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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. |
A and B are positive acute angles satisfying the equations `3cos^(2)A+2cos^(2)B=4and (3sinA)/(sinB)=(2cosB)/(cosA), then A+2B` is equal toA. `pi//4`B. `pi//3`C. `pi//6`D. `pi//2` |
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Answer» Correct Answer - D From the given relation, we have `2 cos^(2) B-1=3(1-cos^(2)A)` or `cos 2B = 3 sin^(2)A` …(1) `(3 sin A)/(sin B)=(2 cos B)/(cos A)` `rArr (3)/(2) sin 2A = sin 2B` `rArr 3 sin 2A = 2 sin 2B` …(2) Now cos (A + 2B) `= cos A cos 2B - sin A sin 2B` `= cos A(3 sin^(2)A)-sin A.(3)/(2)sin 2A` `= 3 cos A sin^(2)A-3 sin^(2)A cos A = 0` `therefore A+2B=90^(@)` |
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| 52. |
If `theta`is eliminated from the equations `x=a"cos"(theta-alpha)`and `y=bcos(theta-beta),`then`sec^2(alpha-beta)`(b) `cos e c^2(alpha-beta)``cos^2(-beta)`(d)`sin^2(alpha-beta)`A. `sec^(2)(alpha-beta)`B. `co sec^(2)(alpha-beta)`C. `cos^(2)(-beta)`D. `sin^(2)(alpha-beta)` |
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Answer» Correct Answer - D `(alpha-beta)=(theta-beta)-(theta-alpha)` `rArr cos(alpha-beta)=cos(theta-beta)cos(theta-alpha)+sin(theta+beta)sin(theta-alpha)` `=(y)/(b)xx(x)/(a)+sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))` `rArr[(xy)/(ab)-cos(alpha-beta)]^(2)=(1-(x^(2))/(a^(2)))(1-(y^(2))/(b^(2)))` or `(x^(2)y^(2))/(a^(2)b^(2))+cos^(2)(alpha-beta)-(2xy)/(ab)cos(alpha-beta)` `1-(y^(2)/(b^(2))-(x^(2))/(a^(2))+(x^(2)y^(2))/(a^(2)b^(2))` or `(x^(2))/(a^(2))+(y^(2))/(b^(2))-(2xy)/(ab)cos(alpha-beta)=sin^(2)(alpha-beta)` |
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| 53. |
Let `x=sin1^@` then find the value of the expression `1/(cos0^@cos1^@)+1/(cos1^@cos2^@)+....+1/(cos44^@cos45^@)`A. xB. `1//x`C. `sqrt(2)//x`D. `x//sqrt(2)` |
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Answer» Correct Answer - B `(1)/(sin1^(@))[(sin(1^(@)-0^(@)))/(cos 0^(@)cos1^(@))+(sin(2^(@)-1^(@)))/(cos1^(@)cos2^(@))+(sin(3^(@)-2^(@)))/(cos2^(@)cos3^(@))+............+(sin(45^(@)-44^(@)))/(cos44^(@)cos45^(@))]` `(1)/(sin1^(@))[tan1^(@)+(tan2^(@)-tan1^(@))+(tan3^(@)-tan2^(@))+(tan4^(@)-tan3^(@))+........+(tan45^(@)-tan44^(@))]` `=(1)/(sin1^(@))=(1)/(x)` |
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| 54. |
`sin^(2)45^(@)+cos^(2)45^(@)=`A. 1B. -1C. 0D. 2 |
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Answer» `sin^(2)45^(@)+cos^(2)45^(@)=((1)/(sqrt(2)))^(2)+((1)/(sqrt(2)))^(2)=(1)/(2)+(1)/(2)=1`. `therefore` (a) is correct . |
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| 55. |
If A,B,C are angles of a triangle, then `2sinA/2cos e c B/2sinC/2-sinAcotB/2-cosAi s`independent of A,B,C(b) function of A,Bfunction of C (d) none of theseA. independent of `A,B,C`B. function of A,BC. function of CD. none of these. |
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Answer» Correct Answer - A `2sin""(A)/(2)co sec"" (B)/(2)(sin((C)/(2))-cos((A)/(2))cos((B)/(2)))-cosA`. `2sin((A)/(2))co sec((B)/(2))xx(cos((A+B)/(2))-cos((A)/(2))cos((B)/(2)))-cosA` `2sin((A)/(2))co sec((B)/(2))(-sin((A)/(2))sin((B)/(2)))-cosA` `=-2sin^(2)((A)/(2))-cosA=-1` |
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| 56. |
If `(x)/(cos theta)=(y)/(cos(theta-(2pi)/(3)))=(z)/(cos(theta+(2pi)/(3))), then x+y+z=`A. 1B. 0C. `-1`D. none of these. |
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Answer» Correct Answer - B We have `(x)/(cos theta)=(y)/(cos(theta-(2pi)/(3)))=(1)/(cos(theta+(2pi)/(3)))` Therefore each ratio is equal to `(x+y+z)/(cos theta+cos(theta-(2pi)/(3)))=(x)/(cos(theta+(2pi)/(3)))=(x+y+z)/(0)` or `x+y+z=0` |
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| 57. |
`"tan"^(2)(pi)/(4) "sin"(pi)/(3) "tan"(pi)/(6) "tan"^(2)(pi)/(3) =`A. 1B. `1(1)/(2)`C. `1(1)/(3)`D. `1(1)/(4)` |
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Answer» `"tan"^(2)(pi)/(2) "sin"(pi)/(3) "tan"(pi)/(6) "tan"^(2)(pi)/(3)` `=tan^(2)45^(@) sin 60^(@) tan30^(@) tan^(2)60^(@) " " [ because pi=180^(@)]` `=(1)^(2)xx(sqrt(3))/(2)xx(1)/(sqrt(3)) xx (sqrt(3))^(2)=1xx(1)/(2) xx 3=(3)/(2) = 1(1)/(2)` `therefore ` (b) is correct. |
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| 58. |
Calculate : `3tan^(2)45^(@)-sin^(2)60^(@)-(1)/(3) cot^(2)30^(@)-(1)/(8) sec^(2)45^(@)`. |
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Answer» `3 tan^(2) 45^(@) -sin^(2)60^(@)-(1)/(3)cot^(2)30^(@)-(1)/(8)sec^(2)45^(@)` `=3xx(1)^(2)-((sqrt(3))/(2))^(2)-(1)/(3)xx(sqrt(3))^(2)-(1)/(8)xx(sqrt(2))^(2)` ` =3-(3)/(4)xx3-(1)/(8)xx2=3-(3)/(4)-1-(1)/(4)=(12-3-4-1)/(4)=(4)/(4)=1.` |
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| 59. |
Fill in the blanks : `sec 45^(@)=ulul(" 1 ")` |
| Answer» `(1)/(sqrt(2)), " since " sec 45^(@)=(1)/(cos45^(@))=(1)/((1)/(sqrt(2)))` | |
| 60. |
Write true or false : The value of `cos theta` can never be greater than 1. |
| Answer» True : Since `cos theta = ("base")/("hypotenuse")` and hypotenuse of a right-angled triangle can never be greater than its base with respect to any angle of it. | |
| 61. |
Write true or false : `sin theta` is a product of sin and `theta` . |
| Answer» False , `sin theta` is a ratio between the two sides of a right-angled triangle with respect to the angle `theta`. | |
| 62. |
Fill in the blanks : `cot 0^(@)= `_____________. |
| Answer» undefined , since `cot0^(@)=(cos 0^(@))/(sin0^(@))=(1)/(0)=`undefined. | |
| 63. |
Calculate : `(4)/(3)cot^(2)30^(@)+3 sin^(2) 60^(@)-2 "cosec"^(2)60^(@)-(3)/(4) tan^(2)30^(@)`. |
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Answer» `(4)/(3)cot^(2)30^(@)+3 sin^(2) 60^(@)-2 "cosec"^(2)60^(@)-(3)/(4) tan^(2)30^(@)` `=(4)/(3) xx (sqrt(3))^(2)+3xx((sqrt(3))/(2))^(2)-2xx((2)/(sqrt(3)))^(2)-(3)/(4)xx((1)/(sqrt(3)))^(2)` `=(4)/(3) xx 3+3xx(3)/(4)-2xx(4)/(3) -(3)/(4)xx(1)/(3)=4+(9)/(4)-(8)/(3)-(1)/(4)` `=(48+27-32-3)/(12)=(75-35)/(12)=(40)/(12)=(10)/(3)=3(1)/(3)` . |
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| 64. |
Prove that `cos 60^(@)=cos^(2)30^(@)-sin^(2)30^(@)`. |
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Answer» `LHS=cos 60^(@)` `=(1)/(2)` `RHS=cos^(2)30^(@)-sin^(2)30^(@)` `=((sqrt(3))/(2))^(2)-((1)/(2))^(2)=(3)/(4)-(1)/(4)=(3-1)/(4)` `=(2)/(4)=(1)/(2)`. `therefore LHS=RHS`. (Proved) |
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| 65. |
If `cos theta=0.6`, then show that `(5 sin theta-3 tan theta)=0`. |
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Answer» `5 sin theta -3 tan theta =5 sin theta -(3 sin theta)/( cos theta)` `=sin theta(5-(3)/( cos theta))` `=sin theta (5-(3)/(0.6)) " " [ because cos theta =0.6]` `=sin theta(5-(30)/(6))=sin theta(5-5)` `=sin theta xx 0=0`. [Proved] |
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| 66. |
Calculate : `sec^(2)45^(@)-cot^(2)45^(@)-sin^(2)30^(@)-sin^(2)60^(@)`. |
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Answer» `sec^(2) 45^(@)-cot^(2)45^(@)-sin^(2)30^(@)-sin^(2)60^(@)` `=(sqrt(2))^(2)-(1)^(2)-((1)/(2))^(2)-((sqrt(3))/(2))^(2)` `=2-1-(1)/(4)-(3)/(4)=(8-4-1-3)/(4)=(0)/(4)=0`. |
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| 67. |
If `a |
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Answer» Correct Answer - A `f(x)=3cos x+5sin(x-pi//6)` `=(1)/(2)cosx+5xx(sqrt(3))/(2)sinx` Then, `-sqrt(((1)/(2))^(2)+((5sqrt(3))/(2))^(2))lef(x)lesqrt(((1)/(2))^(2)+((3sqrt(3))/(2))^(2))` or `-sqrt(19)lef(x)lesqrt(19)` |
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| 68. |
Find the value of `sin^(2)45^(@)-"cosec"^(2)60^(@)+sec^(2)30^(@)`. |
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Answer» `sin^(2)45^(@)-"cosec"^(2)60^(@)+sec^(2)30^(@)` `=((1)/(sqrt(2)))^(2)-((2)/(sqrt(3)))^(2)+((2)/(sqrt(3)))^(2)+((2)/(sqrt(3)))^(2)=(1)/(2)-(4)/(3)+(4)/(3)=(1)/(2)`. Hence the required value `=(1)/(2)`. |
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| 69. |
Calculate : `sin^(2)45^(@)-"cosec"^(2)60^(@)+sec^(2)30^(@)`. |
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Answer» `sin^(2)45^(@)-"cosec"^(2)60^(@)+sec^(2)30^(@)` `=((1)/(sqrt(2)))^(2)-((2)/(sqrt(3)))^(2)+((2)/(sqrt(3)))^(2)` `=(1)/(2)`. |
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| 70. |
If `5 cot theta = 3 ," find the value of "((5sin theta - 3 cos theta )/(4 sin theta + 3 cos theta)).` |
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Answer» `5 cot theta =3implies cot theta =(3)/(5) .` Given expression `=((5sin theta - 3 cos theta )/(4 sin theta+3cos theta ))=((5-3 cot theta))/((4+ 3 cot theta ))` `["dividing num.and denom, by "sin theta ]` `=((5-3xx(3)/(5)))/((4+3xx(3)/(5)))=((5-(9)/(5)))/((4+(9)/(5)))=((16)/(5)xx(5)/(29))=(16)/(29).` |
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| 71. |
`If``cottheta=(15)/8,t h e ne v a l u a t e((2+2sintheta)(1-sintheta)/((1+costheta)(2-2costheta)` |
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Answer» Given expression `=((2+2 sin theta)(1-sin theta))/((1+cos theta )(2-2cos theta ))` `=(2(1+sin theta )(1-sin theta))/(2(1+cos theta )(1-cos theta))` `=((1-sin^(2)theta ))/((1-cos ^(2)theta ))=(cos^(2) theta)/(sin^(2) theta)= cot ^(2)theta` `=( cot theta)^(2)=((15)/(8))^(2)=(225)/(64).` Hence , the value of the given expression is `(225)/(64).` |
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| 72. |
6. Find the positive integers p, q, r, s satisfying `tan (pi/24)=(sqrt(p)-sqrt(q))(sqrt(r)-(s))`A. 6B. 7C. 8D. 9 |
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Answer» Correct Answer - C Using `tan 2A = (2 tan A)/(1-tan^(2)A)` `tan 7.5^(@)=(sin 15^(@))/(1+cos 15^(@))` `=(sin 15^(@))/(1+cos 15^(@))xx(1-cos 15^(@))/(1-cos 15^(@))` `=(sin 15^(@)(1-cos 15^(@)))/(sin^(2)15^(@))` `=(1-cos 15^(@))/(sin 15^(@))` `=(1-(sqrt(3)+1)/(2sqrt(2)))/((sqrt(3)-1)/(2sqrt(2)))` `=(2sqrt(2)-sqrt(3)-1)/(sqrt(3)-1)` `=(2sqrt(2)-sqrt(3)-1)/(sqrt(3)-1)xx(sqrt(3)+1)/(sqrt(3)+1)` `=((2sqrt(2)-sqrt(3)-1)(sqrt(3)+1))/(3-1)` `=(2sqrt(6)+2sqrt(2)-4-2sqrt(3))/(2)` `=(2(sqrt(6)+sqrt(2)-2-sqrt(3)))/(2)` `=sqrt(3)(sqrt(2)-1)-sqrt(2)(sqrt(2)-1)` `= (sqrt(3)-sqrt(2))(sqrt(2)-1)` |
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| 73. |
In a `Delta ABC`, 2 sinA cos B + 2 sin B cos C + 2 sin cos A = sin A + sin B + sin C, then `Delta ABC`isA. isoscelesB. right angledC. acute angledD. none of these |
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Answer» Correct Answer - A Given `sin(A+B)+sin(A-B)+sin (B+C)+sin(B-C)+sin (C+A)+sin(C-A)` `= sin A + sin B + sin C` `rArr sin(A-B)+sin (B-C)+sin(C-A)=0` `rArr 4 sin.(A-B)/(2)sin.(B-C)/(2)sin.(C-A)/(2)=0` `therefore Delta` is isosceles. |
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| 74. |
If `cos A, cosB and cosC` are the roots of the cubic `x^3 + ax^2 + bx + c = 0` where `A, B, C` are the anglesof a triangle then find the value of `a^2 – 2b– 2c`. |
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Answer» Correct Answer - C cos A, cos B and cos C are the roots of the cubic equation `x^(3)+ax^(2)+bx + c=0` `rArr cos A+cos B + cos C = -a` cos A cos B + cos B cos C + cos C cos A = b cos A cos B cos C = -c Now `(cos A+cos B cosC)^(2)=(Sigma cos^(2)A)+2(Sigma cos A cos B)` `therefore cos^(2)A+cos^(2)B+cos^(2)C=a^(2)-2b` `therefore 3+cos 2A+cos 2B+cos 2C=2a^(2)-4b` `therefore 3-1-4 cos A cos B cos C = 2a^(2)-4b` `thererfore 1+2c=a^(2)-2b` `therefore a^(2)-2b-2c =1` |
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| 75. |
If `A+B+C+D = 2pi`, prove that : `cosA +cosB+cosC+cosD=4 cos, (A+B)/2 cos, (B+C)/(2) cos, (C+A)/2`A. `4cos.(A+B)/(2)cos.(B+C)/(2)cos.(C+A)/(2)`B. `4sin.(A+B)/(2)sin.(B+C)/(2)sin.(C+A)/(2)`C. `1-4sin.(A+B)/(2)sin.(B+C)/(2)sin.(C+A)/(2)`D. `-1-4cos.(A+B)/(2)cos.(B+C)/(2)cos.(C+A)/(2)` |
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Answer» Correct Answer - A `L.H.S.=(cos A+ cos B)+(cos C + cos D)` `=2 cos.(A+B)/(2)cos.(A-B)/(2)+2 cos.(C+D)/(2)cos.(C-D)/(2)` `=2cos.(A+B)/(2)cos.(A-B)/(2)cos.(A-B)/(2)+2cos(pi-(A+B)/(2))cos.(C-D)/(2)` `=2 cos.(A+B)/(2)cos.(A-B)/(2)-2cos.(A+B)/(2)cos.(C-D)/(2)` `=2 cos.(A+B)/(2)[cos.(A-B)/(2)-cos.(C-D)/(2)]` `=2 cos.(A+B)/(2)xx 2 sin(A-B+C-D)/(4)sin.(C-D-A+B)/(4)` `=4 cos.(A+B)/(2)sin.(A+C-(B+D))/(4)sin.(B+C-(A+D))/(4)` `=4 cos.(A+B)/(2)sin.(A+C-(2pi-A-C))/(4)sin.(B+C-(2pi-B-C))/(4)` `=4 cos.(A+B)/(2)sin.((A+C)/(2)-(pi)/(2))sin((B+C)/(2)-(pi)/(2))` `=4 cos.(A+B)/(2)[-sin((pi)/(2)-(A+C)/(2))][-sin((pi)/(2)-(B+C)/(2))]` `=4 cos.(A+B)/(2)cos.(A+C)/(2)cos.(B+C)/(2)` |
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| 76. |
`cos e c(360^0)/7+cos e c(540^0)/7=``cos e c(180^0)/7`(b) `cos e c(90^0)/7``sec(180^0)/7`(d) `sec(90^0)/7`A. `co sec((180^(@))/(7))`B. `co sec((90^(@))/(7))`C. `sec( (180^(@))/(7))`D. `sec((90^(@))/(7))` |
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Answer» Correct Answer - A Let `alpha=(180^(@))/(7)` `rArr 180^(@)=7alpha` `rArr 3alpha=180^(@)-4alpha` `rARr sin 3alpha=sin4alpha` `therefore co sec ((360^(@))/(7))+co sec((540^(@))/(7))` `=co sec 2alpha+co sec 3alpha` `=(sin 3alpha+2 alpha)/(sin 3alpha sin 2alpha)` `=(sin 4alpha+sin 2alpha)/(sin 3alpha sin 2alpha)` `=(2sin 3alpha cos alpha)/(sin 3alpha. sin 2alpha)` `=(2cos alpha)/(2sin alpha. cos alpha)` `=co sec alpha` `=co sec ((180^(@))/(7))` |
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| 77. |
If `y=(sqrt(1-sin4x)+1)/(sqrt(1+sin 4x)-1)`, then y can beA. cot xB. `- tan x`C. `-cot((pi)/(4)+x)`D. `tan((pi)/(4)+x)` |
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Answer» Correct Answer - A::B::C::D 0 |
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| 78. |
If sec x + tan x =22/7 , find the value of tanx/2A. the value of tan `(x)/(2)=(29)/(15)`B. the value of tan `(x)/(2)=(29)/(15)`C. the value of cosec x + cot `x=(29)/(15)`D. thevalue of cosec x + cot `x=(15)/(29)` |
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Answer» Correct Answer - B::C `sec x + tan x = (22)/(7)` `rArr (1+sin x)/(cos x)=(22)/(7)` `rArr (1+(2t)/(1+t^(2)))/((1-t^(2))/(1+t^(2)))=(22)/(7)`, where `t= tan.(x)/(2)` `rArr ((1+t)^(2))/(1-t^(2))=(22)/(7)` `rArr (1+t)/(1-t)=(22)/(7)` `rArr t=(15)/(29)` Now cosec x + cot x `=(1+cos x)/(sin x)` `=(2 cos^(2)(x//2))/(2sin(x//2)cos(x//2))` `=cot((x)/(2))=(29)/(15)` |
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| 79. |
If `3 cot theta= 2` , show that `((4 sin theta - 3 cos theta ))/((2 sin theta + 6 cos theta))=(1)/(3)` |
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Answer» `((4sin theta-3 cos theta ))/((2 sin theta +6 cos theta ))=((4-3cot theta ))/((2+ 6 cot theta))" " ["dividing num, and denom , by " sin theta ]` `=((4-3xx(2)/(3)))/((2+6xx(2)/(3)))=(2)/(6)=(1)/(3).` |
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| 80. |
If `u=(1+cos theta)(1+cos 2theta)-sin theta.sin 2theta, v=sin theta(1+cos 2theta)+sin 2theta(1+cos theta)`, then `u^(2)+v^(2)=`A. `4(1+cos theta)(1+cos 2theta)`B. `4(1+sin theta)(1+sin 2theta)`C. `4(1-cos theta)(1-cos 2theta)`D. `4(1-sin theta)(1-sin 2theta)` |
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Answer» Correct Answer - A `u=1+cos theta + cos 2theta + cos (theta + 2theta)` `=(1+cos 3theta)+(cos theta + cos 2theta)` `=2 "cos"^(2)(3theta)/(2)+2 cos.(3theta)/(2)cos.(theta)/(2)` `=2 cos.(3theta)/(2)[cos.(3theta)/(2)+cos.(theta)/(2)]` Similarly `v=2 sin.(3theta)/(2)[cos.(3theta)/(2)+cos.(theta)/(2)]` `therefore u^(2)+v^(2)=4(cos.(3theta)/(2)+cos.(theta)/(2))^(2)` `=4(2cos theta. cos.(theta)/(2))^(2)=16 cos^(2)theta."cos"^(2)(theta)/(2)` `= 4(1+cos theta)(1+cos 2 theta)` |
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| 81. |
If ` sin alpha =(1)/(2) ,` prove that `(3 cos alpha - 4 cos ^(3) alpha )=0.` |
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Answer» `cos ^(2) alpha =(1- sin ^(2) alpha )=(1-(1)/(4))=(3)/(4)implies cos alpha =(sqrt(3))/(2).` `therefore (3 cos alpha -4cos ^(3)alpha)=(3xx(sqrt(3))/(2)-4xx(3sqrt(3))/(8))=((3sqrt(3))/(2) -(3sqrt(3))/(2))=0.` |
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| 82. |
In a triangle ABC, `cos 3A+cos 3B+cos3C=1` and `angleA+angleBltangleC`, then find possible measure of `angleC`. |
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Answer» Correct Answer - `120^(@)` `cos3A+cos3B+cos 3C=1` `rARr 2cos(3(A+B))/(2)cos(3(A-B))/(2)=sin^(2)(3C)/(2)` `rARr -sin(3C)/(2)cos(3(A-B))/(2)=sin^(2)(3C)/(2)` `rArr sin(3C)/(2)[cos(3(A-B))/(2)-cos(3(A+B))/(2))0` `rArr 2sin(3A)/(2)sin(3B)/(2)sin(3C)/(2)=0` |
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| 83. |
If `tanbeta = costheta*tanalpha` , then `tan^2(theta/2)` is equal toA. `(sin(alpha+beta))/(sin(alpha-beta))`B. `(cos(alpha-beta))/(cos(alpha+beta))`C. `(sin(alpha-beta))/(sin(alpha+beta))`D. `(cos(alpha+beta))/(cos(alpha-beta))` |
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Answer» Correct Answer - C `"tan"^(2)(theta)/(2)=(1-cos theta)/(1+cos theta)=(tan alpha - tan beta)/(tan alpha + tan beta)=(sin(alpha-beta))/(sin(alpha+beta))` |
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| 84. |
The value of `sum_(r=1)^(11)tan^(2)((r pi)/(24))` isA. 91B. 85C. 253/3D. none of these |
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Answer» Correct Answer - C `S= sum_(r=1)^(11)tan^(2)((r pi)/(24))` `=("tan"^(2)(pi)/(24)+"tan"^(2)(11pi)/(24))+("tan"^(2)(2pi)/(24)+"tan"^(2)(10pi)/(24))+("tan"^(2)(3pi)/(24)+"tan"^(2)(9pi)/(24))+("tan"^(2)(4pi)/(24)+"tan"^(2)(8pi)/(24))+("tan"^(2)(5pi)/(24+"tan"^(2)(7pi)/(24))+("tan"^(2)(6pi)/(24))` `=("tan"^(2)(pi)/(24)+"cot"^(2)(pi)/(24))+{(2-sqrt(3))^(2)+(2+sqrt(3))^(2)+(sqrt(2)-1)^(2)+(sqrt(2)+1)^(2)+((1)/(sqrt(3)))^(2)+(sqrt(3))^(2)}+("tan"^(2)(5pi)/(24)+"cot"^(2)(5pi)/(24))+1` Now `(tan^(2) theta + cot^(2)theta)=2+4 cot^(2)2theta` `S=2+4"cot"^(2)(pi)/(12)+(70)/(3)+2+4"cot"^(2)(5pi)/(12)+1` `=(85)/(3)+4[(2+sqrt(3))^(2)+(2-sqrt(3))^(2)]` `therefore = (85)/(3)+4(2)(4+3)=(253)/(3)` |
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| 85. |
A function `f(theta)=sin^(2)theta+3sin theta cos theta+5cos^(2)theta` is defined `AA theta in R`. Another function`g(theta)=f((pi)/(2)-theta)` at the point `theta` where `f(theta)` is minimum, then thevalue of `(1)/(g(theta))` isA. 2B. 5C. 3D. 4 |
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Answer» Correct Answer - A `f(theta)=sin^(2)theta+3 sin theta cos theta cos theta + 5 cos^(2) theta` `=1+4 cos^(2) theta + (3)/(2) sin 2 theta` `3+2 cos 2theta + (3)/(2) sin 2 theta` Minimum of `f(theta)` and `f((pi)/(2)-theta)` will be same because of the type of function a `sin theta+b cos theta` and is `3-(5)/(2)=(1)/(2)` `therefore g(theta)=(1)/(2)` `therefore (1)/(g(theta))=2` |
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| 86. |
The number of integers in the range of `3 sin^(2)x+3sin x cos x+7cos^(2)x` isA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - C `3 sin^(2)x+3 sin x.cos x+7 cos^(2)x` `=(3(1-cos 2x))/(2)+(3sin 2x)/(2)+(7(1+cos 2x))/(2)` `=(3 sin 2x 2x + 4 cos 2x)/(2)+5` Now `-5 le 3 sin 2x + 4 cos 2x le 5` `therefore (3 sin 2x + 4 cos 2x)/(2)+5 in [(5)/(2),(15)/(2)]` |
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| 87. |
If ` ( cos x - cos alpha)/(cos x - cos beta) = ( sin^2 alpha cos beta)/(sin^2 beta cos alpha)` then cos x =A. `cos x=(cos alpha+cos beta)/(1-cos alpha cos beta)`B. `cos x=(cos alpha+cos beta)/(1-cos alpha cos beta)`C. `tan.(x)/(2)=tan.(alpha)/(2)tan.(beta)/(2)`D. `tan.(x)/(2)=-tan.(alpha)/(2)tan.(beta)/(2)` |
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Answer» Correct Answer - A::C::D From the given redults `cos x sin^(2)bet cos alpha - cos^(2) alpha sin^(2)beta` `= sin^(2) alpha cos beta cos x - sin^(2) alpha cos^(2) beta` `rArr os x[cos alpha sin^(2)beta - sin^(2) alpha cos beta]` `= cos^(2)alpha sin^(2)beta-sin^(2)alpha cos^(2) beta` `rArr cos x=(cos^(2)alpha[1-cos^(2)beta]-(1-cos^(2)alpha)cos^(2)beta)/(cos alpha[1-cos^(2)beta]-[1-cos^(2)alpha[cos beta)` `=(cos^(2)alpha-cos^(2)beta)/((cos alpha-cos beta)(cos alpha cos beta+1))` `=(cos alpha + cos beta)/(1+cos alpha cos beta)` `rArr (co x)/(1)=(cos alpha + cos betA)/(1+cos alpha cos beta)` `rarr (1-cos x)/(1+cos x)=(1+cos alpha cos beta - cos alpha - cos beta)/(1+cos lapha cos beta + cos alpha + cos beta)` `=((1-cos alpha)(1-cos beta))/((1+cos alpha)(1+cos beta))` `rArr "tan"^(2)(x)/(2)="tan"^(2)(alpha)/(2)."tan"^(2)(beta)/(2)` `rArr tan.(x)/(2)= pm tan.(alpha)/(2)tan.(beta)/(2)` |
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| 88. |
If `0 leq theta leq pi and sin0/2=sqrt(1+sintheta)-sqrt(1-sintheta)` ,then the possible value of `tan theta` , is -A. `(4)/(3)`B. 0C. `(-3)/(4)`D. `(-4)/(3)` |
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Answer» Correct Answer - B::D `sin.(theta)/(2)=sqrt(1+sin theta)-sqrt(1-sin theta)` `rArr sin.(theta)/(2)=|cos.(theta)/(2)+sin.(theta)/(2)|-|cos.(theta)/(2)-sin.(theta)/(2)|` If `0 le theta le (pi)/(2), sin.(theta)/(2)=2 sin.(theta)/(2) rArr theta = 0` If `(pi)/(2) lt theta le pi, sin.(theta)/(2) = 2 cos.(theta)/(2)` `rArr tan.(theta)/(2)=2` `rArr tan theta = (2 tan.(theta)/(2))/(1-"tan"^(2)(theta)/(2))=(4)/(1-4)=(-4)/(3)` |
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| 89. |
if (1+tan`alpha`)(1+tan4`alpha`) =2 where `alpha` `in` (0 , `pi/16`) then `alpha ` equal toA. `(pi)/(20)`B. `(pi)/(30)`C. `(pi)/(40)`D. `(pi)/(60)` |
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Answer» Correct Answer - A `(1+tanA)(1+tanB)=2` `rArr tan A+tan B=1-tan A tan B` `rARr tan(A+B)=1,i.e, A+B=(pi)/(4)` or `alpha+4alpha=(pi)/(4)` i.e. `alpha=(pi)/(20)` |
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| 90. |
If `cos28^0+sin28^0=k^3, t h e ncos17^0`is equal to`(k^3)/(sqrt(2))`(b) `-(k^3)/(sqrt(2))`(c) `+-(k^3)/(sqrt(2))`(d) none of theseA. `(k^(3))/sqrt(2)`B. `-(k^(3))/sqrt(2)`C. `+-(k^(3))/sqrt(2)`D. none of these. |
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Answer» Correct Answer - A `cos17^(@)=cos(45^(@)-28^(@))` `cos45^(@)cos28^(@)+sin45^(@)sin28^(@)` `=(cos 28^(@)+sin 28^(@))/(sqrt(2))=(k^(3))/(sqrt(2))` |
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| 91. |
If A,B,C, are the angles of a triangle such that `cotA/2=3tanC/2,`then `sinA ,sinB ,sinC`are in`AdotPdot`(b) `GdotPdot`(c) `HdotPdot`(d) none of theseA. APB. GPC. HPD. none of these. |
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Answer» Correct Answer - A Given `cot((A)/(2))cot((C)/(2))=3` `rArr (cos((A)/(2))cos((C)/(2)))/(sin((A)/(2))sin((C)/(2)))=3` `rArr(cos((A-C)/(2)))/(cos((A+C)/(2)))=2` `rArr 2sinB=sinA+sinC` |
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| 92. |
If A > 0, B > 0, and A + B = `pi/3` then the maximum value of tan A tan B is |
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Answer» Correct Answer - 3 `A+b= pi//3` `rArr tan (A+B) = sqrt3` `rArr (tanA + tanB)/(1-tan A tan B) = sqrt3` `rArr = (tan A + (y)/(tan A)) /( 1-y) = sqrt3` ` rArr tan^(2)A+ sqrt3 (y-1) tan A +y =0 ` For real values of `tan A`, `" " 3(y-1)^(2) - 4y =0` `rArr 3y^(2) - 10y + 3 ge 0` `rArr (y-3)(3y-1) ge 0` `rArr y le (1)/(3) or y ge 3` But A, B `gt` 0 and A +B `= pi//3` `rArr A, B lt pi//3` `rArr tan A tan B lt 3` so, ` ge 3` is not a possibility. Therefore, `y le (1)/(3)` i.e., max. value of `y ` is `1//3` |
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| 93. |
Let `f(theta)=cottheta/(1+cottheta)` and `alpha+beta=(5pi)/4` then the value ` f(alpha)f (beta)` isA. `(1)/(2)`B. `-(1)/(2)`C. 2D. none of these. |
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Answer» Correct Answer - A `f(beta)=f((5pi)/(4)-alpha)=(cot((5pi)/(4)-alpha))/(1+cot((5pi)/(4)-alpha))` `=(1)/(1+(1-tanalpha)/(1+tan alpha))=(1+tan alpha)/(2)` As `f(x)=(cotalpha)/(1+cotalpha)=(1)/(1+tan alpha)`, we have `f(alpha)f(beta)=(1)/(2)` |
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| 94. |
If `tan^2(pi-A)/4+tan^2(pi-B)/4+tan^2(pi-C)/4=1`, then ` A B C`isequilateral (b)isosceles (c) scalene(d) none of theseA. equilateralB. isoscelesC. scaleneD. none of these. |
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Answer» Correct Answer - A Let `alpha=(pi-A)/(4),beta(pi-B)/(4),gamma=(pi-C)/(4)` `rArr alpha+beta+gamma=(pi)/(2)` `sum tan alpha tan beta=1` `rArr sum tan^(2)alpha=1sum tan alpha tan beta` `rArr tan alpha=tan beta=tan gamma` |
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| 95. |
If `cos(alpha-beta)=3sin(alpha+beta),t h e n1/(1-3sin2alpha)+1/(1-3sin2beta)=``1/2`(b) `(-1)/2`(c) `1/4`(d) `(-1)/4`A. `(1)/(2)`B. `(-1)/(2)`C. `(1)/(4)`D. (-1)/(4)` |
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Answer» Correct Answer - D Put, `beta=0rArr tan alpha=(1)/(3)` `rArr sin 2alpha=(2((1)/(3)))/(1+(1)/(9))=(3)/(5)` `therefore (1)/(1-3sin 2alpha)+(1)/(1-3sin2beta)` `=(1)/(1-3sin2alpha)+1=(1)/(1-(9)/(5))+1=(-1)/(4)`. |
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| 96. |
The value of `[ ( sin ""(pi)/(9)) (4+ sec""(pi)/(9))]^(2)` is _______. |
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Answer» Correct Answer - 3 `(sin ""(pi)/(9)) (4+sec""(pi)/(9))` ` " " = sin 20^(@) ( 4+ (1)/( cos 20^(@)))` ` " " = ( 4 sin 20^(@) cos 20^(@) + sin 20^(@))/( cos 20^(@))` `" " = ( 2 sin 40^(@) + sin 20^(@))/( cos 20^(@))` `" " = (sin 40^(@) + ( sin 40^(@) + sin 20^(@))/(cos 20^(@))` ` ( sin 40^(@) + 2 sin 30^(@) cos 10^(@))/(cos 20^(@))` `= ( sin 40^(@) + cos 10^(@))/( cos 20^(@))` `= (cos 50^(@) + cos 10^(@))/(cos 20^(@))` `= ( 2 cos 30^(@) cos 20^(@))/(cos 20^(@))` ` = sqrt3` |
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| 97. |
`((sin 33^(@))/( sin 11^(@) sin 49^(@) sin 71^(@)))^(2) + (( cos 33^(@))/(cos 11^(@) cos 49^(@) cos 71^(@)))^(2) ` is equal to ______. |
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Answer» Correct Answer - 32 Given expression is `(( sin 33^(@))/(sin 11^(@) sin ( 60^(@) - 11^(@)) sin (60+11^(@))))^(2) + (( cos 33^(@))/( cos 11^(@) cos ( 60^(@) - 11^(@)) cos ( 60^(@) + 11^(@)))) ^(2)` `= (( sin 33^(@))/( (1)/(4) sin 33^(@)))^(2) + ( ( cos 33^(@))/( (1)/(4) cos 33^(@)))^(2)` ` = 16 + 16 = 32` |
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| 98. |
The expression `(1+sin22^@sin33^@sin35^@)/(cos^2 22^@+cos^2 33^@+cos^2 35^@)` simplifies to |
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Answer» Correct Answer - 0.5 Let `(A)/(2) = 22^(@), (B)/(2) = 33^(@), (C )/(2) = 35^(@)` `rArr A +B+C= 180^(@)` In `DeltaABC, cos^(2) ""(A)/(2) + cos^(2)""(B)/(2) +cos^(2) ""(C )/(2)` `= 2+ 2 sin ""(A)/(2)* sin""(B)/(2)* sin ""(C )/(2)` Hence, `(1+ sin 22^(@) sin 33^(@) sin 35^(@))/(cos ^(2) 22^(@) + cos ^(2) 33^(@) + cos ^(2) 35^(@)) = (1)/(2)` |
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| 99. |
If `sin(x +3alpha)=3 sin (alpha-x)`, thenA. `tan x = tan alpha`B. `tan x = tan^(2)alpha`C. `tan x = tan^(3)alpha`D. `tan x=3 tan alpha` |
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Answer» Correct Answer - C `sin (x+3alpha)=3 sin (alpha - x)` `rArr sin x cos 3alpha + sin 3 alpha cos x = 3 (sin alpha cos x - sin x cos alpha)` `rArr sin x(cos 3alpha + 3 cos alpha) = cos x(3 sin alpha - sin 3 alpha)` `rArr tan x = (4sin^(3)alpha)/(4 cos^(3)alpha)=tan^(3)alpha` |
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| 100. |
If `t a n^2theta=2t a n^2varphi+1`, prove that `cos2theta+s in^2varphi=0.`A. `-1`B. 0C. 1D. none of these. |
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Answer» Correct Answer - B `tan^(2)theta=2tan^(2)phi+1` or `1+tan^(2)theta=2(1+tan^(2)phi)` `rArr sec^(2)theta=2sec^(2)phi` `rArr cos^(2)phi=2cos^(2)phi=1=+cos2theta` `rArr cos 2theta=cos^(2)phi-1=-sin^(2)phi` `rArr sin^(2)phi+cos2theta=0` |
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