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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. |
Solve: `3tan(theta-15^(@))=tan(theta+15^(@))` |
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Answer» `3tan(theta-15^(@))=tan(theta+15^(@))` `rArr 3sin(theta-15^(@))/(cos(theta-15^(@)))=(sin(theta +15^(@))cos(theta-15^(@))` `rArr 3.{2cos(theta+15^(@))sin(theta-15^(@))}=2sin(theta+15^(@)).cos(theta-15^(@))` `rAr 3(sin2theta-sin30^(@))=sin2theta+sin30^(@)` `=4 xx 1/2` `rArr sin2theta=1` `=sinpi/2` `therefore 2theta=npi+(-1)^(n)pi/2` `rArr theta=(npi)/(2)+(-1)^(n)pi/4` Ans. |
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| 52. |
The solution of `costheta.cos2theta.cos3theta=1/4,0 lt theta lt pi/4` is:A. `pi/8`B. `pi/6`C. `pi/9`D. `pi/12` |
| Answer» Correct Answer - a | |
| 53. |
`tantheta+tan(pi/3+theta)+tan((2pi)/(3)+theta)=3` |
| Answer» `theta=1/3(npi+pi/4)` | |
| 54. |
The maximum value of `5costheta+3cos(theta+pi/3)+3` is:A. 5B. 11C. 10D. None of these |
| Answer» Correct Answer - c | |
| 55. |
Solve: `sin3alpha=4sinalpha. Sin(theta+alpha).sin(theta-alpha)` where `alpha ne n pi, n in I` |
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Answer» `sin3alpha=4sinalpha. Sin(theta+alpha).sin(theta-alpha)` `rArr 3-4sin^(2)alpha=4sin^(2)theta-4sin^(2)alpha` `rArr 4sin^(2)theta=3` `rArr sin^(2)=3/4` `= (sqrt(3)/(2))^(2)=sin^(2)pi/3` `rArr theta=npi+-pi/3`. Ans. |
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| 56. |
Solve: `tan4theta=-cot(pi/6+theta)` |
| Answer» `theta=1/3(npi+(2pi)/(3))` | |
| 57. |
Prove that: `sin(-420^(@))cos390^(@)+cos(-660^(@))sin330^(@)=1` |
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Answer» LHS `=sin(-420^(@))cos390^(@) + cos(-660^(@))sin330^(@)` `-sin420^(@)cos390^(@)+cos660^(@)sin330^(@)` `=-sin(360^(@)+60^(@))cos(360^(@)+30^(@))+cos(360^(@)+300^(@))sin(360^(@)+30^(@))` `=-sin60^(@)cos30^(@)+cos300^(@)(-sin30^(@))` `=-sqrt(3)/2 xx sqrt(3)/2-1/2cos(360^(@))-60^(@)` `=-3/4-1/2 cos60^(@)` `=-3/4 -1/2 xx 1/2` `-3/4 -1/4 =-1` =RHS Hence proved. |
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| 58. |
Find the value of: `sin75^(@)` |
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Answer» `sin75^(@)` `=sin(45^(@)+30^(@))` `=sin45^(@)cos30^(@)+cos45^(@)sin30^(@)` `=1/sqrt(2). sqrt(3)/2+ 1/sqrt(2).1/2 = (sqrt(3)+1)/(2sqrt(2))` Ans. |
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| 59. |
`tan(pi/4+theta)+tan(pi/4-theta)=4` |
| Answer» `theta=npi+-pi/6` | |
| 60. |
Find the general value of `theta` from the equation `tantheta+tan2theta+tan3theta=tantheta*tan2theta*tan3theta.` |
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Answer» `tantheta+tan2theta+tan3theta=tanthetatan2theta tan3theta` `rArr tantheta+tan2theta=-tan3theta+tanthetatan2theta tan3theta=-tan3theta(1-tanthetatan2theta)` `rArr (tantheta+tan2theta)/(1-tanthetatan2theta)=-tan3theta` `rArr tan(theta+2theta)=-tan3theta` `rArr tan3theta=-tan3theta` `rArr 2tan3theta=0` `rArr tan3theta=0` `rArr 3theta=npi` `rArr theta=(npi)/(3)`. Ans. |
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| 61. |
Find the values of `sin75^(@)`. |
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Answer» `sin75^(@)=sin(45^(@)+30^(@))` `=sin45^(@)cos30^(@)+cos45^(@)sin30^(@)` `=1/sqrt(2).sqrt(3)/2+1/sqrt(2).1/2=(sqrt(3)+1)/(2sqrt(2))`. Ans. |
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| 62. |
`tantheta+tan2theta+sqrt(3)tanthetatan2theta=sqrt(3)` |
| Answer» `theta=1/3(npi+pi/4)` | |
| 63. |
`(cosx-cosy)^(2)+(sinx-siny)^(2)=?`A. `4sin^(2)(x-y)/(2)`B. `4cos^(2)(x-y)/(2)`C. `4sin(x+y)/(2)sin(xy)/(2)`D. `4cos(x+y)/(2)cos(x-y)/(2)` |
| Answer» Correct Answer - A | |
| 64. |
If `A+B=45^(@)`, then `(1+tanA)(1+tanB)=?`A. 2B. `-2`C. 1D. 0 |
| Answer» Correct Answer - A | |
| 65. |
If `2costheta=x+1/x`, then `2cos2theta=`?A. `x^(2)-1/x^(2)`B. `x^(2)+1/x^(2)`C. `x-1/x`D. None of these |
| Answer» Correct Answer - B | |
| 66. |
Solve: `sin^(2)theta+2costheta+1/4=0` |
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Answer» `sin^(2)theta+2costheta+1/4=0` `4sin^(2)theta+8costheta+1=0` `rArr 4(1-cos^(2)theta)+8costheta+1=0` `rArr 4cos^(2)theta-8costheta-5=0` `rArr 4cos^(2)theta-10costheta+2costheta-5=0` `rArr 2costheta(2costheta-5)+1(2costheta-5)=0` `rArr (2costheta-5)(2costheta+1)=0` `rArr 2costheta-5=0` or `2costheta+1=0` Now `2costheta-5=0` `rArr costheta=-1/2=-cospi/3` `=cos(pi-pi/3)=cos(2pi)/(3)` `therefore theta=2npi +- (2pi)/(3)` Ans. |
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| 67. |
Solve: `cos^(2)theta-sinthetacostheta-1/2=0` |
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Answer» `cos^(2)theta-sinthetacostheta-1/2=0` `rArr 2cos^(2)theta-2sinthetacostheta-1=0` `rArr (2cos^(2)theta-1)=2sinthetacostheta` `rArr cos2theta=sin2theta` `rArr tan2theta=1` `rArr tan2theta=tanpi/4` `rArr 2theta=npi+pi/4` `rArr theta=(npi)/2+pi/8` Ans. |
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| 68. |
In `DeltaABC`, prove that: `a(cos^(2)C/3-cos^(2)B/2)=(b-c).cos^(2)A/2` |
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Answer» LHS= `a(cos^(2)C/2-cos^(2)B/2)` `=a[(s(s-c))/(ab)-(s(s-b))/(ac)]` `=a.s/a[(s-c)/(b)-(s-b)/( C)]` ` |
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| 69. |
`sin5theta+sin3theta+sintheta=0` |
| Answer» `theta=(npi)/(3), theta=npi+-pi/3` | |
| 70. |
Prove that: `sin5A=5sinA-20sin^(3)A+16sin^(5)A` |
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Answer» LHS `=sin5A=sin(3A+2A)` `=(3sinA-4sin^(3)A)(1-2sin^(2)A)+(4cos^(3)A-3cosA)2sinAcosA` `=(3sinA-4sin^(3)A)(1-2sin^(2)A)+(4cos^(2)A-3)2sinAcos^(2)A` `=(3sinA-4sin^(3)A)(1-2sin^(2)A)+[4(1-sin^(2)A)-3]2sina(1-sin^(2)A` `=(3sinA-4sin^(3)A)(1-2sin^(2)A)+(2sinA-2sin^(3)A)(1-4sin^(2)A)` `=3sinA-6sin^(3)A-4sin^(3)A+8sin^(5)A +2sinA-8sin^(3)A-2sin^(3)A+8sin^(5)A` `5sinA-20sin^(3)A+16sin^(5)A` = RHS Hence Proved. |
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| 71. |
If arcs of same length in two circles subtend angles of `60^0a n d75^0`at their centers, find the ratios of their radii. |
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Answer» Let radius of first circle be `r_(1)` and radius of second circle be `r_(2)` cm. Angle subtended by arc of first circle at center `theta_(1)=60^(@)` then length of arc `l_(1) = r_(1)theta_(1)` Angle subtended by arc of second circle at center `theta_(2)=75^(@)` then length of arc `l_(2)=r_(2)theta_(2)`. `therefore` According to the problems, length of arcs are same. `therefore l_(1)=l_(2)` `rArr r_(1)0_(1)= r_(2)theta_(2)` `rArr r_(1)/r_(2)=r_(2)theta_(2)` `rArr r_(1)/r_(2) = theta_(2)/theta_(1) = 75^(@)/60^(@)` `=5/4 =5:4` Therefore, ratio of radii of circles `r_(1):r_(2)=5:4` Ans. |
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| 72. |
`sqrt(3)" cosec "20^(@)-sec20^(2)=?`A. 1B. 2C. 4D. None of these |
| Answer» Correct Answer - c | |
| 73. |
Find the value of other five trigonometric function `cotx=3/4`, x lies in third quadrant. |
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Answer» Given that `cotx=3/4` and x lies in third quadrant. `cotx=3/4 rArr tan x=1/(cotx)` `rArr tanx=1/(3//4) = 1 xx 4/3=4/3` `therefore` x lies in third quadrant so, all ratio except tan and cot will be negative. `therefore "cosec"^(2)x = 1+(3/4)^(2) = 1+9/16=25/16` `rArr "cosec"x = -5/4` `rArr sinx=1/(-5/4)= 1 xx (-4/5) =-4/5` `cosx=(cosx)/(sinx). sinx` `=cotx. sinx` `=(3/4). (-4/5)` `=-12/20 = -3/5` `secx = 1/(cosx)=1/(-3/5) = 1 xx -5/3 -5/3` Therefore, `sinx=-4/5, cosx=-3/5` `tanx=4/3, "cosec"x=-5/4, secx =-5/3` Ans. |
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| 74. |
Find the values of trignometric functions in Questions 6 to 10. `sin765^(@)` |
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Answer» `sin765^(@) = sin(2 xx 360^(@) + 45^(@))` `=sin45^(@)=1/sqrt(2)` Therefore, `sin765^(@)=1/sqrt(2)` Ans. |
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| 75. |
`sin(-11pi)/(3)` |
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Answer» `sin(-11pi)/(3) = -sin(11pi)/(3)` `=-sin(2 xx 2pi -pi/3)` `=-(-sinpi/3) = sinpi/3 = sqrt(3)/2` Therefore, `sin(-11pi)/(3) = sqrt(3)/2`. Ans. |
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| 76. |
`sin3x+sin2x-sinx=4sinxcosx/2cos(3x)/(2)` |
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Answer» LHS `=sin3x+sin2x-sinx` `=sin3x+2cos(2x+x)/(2)sin(2x-x)/(2)` `=2sin(3x)/(2)cos(3x)/(2)+2cos(3x)/(2)sinx/2` `=2cos(3x)/(2) [ sin(3x)/(2)+sinx/2]` `=2cos(3x)/(2)[2sin((3x)/(2)+x/2)sin(2x-x)/(2)]` `=2cos(3x)/(2)[sin(3x)/(2)+sinx/2]` `=2cos(3x)(2)[2sin(2x)/(2)cosx/2]` `=2cos(3x)/(2)[2sinxcosx/2]` `=4sinxcosx/2cos(3x)/(2)`= RHS Hence Proved. |
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| 77. |
If `cotA=5/12` and A lies in 3rd quadrant, then find the values of five other trignometric ratios. |
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Answer» `cotA = 5/12` `rArr tanA=1/(cotA) = 12/15`. Now, `"cosec^(2)A=1+cot^(2)A` `=1+(5/12)^(2)=169/144` `rArr "cosec"A = -13/12` `therefore` cosecA is negative in 3rd quadrant sinA`=1/("cosec"A)=-12/13` `cosA=(cosA)/(sinA).sinA` `=cotA. sinA` `=5/12(-12/13)=-5/13` `secA=1/(cosA)=-13/5` |
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| 78. |
Find the values of the following : i) `tan ((19pi)/3)` ii) `sec((-22pi)/3)` |
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Answer» `tan (19pi)/(3) = tan(19 xx 180^(@))/(3)` `tan(1140^(@))` `tan(3xx 360^(@) + 60^(@))=sqrt(3)` Ans. |
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| 79. |
Evaluate: `tan (13pi)/(12)` |
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Answer» `tan(13pi)/(12)=tan(pi+pi/12)` `=tanpi/12=tan(pi/3-pi/4)` `=(tanpi/3-tanpi/4)/(1+tanpi/3tanpi/4)=(sqrt(3)-1)/(1+sqrt(3))` Ans. |
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| 80. |
`sinpi/10+sin(13pi)/(10)=?`A. 1B. `1/2`C. `-1/2`D. None of these |
| Answer» Correct Answer - C | |
| 81. |
The interior angles of a polygon are in A.P. The smallest anlge is `((2pi)/(3))^(c )` and common difference is `5^(@)`. Find the number of sides in the polygon. |
| Answer» Correct Answer - 9 | |
| 82. |
If tan A `=a/b`, then find the value of `("a sin A- b cos A")/("a sin A + b cos A")`. |
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Answer» tan A = `a/b` Now `("a sinA-b cosA")/("a sin A+bcosA")=((a sinA)/(cosA)-b)/((a sinA)/(cosA)+b)` (Divide `N^( r)` and `D^(r )` by cos A) `=("a tanA-b")/("a tan A+b")= (a.a/b-b)/(a.a/b+b)` `=((a^(2)-b^(2))/b)/((a^(2)+b^(2))/b)= (a^(2)-b^(2))/(a^(2)+b^(2))`. Ans |
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| 83. |
Find the angle between the large hand and small hand of a clock at the time `4:30`. |
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Answer» For small hand Angle subtends in 12 hours `=360^(@)` 4 hrs 30 min = 4 hrs + `30/60` hrs `=9/2` hrs Angle subtends in `9/2` hours `=9/2 xx 360^(@)/12=135^(@)` For large hand 1 hour = angle subtends in 60 min. = `360^(@)` `rArr` Angle subtends in 30 min. =`180^(@)` `therefore` Angle between large hand and small hand at the time `4:30` `=180^(@)- 135^(@)=45^(@)`. |
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| 84. |
In `DeltaABC, a=x^(2)-1, b=2x+1, angleC=120^(@)`, find c. |
| Answer» Correct Answer - `(x^(2)+x+1)` | |
| 85. |
Prove that: `sinA.sin(60^(@)+A).sin(60^(@)-A)=1/4sinA` |
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Answer» LHS `=sinA.sin(60^(@)+A).sin(60^(@)-A)` `=1/2sinA.[2sin(60^(@)+A)-(60^(@)-A)]` `1/2sinA[cos{(60^(@)+A)-(60^(@)-A)}]-cos{(60^(@)+A)+(60^(@)-A)}` `=1/2sinA[cos2A-cos120^(@)]` `=1/2[cos2AsinA-cos(90^(@)+30^(@)).sinA]` `=1/4[2cos2AsinA+2sin30^(@)sinA]` `=1/4[sin(2A+A)-sin(2A-A)=2.1/2.sinA]` `=1/4[sin3A-sinA+sinA]` `=14sin3A=R.H.S` Hence Proved. |
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| 86. |
The sides of a triangle `ABC` are in the ratio `3:4:5`. If the perimeter of triangle `ABC` is `60,` then its lengths of sides are: |
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Answer» Let `A=3x, B=4x, C=5x` but `A+B+C=180^(@)` `rArr 3x+4x+5x=180^(2)` `rArr 12x=180^(@)` `rArr x=15^(@)` `therefore A=3 xx 15^(@)=45^(@)` `B=4 xx 15^(@)=60^(2)` `C=5 xx 15^(2)=75^(@)` Now sinA=`sin45^(@)=1/sqrt(2)` `sinB=sin60^(2)=sqrt(3)/2` `sinC=sin75^(@)=(sin45^(@)+30^(@))` `=sin45^(2)cos30^(2)+cos45^(2)sin30^(@)` `=1/sqrt(2) xx sqrt(3)/2 + 1/sqrt(2)xx 1/2=(sqrt(3)+1)/(2sqrt(2))` `therefore a:b:c = sinA:sinB:sinC` `=1/sqrt(2):sqrt(3)/2:(sqrt(3)+1)/(2sqrt(2))` `=2:sqrt(6):(sqrt(2)+1)`. Ans. |
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| 87. |
In `Delta ABC`, prove that: `(a+b+c)/(a-b+c)=cotA/2cotC/2` |
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Answer» RHS= `cotA/2.cotC/2` `=sqrt(s(s-a))/(s-b)(s-c).sqrt((s(s-c))/(s-a)(s-b)` `=sqrt(s^(2))/((s-b)^(2))=s/(s-b)=(2s)/(2s-2b)` `=(a+b+c)/(a+b+c-2b)=(a+b+c)/(a-b+c)`= LHS Hence Proved. |
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| 88. |
Prove that: `sec(45^(@)+A)sec(45^(@)-A)=2secA` |
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Answer» LHS= `sec(45^(@)+A).sec(45^(@)-A)` `=2/((cos45^(@)+A)+(45^(@)-A))=2/(0+cos2A)=2/(cos2A)=2sec2A`=RHS Hence Proved. |
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| 89. |
`cos(pi/4-x)cos(pi/4-y)-sin(pi/4-x)sin(pi/4-y)=sin(x+y)` |
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Answer» LHS. `=cos(pi/4-x) cos(pi/4-y)-sin(pi/4-x)sin(pi/4-y)` `=cos[pi/2-(x+y)]` `=sin(x+y)`= RHS Hence Proved. |
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| 90. |
Solve `cos3x+cosx-cos2x=0` |
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Answer» Given equations: `cos3x+cosx-cos2x=0`, or `2cos(3x+x)/(2)cos(3x-x)/(2)-cos2x=0,` or `2cos2xcosx-cos2x=0` or `cos2x.(2cosx-1)=0`, `rArr cos2x=0` or `2cosx-1=0` If `cos2x=0`, then `2x=(2n+1)pi/2 rArr x=(2n+1)pi/4` where `n in z` If `2cosx-1=0`, then `cosx=1/2 rArr cosx=cospi/3` `rArr x=2npi+-pi/3` Therefore, the general solution of given equation is `x=(2n+1)pi/2`. or `2npi+- pi/3, n in Z` Ans. |
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| 91. |
`sec^(2)2x=1-tan2x` |
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Answer» Given equation is `sec^(2)2x=1-tan2x` or `1+tan^(2)2x=1-tan2x` or `tan^(2)2x+tan2x=0` or `tan2x(1+tan2x)=0` `rArr tan2x=0` or `1+tan2x=0` If `tan2x=0`, then `2x=npi` `rArr x=(npi)/(2)` where `n in Z` and `tan2x = 1=tan(pi-pi/4)` `rArr tan2x=tan(3pi)/(4)` `rArr 2x=npi + (3pi)/(4)` `rArr x=(npi)/(2) + (3pi)/(8)` where `n in Z` Therefore, the general solution of given equations is `x=(npi)/(2)` or `(npi)/(2)+(3pi)/(8), n in Z` Ans. |
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| 92. |
`(sinx+sin3x)/(cosx+cos3x) = tan2x` |
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Answer» LHS `=(sinx+sin3x)/(cosx+cos3x)` `=(2sin(x+3x)/(2)cos(x-3x)/(2))/(2cos(x+3x)/(2)cos(x-3x)/(2)) = (sin(x+3x)/(2))/(cos(x+3x)/(2))` `=(sin2x)/(cos2x)=tan2x`=RHS Hence Proved. |
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| 93. |
Solve: `cos4x=cos2x` |
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Answer» Given equation: `cos4x=cos2x` `rArr 4x=(2npi+-2x)` then, taking `+` sign `4x = 2npi+2x` or `4x-2x=2npi` `rArr x=npi, n in Z` or `4x+2x=2npi` or `6x=2npi rArr x=(npi)/(3), n in Z` Therefore, general solution of given equation is `x=npi`. or `x=(npi)/(3), n in Z` Ans. |
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| 94. |
(sin 3x + sin x) sin x + (cos 3x - cos x) cos x =0 |
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Answer» LHS `=(sin3x+sinx)sinx+(cos3x-cosx)cosx` `=[2sin(3x+x)/(2)cos(3x-x)/(2)]sinx+[2sin(3x+x)/(2)sin(x-3x)/(2)]cosx` `=[2sin2xcosx]sinx+[2sin2xsini(-x)]cosx` `=2sin2xsinx.cosx-2sin2x sinx cosx` =0 = RHS Hence Proved. |
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| 95. |
`cotxcot2x-cot2xcot3x-cot3xcotx=1` |
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Answer» `therefore cot3x=cot(2x+x)` or `cot3x=(cotxcot2x-1)/(cot2x+cotx)` or `cot3x(cot2x+cotx)=cotxcot2x-1` or `-cotxcot2x+cot2xcot3x+cot3xcotx=-1` or `-(cotxcot2x-cot2xcot3x-cot3xcotx)=-1` Therefore, `cotxcot2x-cot2xcot3x-cot3xcotx=1`. Hence Proved. |
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| 96. |
`(sinx-sin3x)/(sin^(2)x-cos^(2)x) = 2sinx` |
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Answer» LHS` = (sinx-sin3x)/(sin^(2)x-cos^(2)x)` `=((2cos(x+3x))/(2)sin(x-3x)/(2))/(-(cos^(2)x-sin^(2)x))` `=(2cos2xsin(-x))/(-cos2x)` `=(-2cos2xsinx)/(-cos2x)` `(-2cos2xsinx)/(-cos2x)` =2sinx = RHS Hence Proved. |
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| 97. |
`cotx=-sqrt(3)` |
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Answer» `cotx=-sqrt(3)=-cotpi/6` `rArr cotx=cot(pi-pi/6)`or `cot(2pi-pi/6)` `rArr cotx=cot(5pi)/(6)` or `cot(11pi)/(6)` `rArr x=(5pi)/(6)` or `(11pi)/(6)` Therefore, for the equation `cotx=-sqrt(3)` Principle solution is `x=(5pi)/(6)` or `(11pi)/(6)` Ans. and general solution is `x=npi + (5pi)(6), n in Z`, Ans. |
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| 98. |
`2cos(pi/13)cos(9pi/13)+cos(3pi/13)+cos(5pi/13)=0` |
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Answer» LHS `2cospi/13 cos(9pi)/(13)+cos(3pi)/(13)+cos(5pi)/(13)=0` `=2cospi/13 cos(9pi)/(13) + 2cos((5pi)/(13)+(3pi)/(13))/(2) cos((5pi)/(13)-(3pi)/(13))/(2)` `=2cospi/13cos(9pi)/(13)+2cos(4pi)/(13)cospi/13` `=2cospi/13(cos(9pi)/(13) + cos(4pi)/(13))` `=2cospi/13[2cos((9pi)/(13)+(4pi)/(13))/(2) cos((9pi)/(13)-(4pi)/(13))/(2)]` `=2cospi/13[(2cospi/2cos(5pi)/(26))]` `=4cospi/13 cos(5pi)/(26) xx cospi/2` `=4cospi/13 cos(5pi)/(26) xx 0` `(therefore cospi/2=0)` =0 = RHS Hence Proved. |
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| 99. |
`secx=2` |
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Answer» `secx = 2=secpi/3` or `sec(2pi-pi/3)` `rArr x=pi/3` or `(2pi-pi/3)` `rArr x=pi/3`or `(5pi)/(3)` Therefore, principal solution of equation `secx=2` is `x=pi/3` or `(5pi)/(3)`. `secx=2` `1/(cosx)=2` `cosx=1/2=cos60^(@)` `cosx=cospi/3` and general solution is `x=2npi+-pi/3, n in Z` Ans. |
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| 100. |
`sinx+sin3x+sin5x=0` |
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Answer» Given equation: `sinx+sin3x+sin5x=0`, or `sin3x+(sin5x+sinx) =0`, or `sin3x+2sin(5x+x)/(2)cos(5x-x)/(2)=0`, or `sin3x+2sin3xcos2x=0`, or `sin3x(1+2cos2x)=0`, `rArr sin3x=0` or `1+2cos2x=0`, `sin3x=0` `3x=npi` `x=(npi)/(3)` If `1+2cos2x=0` `rArr cos2x=-1/2= cos(pi-pi/3)=cos(2pi)/(3)` `rArr x=npi+-pi/3` where `n in Z` Therefore, the general solution of given equation is `x=(npi)/(3)` or `npi +- pi/3, n in Z`. Ans. |
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