)/(4)`C. ``-(pi)/(4)`D. `(pi)/(2)` |
Answer» Correct Answer - D | |
67. |
The value of `cos(tan^-1 (tan 2))` isA. `1//sqrt(5)`B. `-1//sqrt(5)`C. cos 2D. `-cos 2` |
Answer» Correct Answer - D | |
68. |
If `sin^(-1)((2x)/(1+x^(2)))+cos^(-1)((1-x^(2))/(1+x^(2)))=4 tan^(-1) x` thenA. `x in -(-oo,-1)`B. `xin (1,oo)`C. `x in [0,1]`D. `x in [-1,0)` |
Answer» Correct Answer - C | |
69. |
If `tan^(-1) x + tan^(-1)y + tan^(-1)z= pi then x + y + z` is equal toA. xyzB. 0C. 1D. 2 xyz |
Answer» Correct Answer - A | |
70. |
Prove that: `2tan^(-1)(1/2)+tan^(-1)(1/7)=tan^(-1)(31/17)` |
Answer» We know, `tan^-1x+tan-^-1y = tan^-1((x+y)/(1-xy))` Here, `2tan^-1(1/2)` can be written as: `tan^-1(1/2)+tan^-1(1/2) = tan^-1((1/2+1/2)/(1-1/2*1/2)) =tan-1(4/3) ` `L.H.S. = 2tan^-1(1/2)+tan^-1(1/7)` `=tan^-1(4/3)+tan^-1(1/7) = tan^-1((4/3+1/7)/(1-4/3*1/7))` `=tan^-1((31/21)/(17/21)) = tan^-1(31/17) = R.H.S.` |
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71. |
Find the value of the following: `cos^(-1)(cos((13pi)/6))` |
Answer» `cos^-1(cos13pi/6)` can be written as `cos^-1(cos(2pi+pi/6))` As, `cos(2pi +x) = cos x`, So, our expression becomes `=cos^-1(cos(pi/6)) = pi/6` |
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72. |
Show that `sin^(-1)(12)/(13)+cos^(-1)4/5+tan^(-1)(63)/(16)=pi`. |
Answer» Let, `sin^-1(12/13) = x and cos^-1(4/5) = y and tan^-1(63/16)=z->(1)` Then,`sin x = 12/13 and cos y = 4/5` `tan x = 5/12 and tan y = 3/4` `x = tan^-1(5/12) and y = tan^-1(3/4)->(2)` From (1) and (2), `sin^-1(12/13) + cos^-1(4/5) = tan^-1(12/5)+tan^-1(3/4) ` We know, `tan^-1x+tan-^-1y = tan^-1((x+y)/(1-xy))` `tan^-1(12/5)+tan^-1(3/4) = tan^-1((12/5+3/4)/(1-12/5**3/4))` `=tan^-1((63/20)/(-4/5)) = tan^-1(-63/16)->(3)` From (1),`tan^-1(63/16) = z`, So, `tan z = 63/16` `=>-tan z = -63/16` `=>tan(pi - z) = -63/16` `=>tan^-1(-63/16) = pi-z->(4)` So, From(1), (3) and (4), `L.H.S. = pi-z+z = pi = R.H.S.` |
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73. |
Prove that:`sin^(-1)(8/17)+sin^(-1)(3/5)=tan^(-1)(77/36)` |
Answer» Let, `sin^-1(8/17) = x` and `sin^-1(3/5) = y->(1)` Then, `sin x = 8/17 and sin y = 3/5` If we create right angle triangles for x and y, we get `tan x = 8/15 and tan y = 3/4` `x=tan^-1(8/15) and y = tan^-1(3/4)->(2)` From (1) and (2),`L.H.S = tan^-1(8/15)+tan^-1(3/4)` We know, `tan^-1x+tan-^-1y = tan^-1((x+y)/(1-xy))` So,`=tan^-1(8/15+3/4)/(1-(8/15)*(3/4))` `=tan^-1(77/36) = R.H.S.` |
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74. |
If `a_(1),a_(2),a_(3),….a_(n)` is a.p with common difference d then `tan{tan^(-1)((d)/(1+a_(1)a_(2)))+tan^(-1)((d)/(1+a_(2)a_(3))) +..+ tan^(-1)((d)/(1+a_(n-1)a_(n)))}` is equal toA. `((n-1)d)/(a_(1)+a_(n))`B. `((n-1)d)/(1+a_(1)a_(n))`C. `(nd)/(1+a_(1)a_(n))`D. `(a_(n)-a_(1))/(a_(n)+a_(1))` |
Answer» Correct Answer - B | |
75. |
Two angles of a triangle are `cot^-1 2` and `cot^-1 3,` then the third angle isA. `(pi)/(4)`B. `3(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(3)` |
Answer» Correct Answer - B | |
76. |
Let a, b and c be positive real numbers. Then prove that `tan^(-1) sqrt((a(a + b + c))/(bc)) + tan^(-1) sqrt((b (a + b + c))/(ca)) + tan^(-1) sqrt((c(a + b+ c))/(ab)) = pi`A. `pi//4`B. `pi//2`C. `pi`D. 0 |
Answer» Correct Answer - C | |
77. |
If `sin^(-1)(x/5)+cose c^(-1)(5/4)=pi/2`then a value of x is:A. 4B. 5C. 1D. 3 |
Answer» Correct Answer - D | |
78. |
`sin(2sin^(-1) 0.8)=`A. `sin 1.2^(@)`B. `sin 1.6^(@)`C. 0.478D. 0.96 |
Answer» Correct Answer - D | |
79. |
The value of `cos[1/2 cos^(-1){cos(sin^(-1)((sqrt63)/(8)))}]` isA. `3/16`B. `3/8`C. `3/4`D. `3/2` |
Answer» Correct Answer - C | |
80. |
The value of `sin^(-1)`(sin 10) isA. `3pi-10`B. `10-3pi`C. `3pi+10`D. `4pi-10` |
Answer» We know that `sin^(-1)(sin theta)=theta if -(pi)/(2)letheta le(pi)/(2)` Here `theta =10` radians which does not lie between `-(pi)/(2) and (pi)/(2)` But `3pi-theta` lies between `-(pi)/(2)and (pi)/(2)` also `sin(3pi-10)=sin10` `therefore sin^(-1)(sin10)=sin^(-1)sin(3pi-10)=3pi-10` |
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81. |
Find the value of `sin^(-1)(sin5)+cos^(-1)(cos10)+tan^(-1){tan(-6)}+cot^(-1){cot(-10)}dot` |
Answer» `sin^-1(sin5)+cos^-1(cos10)+tan^-1(tan(-6))+cot^-1(cot(-10))` `=sin^-1(sin(5-2pi))+cos^-1(cos(4pi-10))-tan^-1(tan(6-2pi))+pi - cot^-1(cot(10-3pi))` `=5-2pi+4pi-10-6+2pi+pi-10+3pi` `=8pi-21` `:. sin^-1(sin5)+cos^-1(cos10)+tan^-1(tan(-6))+cot^-1(cot(-10)) = 8pi-21` |
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82. |
Which of the following angles is greater ? `theta_(1) = sin^(-1).(4)/(5) + sin^(-1).(1)/(3) and theta_(2) = cos^(-1).(4)/(5) + cos^(-1).(1)/(3)` |
Answer» `theta_(1) + theta_(2) = sin^(-1).(4)/(5) + sin^(-1).(1)/(3) + cos^(-1).(4)/(5) + cos^(-1).(1)/(3)` `= (sin^(-1). (4)/(5) + cos^(-1).(4).(5)) + (sin^(-1).(1)/(3) + cos^(-1).(1)/(3))` `= (pi)/(2) + (pi)/(2) = pi` `theta_(2) = cos^(-1) [((4)/(5)) ((1)/(3)) - sqrt(1 - ((4)/(5))^(2)) sqrt(1 - ((1)/(3))^(2))]` `= cos^(-1) [(4)/(5).(1)/(3) - (3)/(5).(sqrt8)/(3)]` Now, `(4 - 6 sqrt2)/(15) lt 0` Therefore, `theta_(2)` is obtuse and hence `theta_(1)` is acute. `:. theta_(1) lt theta_(2)` |
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83. |
If `3tan^(-1)(1/(2+sqrt(3)))-tan^(-1)1/x=tan^(-1)1/3,`then `x`is equal to1 (b) 2(c) 3 (d)`sqrt(2)` |
Answer» `3tan^-1(1/(2+sqrt3)) - tan^-1 (1/x) = tan^-1 (1/3)` `=>3tan^-1(1/(2+sqrt3)**(2-sqrt3)/(2-sqrt3)) = tan^-1 (1/x) + tan^-1 (1/3)` `=>3tan^-1(2-sqrt3) = tan^-1 (1/x) + tan^-1 (1/3)` Now, we know, `tan15^@ = 2-sqrt3=>15 = tan^-1(2-sqrt3)` `:. 3*15^@ = tan^-1((1/3+1/x)/(1-(1/3)(1/x)))` `=>tan45^@ = (x+3)/(3x-1)` `=>1(3x-1) = x+3` `=>2x = 4` `=> x = 2.` |
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84. |
The value of `cos^(-1)(cos10)` isA. `4pi-10`B. `10-4pi`C. `3pi-10`D. `10-3pi` |
Answer» We know that `cos^(-1)(cos theta)=theta if theta le theta le pi` Here `theta =10` radians clearly it does not lie between 0 and pi However `(4pi-10)` lies between 0 and `pi` such that `cos (4pi-10) =cos10` `therefore cos^(-1)(cos 10)=cos^(-1)(cos(4pi-10)-4pi-10` |
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85. |
`sin^(-1)(5/x) +sin^(-1)(12/x) =pi/2` |
Answer» `sin^(-1)(5/x)=pi/22-sin^(-1)(12/x)` `sin^(-1)(5/x)=cos^(-1)(12/x)=0` `sin^(-1)(5/x)=theta` `sintheta=5/x` `cos^(-1)(12/x)=theta` `costheta=12/x` `x=sqrt(12^2+5^2` `x=sqrt169` `x=13`. |
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86. |
Solve `sin^(-1)(1-x)-2sin^(-1)x=pi/2` |
Answer» `sin^(-1)(1-x)-2sin^(-1)x=pi/2` Let `x=siny` `sin^(-1)(1-siny)=2y=pi/2` `sin^(-1)(1-siny)=(pi/2+2y)` `1-siny=cos2y` `1-siny=1-2sin2y` `2sin^2y=siny` `siny=0,siny=1/2` `siny=0,x=0` `sin=1/2,x=1/2` `sin^(-1)1=pi/2` `sin^(-1)1/22-2sin^(-1)1/2` `-sin^(-1)1/2` `-x/6!=pi/2`. |
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87. |
Prove that : `tan^(-1)1/2+tan^(-1)1/5+tan^(-1)1/8=pi/4`A. `pi/4`B. `pi/3`C. `tan^(-1)(12/5)`D. `tan^(-1)(4/3)` |
Answer» Correct Answer - A | |
88. |
Taking only principal values, the values of `cos^(-1)(-1/2)+sin^(-1)(-sqrt(3)/2)` is equal to |
Answer» Here, we will use, `cos^-1(-x) = pi - cos^-1x` `sin^-1(-x) = -sin^-1x` Now, `cos^-1(-1/2) + sin^-1(-sqrt3/2)` `=pi - cos^-1(1/2) +(-sin^-1(sqrt3/2))` `=pi - pi/3 - pi/3 = pi - (2pi)/3 = pi/3` `:. cos^-1(-1/2) + sin^-1(-sqrt3/2) = pi/3` |
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89. |
If `0 le x le 1 then cos^(-1)(2x^(2)-1)` equalsA. `2cos^(-1)x`B. `2sin^(-1)x`C. `pi-2cos^(-1)x`D. `pi+2cos^(-1)x` |
Answer» Correct Answer - A | |
90. |
If `(1)/sqrt(2) le x le 1 then sin^(-1) 2xsqrt(1-x^(2))` equalsA. `2 sin^(-1)x`B. `pi-2 sin^(-1)x`C. `-pi--2sin^(-1)x`D. none of these |
Answer» Correct Answer - B | |
91. |
`sin^(-1) (5/x) +sin^(-1) (12/x) =pi/2`A. -13B. 13C. 15D. 17 |
Answer» Correct Answer - B | |
92. |
`cos^(-1) (cos (2 cot^(-1) (sqrt2 -1)))` is equal toA. `sqrt2 -1`B. `(pi)/(4)`C. `(3pi)/(4)`D. none of these |
Answer» Correct Answer - C `cos^(-1) (cos (2 cot^(-1) (sqrt2 -1))) = cos^(-1) (cos (2(67.5^(@))))` `= cos^(-1) (cos (135^(@)))` `= 135^(@) = (3pi)/(4)` |
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93. |
Satement-1: if `1/2lexle1`then `cos^(-1)x-sin^(-1){x/2+sqrt(3-3x^(2))/(2)}` is equal to `(pi)/(5)` Statement-2: `sin^(-1)(2xsqrt(1-x^(2))=2sin^(-1)x if x in -(1)sqrt(2),(1)sqrt(2))`A. Statement-1 is is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
Answer» Let x = cos `theta` then `1/2 le x le 1 rarr 1/2 le cos theta le 1 rarr 0 le theta le (pi)/(3)` `therefore -theta-sin^(-1){1/2cos theta +sqrt(2)/(2)sin theta}` `=theta -sin^(-1){sin(pi)/(6)cos theta +cos (pi)/(6) sin theta}` `=theta -sin^(-1){sin(theta+(pi)/(6))}` `=theta -(theta +(pi)/(6))=-(pi)/(6)` so statement 1 true let x =sinx `theta` then `(1)/sqrt(2) le x le (1)sqrt(2) rarr -(1)/sqrt(2) le sin theta le (1)/sqrt(2) rarr =(pi)/(4) le theta le (pi)/(4)` `there sin^(-1)2xsqrt(1-x^(2))=sin^(-1)(sin 2 theta) =2 sin^(-1)x` so statement 2 true we have `1/2 le x le 1 rarr x^(2)+3/4gt 1` using : `sin^(-1) x + sin^(-1)y =pi -sin^(-1)xsqrt(1-y^(2))+ysqrt(1-x^(2))` when `0 lt x,y le 1 and x^(2) a+ y^(2) gt1` we have `therefore sin^(-1)((x)/(2)+(sqrt(3-3x^(2))/(2))=pi-sin^(-1)x-(pi)/(3)=2pi)/(3)-=sin^(-1)x` `=cot^(-1)x=(2x)/(3)+sin^(-1)x=(pi)/(2)-(2pi)/(3)=-(pi)/(6)` so statement -1 is true statement -2 true (see theory ) |
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94. |
`tan^(-1)(tansqrt(1-theta))=sqrt(1-theta)` whenA. `-(pi)/(2) lt theta (pi)/(2)`B. `theta gt (4-pi)/(4)`C. `thetalt (4-pi)/(4)`D. `(4-pi^(2))/(4) lt le 1` |
Answer» Clearly`sqrt(1-theta)` is real if `theta le1` Now `tan^(-1)tansqrt(1-theta)=sqrt(1-theta)` `rarr 0lesqrt(1-theta)lt (pi)/(2) and theta le1` `rarr 0le1-theta lt pi^(2)/(4)and theta le 1` `rarr -1le-theta le(pi^(2))/(4)-1 and theta le1` `rarr 1-(pi^(2))/(4)lt theta le1` and `theta le1 rarr (4-pi^(2))/(4)lt theta le1` |
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95. |
If `theta in [(pi)/(2),3(pi)/(2)] then sin^(-1)(sin theta)` equalsA. `theta`B. `pi - theta`C. `2pi - theta`D. `-pi + theta` |
Answer» We have `theta in [(pi)/(2),(3pi)/(2)]rarr-theta in[-(3pi)/(2),(pi)/(2)]rarr pi -thetha in[-(pi)/(2)-(pi)/(2)]` Aslo sin`(pi-theta) =sin theta` `therefore sin^(-1)(sin-theta)=sin^(-1)sin(pi 0 theta)=pi-theta` |
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96. |
If `sin^(- 1)(3/x)+sin^(- 1)(4/x)=pi/2` then `x=`A. 3B. 5C. 7D. 11 |
Answer» We have `sin^(-1)(3)/(x)+sin^(-1)(4)/(x)=(pi)/(2)` `rarr sin^(-1)(3)/(x)=(pi)/(2)-sin^(-1)(4)/(x)` `rarr sin^(-1)(3)/(x)=cos^(-1)(4)/(x)` `rarr sin^(-1) (3)/(x)=sin^(-1)(sqrt(x^(2)-16)/(x))` `rarr(3)/(x)=sqrt(x^(2))=-16)/(x)rarrx=5` |
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97. |
`sin(2sin^(-1) 0.8)=` |
Answer» Correct Answer - `0.96` `sin {2 sin^(-1) (0.8)} = sin {sin^(-1) (2 xx 0.8 sqrt(1 - (0.8)^(2)))}` `= sin (sin^(-1) 0.96) = 0.96` |
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98. |
If `sin^(-1) ((2a)/(1+a^2))+ sin^(-1) ((2b)/(1+b^2)) = 2 tan^(-1)x` then `x=`A. `(a-b)/(1+ab)`B. `(b)/(1+ab)`C. `(b)/(1-ab)`D. `(a+b)/(a-ab)` |
Answer» Correct Answer - D | |
99. |
Which of the following is the solution set of the equat where `x in (0, 1)`, is equal to |
Answer» Correct Answer - B `tan (sin^(-1) (cos (sin^(-1) x))) tan(cos^(-1) (sin (cos^(-1) x)))` `= tan(sin^(-1) (cos (cos^(-1) sqrt(1 -x^(2)))))` `tan {cos^(-1) (sin (sin^(-1) sqrt(1 -x^(2))))}` `= tan (sin^(-1) sqrt(1 -x^(2)) tan (cos^(-1) sqrt(1 - x^(2)))` `= tan (cos^(-1) x) tan (sin^(-1) x)` `= tan(cos^(-1) x) tan (pi//2 - cos^(-1 x)` `= tan (cos^(-1) x) cot (cos^(-1) x) = 1` |
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100. |
If `sin^(-1) : [-1, 1] rarr [(pi)/(2), (3pi)/(2)] and cos^(-1) : [-1, 1] rarr [0, pi]` be two bijective functions, respectively inverse of bijective functions `sin : [(pi)/(2), (3pi)/(2)] rarr [-1, 10 and cos : [0, pi] rarr [-1, 1] " then " sin^(-1) x + cos^(-1) x` isA. `(pi)/(2)`B. `pi`C. `(3pi)/(2)`D. not a constant |
Answer» Correct Answer - D Let `sin^(-1) x = theta` `rArr x = sin theta, (pi)/(2) le theta le (3pi)/(2)` Now, `cos^(-1) x = cos^(-1) (sin theta)` `= cos^(-1) (-cos ((3pi)/(2) - theta))` `= pi - cos^(-1) (cos ((3pi)/(2) - theta))` `= pi -((3pi)/(2) - theta), " as " 0 le (3pi)/(2) - theta le pi` `= theta -(pi)/(2) = sin^(-1) x - (pi)/(2)` Hence, `sin^(-1) x + cos^(-1) x = 2 sin^(-1) x - (pi)/(2)` |
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