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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. |
सिद्ध कीजिए `tan^(-1)sqrt(x)=(1)/(2)cos^(-1) ""((1-x)/(1+x)), x in [0,1]` |
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Answer» Putting, `x=tan^(2)theta`, we get RHS `=1/2cos^(-1)(1-x)/(1+x)` `=1/2cos^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))` `=1/2cos^(-1)(cos2theta)` `=(1/2 xx 2theta) = theta=tan^(-1)sqrt(x)`= LHS `[therefore x=tan^(2)theta rArr tantheta=sqrt(x) rArr theta=tan^(-1)sqrt(x)]` Hence, `tan^(-1)=1/2cos^(-1)((1-x)(1+x))`. |
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| 52. |
Prove that `tan^(-1)(x/sqrt(a^(2)-x^(2)))=sin^(-1)x/a.` |
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Answer» Putting, `x=asintheta`, we get LHS`=tan^(-1)(x/sqrt(a^(2)+x^(2))` `=tan^(-1)(asintheta)/sqrt(a^(2)-a^(2)sin^(2)theta)`. `=tan^(-1)(asintheta)/(a costheta)=tan^(-1)(tantheta)` `=theta=sin^(-1)x/a`=RHS. `therefore tan^(-1)(1/(sqrt(x^(2)-1))=(pi/2-sec^(-1)x)`. |
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| 53. |
The value of `cos^(-1)(cos(14pi/6))` isA. `(13pi)/6`B. `(7pi)/6`C. `(5pi)/6`D. `2pi/6` |
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Answer» Correct Answer - D Let `cos^(-1)(cos(14pi)/6)=x,` where `x in [0,pi]`. Then, `cosx=cos(13pi)/6=cos(2pi+2pi/6)=cos2pi/6 rArr x=2pi/6`. |
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| 54. |
Find the value of `tan^(-1)(x/y)-tan^(-1)((x-y)/(x+y))` |
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Answer» We have, `tan^(-1)(x/y)-tan^(-1)((x-y)/(x+y))` `=tan^(-1)(x/y)-tan^(-1)(x-y)/(x+y)`. `=tan^(-1)(x/y)-tan^(-1){(x/y-1)/(x/y+1))}` `=tan^(-1)(1)=pi/4`. |
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| 55. |
If `sin(sin^(-1)(1/5)+cos^(-1)(x))=1` Find the value of x |
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Answer» We have, `sin(sin^(-1)1/5+cos^(-1)x)=1` `rArr sin^(-1)1/5+ cos^(-1)x=sin^(-1)1=pi/2`. `rArr cos^(-1)x=(pi/2-sin^(-1)1/5)` `rArr cos^(-1)x=cos^(-1)1/5 rArr x=1/5`. Hence, `x=pi/4`. |
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| 56. |
Find the value of: `tan^(-1)(1)+cos^(-1)(-1/2)+sin^(-1)(-1/2)`A. `pi`B. `(2pi)/3`C. `(3pi)/4`D. `pi/2` |
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Answer» Correct Answer - C `tan^(-1)1+cos^(-1)(-1/2)+sin^(-1)(-1/2)=pi/4+(pi-cos^(-1)1/2)=-sin^(-1)1/2` `(pi/4+pi-pi/3-pi/6)=(3pi)/4`. |
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| 57. |
`cos^-1 (1/2) + 2 sin^-1 (1/2).`A. `(2pi)/3`B. `(3pi)/3`C. `2pi`D. none of these |
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Answer» Correct Answer - A `cos^(-1)1/2+2sin^(-1)1/2=pi/3+(2 xx pi/6)=(pi/3+pi/3)=(2pi)/3`. |
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| 58. |
Find the principal value of: `sin^(-1)(-1/2)`A. `-pi/6`B. `(5pi)/6`C. `(7pi)/6`D. none of these |
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Answer» Correct Answer - A Let `sin^(-1)(-1/2)=x`, where `x in [-pi/2,pi/2]`. Then, `sinx=-1/2=-sinpi/6=sin(-pi/6) rArr x=-pi/6`. |
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| 59. |
Find the principal value of: `cos^(-1)(-1/(sqrt(2)))`A. `-pi/4`B. `pi/4`C. `(3pi)/4`D. `(5pi)/4` |
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Answer» Correct Answer - C Let `cos^(-1)(-1/sqrt(2))=x`, where `x in [0,pi]` Then, `cosx=-1/sqrt(2)=-cospi/4=cos(pi-pi/4)=cos(3pi)/4 rArr x=(3pi)/4` |
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| 60. |
Solve `tan^(-1)2x+tan^(-1)3x=pi/4`.A. `1/2` or `-2`B. `1/3` or `-3`C. `1/4` or `-2`D. `1/6` or `-1` |
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Answer» Correct Answer - D `tan^(-1)3x+tan^(-1)3=tan^(-1)2x=pi/4 rArr tan^(-1)((3x+2x)/(1-6x^(2))=pi/4`. `therefore (5x)/(1-6x^(2))=tanpi/4 =1 rArr 6x^(2)+5x-1=0` `rArr (x+1)(6x-1)=0`. `rArr x=-1` or `x=1/6`. |
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| 61. |
Prove the following results:`tan(cos^(-1)4/5+tan^(-1)2/3)=(17)/6`A. `13/6`B. `17/6`C. `19/6`D. `23/6` |
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Answer» Correct Answer - B `cos^(-1)x=tan^(-1)sqrt(1-16/25)/(4/5)=tan^(-1)3/4`. `therefore cos^(-1)4/5+tan^(-1)2/3=tan^(-1)3/4+tan^(-1)2/3=tan^(-1)(3/4+2/3)/(1-3/4 xx 2/3)=tan^(-1)(3/4+2/3)/(1-3/4 xx 2/3) = tan^(-1)17/6` `therefore` given exp. `=tan{tan^(-1)17/6}=17/6` `therefore` given exp. `=tan{tan^(-1)17/6}=17/6`. |
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| 62. |
If `sin^(-1)x+sin^(-1)y=(2pi)/3`, then `cos^(-1)x+cos^(-1)y` is equal toA. `pi/6`B. `pi/3`C. `pi`D. `(2pi)/3` |
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Answer» Correct Answer - B `sin^(-1)x+sin^(-1)y=(2pi)/3 rArr (pi/2-cos^(-1)x)+(pi/2-cos^(-1)y)=(2pi)/3`. `therefore cos^(-1)x+cos^(-1)y=(pi-(2pi)/3)=pi/3`. |
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| 63. |
Find the value of: i) `sin^(-1)(sinpi/3)` ii) `cos^(-1)(cos(2pi)/3))`, iii) `tan^(-1)(tanpi/4)` |
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Answer» We have, i) `sin^(-1)(sinpi/3)=pi/3`, since `pi/3 in [-pi/2, pi/2]`. ii) `cos^(-1)(cos(2pi)/3)=(2pi)/3`, Since `(2pi)/3 in [0,pi]` iii) `tan^(-1)(tanpi/4)=pi/4`, since `pi/4 in (-pi/2,pi/2)` |
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| 64. |
Prove that : `cot^(-1)7+cot^(-1)8+cot^(-1)18=cot^(-1)3` |
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Answer» We have LHS `=cot^(-1)7+cot^(-1)8+cot^(-1)18` `=(tan^(-1)1/7+ tan^(-1)1/8)+tan^(-1)1/18` `=tan^(-1)((1/7+1/8)/(1-1/7 xx 1/8)) + tan^(-1)1/18=tan^(-1)((15/56)/(55/56))+tan^(-1)1/18` `=(tan^(-1)3/11+tan^(-1)1/18)=tan^(-1)((3/11+1/18)/(1-3/11 xx 1/18))` `=tan^(-1)((1/7+1/8)/(1-1/7 xx 1/80) + tan^(-1)1/18=tan^(-1)((15/56)/(55/56))+tan^(-1)1/18` `=(tan^(-1)3/11 + tan^(-1)1/18)=tan^(-1)((3/11+1/18)/(1-3/11 xx 1/18))` `=tan^(-1)((65/198)/(195/198))=tan^(-1)1/3=cot^(-1)3` RHS. `therefore` LHS=RHS. |
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| 65. |
`Sin[pi/3-sin^(-1)(-1/2)]`A. 1B. 0C. `-1/2`D. none of these |
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Answer» Correct Answer - A `sin{pi/3-sin^(-3)(-1/2)}=sin{pi/3+sin^(-1)1/2}=sin(pi/3+pi/6)=sinpi/2=1`. |
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| 66. |
Prove that `ta n^-1 1/3 + ta n ^-1 1/5 + ta n ^-1 1/7 + ta n ^-1 1/8 = pi/4` |
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Answer» We have, LHS `=(tan^(-1)1/3+tan^(-1)1/5)+(tan^(-1)1/7+tan^(-1)1/8)`. `=tan^(-1)((1/3 +1/5))/(1-1/3 xx 1/5) + tan^(-1)((1/7+1/8)/(1-1/7 xx 1/8))` `=tan^(-1)((8/15)/(14/15)) + tan^(-1)((15/56)/(55/56)` `=tan^(-1)8/14+tan^(-1)15/55=tan^(-1)4/7+tan^(-1)3/11` `=tan^(-1)((4/7+3/11)/(1-4/7 xx 3/11))= tan^(-1)((65/77)/(65/77)) = tan^(-1)1=pi/4`= RHS. `therefore` LHS=RHS. |
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| 67. |
Evaluate: `sin(1/2cos^(-1)4/5)`A. `1/sqrt(5)`B. `2/sqrt(5)`C. `1/sqrt(10)`D. `2/sqrt(10)` |
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Answer» Correct Answer - C Let `cos^(-1)4/5=x`, where `x in [0,pi]`. Then, `cosx=4/5`. Since, `x in [0,pi] to 1/2 x in [0,pi/2] rArr sin1/2 x gt 0`. `sin(1/2cos^(-1)4/5) = sin1/2x =sqrt((1-cosx)/2)=sqrt((1-4/5)/(2))=1/sqrt(10)`. |
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| 68. |
Evaluate: `cos(sin^(-1)3/5+sin^(-1)5/(13))` |
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Answer» Let `sin^(-1)=3/5=A` and `sin^(-1)5/13=B`. Then, `A,B in [-pi/2,pi/2] rArr cosA gt 0` and `cosB gt 0`. `therefore sinA=3/5` and `sinB=5/13` `rArr cosA=sqrt(1-sin^(2)A)=sqrt(1-9/25)=sqrt(16/25)=4/5` and `cosB=sqrt(1-sin^(2)B) = sqrt(1-25/169)=sqrt(144/169)=12/13`. `therefore cos(sin^(-1)3/5+sin^(-1)5/13)=cos(A+B)=cosAcosB-sinAsinB` `=(4/5 xx 12/13) -(3/5 xx 5/13)` `=(48/65 -15/65) = 33/65`. |
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| 69. |
Prove that `tan^(-1)3/4+tan^(-1)3/5-tan^(-1)8/19=pi/4`. |
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Answer» We have, LHS `={tan^(-1)3/4+tan^(-1)3/5}-tan^(-1)8/19` `=tan^(-1){((3/4+3/5)/(1-3/4 xx 3/5))}-tan^(-1) 8/19` `=tan^(-1)(27/1)-tan^(-1)8/19` `=tan^(-1)((27/11-8/19))/(1+27/11 xx 8/9))=tan^(-1)(425/425) = tan^(-1)=pi/4` =RHS. `therefore` LHS=RHS. |
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| 70. |
Prove that `tan^-1(1/7)+tan^-1(1/13)=tan^-1(2/9)` |
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Answer» We know that `tan^(-1)x+tan^(-1)y=tan^(-1)(x+y)/(1-xy), xy lt 1`. Let `x=1/7` and `y=1/13`. Then, `xy=1/91 lt 1`. `therefore tan^(-1)1/7 +tan^(-1){(1/7+1/13)/(1-1/7 xx 1/13)}=tan^(-1)(20/90)=tan^(-1)2/9`. |
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| 71. |
if `tan^(-1)(4/3)=theta`, find the value of `costheta`. |
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Answer» `tan^(-1)4/3=theta`, where `theta in (-pi/2,pi/2)`. `therefore tantheta=4/3`. We know that `costheta gt 0`, when `theta=(-pi/2,pi/2)`. `therefore costheta=1/(sectheta)=1/sqrt(1+tan^(2)theta)=1/sqrt(1+16/9)=3/5`. |
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| 72. |
If `cot^(-1)(-1/5)=theta`, find the values of `sintheta`. |
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Answer» Given: `cot^(-1)(-1/5)=theta`, where `theta in (0,pi)`. `therefore cot(theta)=-1/5`. `sintheta gt 0` in `(0,pi)`. `sintheta =1/("cosec"theta)=1/sqrt(1+cot^(2)theta)=1/sqrt(1+1/25)=5/sqrt(26)`. |
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| 73. |
Prove that: i) `sin^(-1)(3x-4x^(3))=3sin^(-1)x, |x| le 1/2` ii) `cos^(-1)(4x^(2)-3x)=3cos^(-1)x,1/2 le x le 1` iii) `tan^(-1)""(3x-x^(3))/(1-3x^(2))=3tan^(-1)x, |x| lt 1/sqrt(3)` iv) `tan^(-1)x+tan^(-1)""(2x)/(1-x^(2))=tan^(-1)""(3x-x^(3))/(1-3x^(2))` |
| Answer» i) Put `x=sintheta`, ii) Put `x=costheta`, iii) Put `x=tantheta`, iv) Use `tan^(-1)A+tan^(-1)B` | |