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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.
| 551. |
`sin(sin^(-1)((1)/(3))+sec^(-1)(3))+cos(tan^(-1)(1/2)+tan^(-1)2)`=A. `1`B. `2`C. `0`D. `-1` |
| Answer» Correct Answer - A | |
| 552. |
If : `(tan 3 A+tan A)/(tan 3A - tanA)=(sin(mA)/(sin(nA)))` then `(m,n)=` (a) (2,3) (b)(3,4) (c)(4,2) (d) (1,2)A. (2,3)B. (3,4)C. (4,2)D. (1,2) |
| Answer» Correct Answer - C | |
| 553. |
If `cos^(2)A+cos^(2)B+cos^(2)C=1`, then `triangle ABC` isA. equilateralB. isoscelesC. right angledD. right angled isosceles |
| Answer» Correct Answer - D | |
| 554. |
In `triangleABC, (a-b)/(a+b)=`A. `cot((A-B)/(2))cot((A+B)/(2))`B. `tan((A-B)/(2))cot((A+B)/(2))`C. `cot((A-B)/(2))tan((A+B)/(2))`D. `tan((A-B)/(2))tan((A+B)/(2))` |
| Answer» Correct Answer - B | |
| 555. |
In `triangleABC`, if `a=sqrt(3)+1, b=sqrt(3)-1 and angleC=60^(@)`, then c=A. `sqrt(6)`B. `-sqrt(6)`C. `2`D. `4` |
| Answer» Correct Answer - A | |
| 556. |
Solve, `(sec^- 1)x/a-(sec^- 1)x/b=sec^- 1 b-sec^(- 1)a.`A. `1`B. `pq`C. `(p)/(q)`D. `(q)/(p)` |
| Answer» Correct Answer - B | |
| 557. |
Find the value of `tan^(-1)(x/y)-tan^(-1)((x-y)/(x+y))`A. `(pi)/(4)`B. `(pi)/(3)`C. `(pi)/(2)`D. `(pi)/(4) or (-3pi)/(4)` |
| Answer» Correct Answer - A | |
| 558. |
In `triangleABC`, If `(b+c)/(11)=(c+a)/(12)=(a+b)/(13)`, then `cosC=`A. `(5)/(7)`B. `(7)/(5)`C. `(16)/(17)`D. `(17)/(16)` |
| Answer» Correct Answer - A | |
| 559. |
In ` A B C ,=`if `(a+b+c)(a-b+c)=3a c ,`then find `/_Bdot`A. `angleB=30^(@)`B. `angleB=60^(@)`C. `angleB=100^(@)`D. `angleB=90^(@)` |
| Answer» Correct Answer - B | |
| 560. |
In `triangleABC,` if `a=2, b=1, c=sqrt(3)`, then `angleA=`A. `90^(@)`B. `60^(@)`C. `30^(@)`D. `45^(@)` |
| Answer» Correct Answer - A | |
| 561. |
In any `DeltaABC`, prove that `a(bcosC-c cosB)=(b^(2)-c^(2))`A. `c^(2)-b^(2)`B. `b^(2)-c^(2)`C. `2(c^(2)-b^(2))`D. `2(b^(2)-c^(2))` |
| Answer» Correct Answer - B | |
| 562. |
In `triangleABC`, if `a=sqrt(3)+1, b=sqrt(3)-1 and angleC=60^(@)`, then `angleA`=A. `45^(@)`B. `105^(@)`C. `15^(@)`D. `60^(@)` |
| Answer» Correct Answer - B | |
| 563. |
In `triangleABC,` if `a=2, b=1, c=sqrt(3)`, then `angleB=`A. `90^(@)`B. `60^(@)`C. `30^(@)`D. `45^(@)` |
| Answer» Correct Answer - C | |
| 564. |
In `triangleABC` , if a=2,b=3 and `sinA=2/3`, then `angleB=`A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - D | |
| 565. |
In `triangleABC`, If `(1)/(b+c)+(1)/(c+a)=(3)/(a+b+c)`, then `cosC=`A. `(1)/(2)`B. `(-1)/(2)`C. `(sqrt(3))/(2)`D. `(1)/(sqrt(2))` |
| Answer» Correct Answer - A | |
| 566. |
`(tan^(- 1)(sqrt(3))-sec^(- 1)(-2))/(cosec^(- 1)(-sqrt(2))+cos^(- 1)(-1/2)) =`A. `(4)/(5)`B. `(-4)/(5)`C. `(3)/(5)`D. `0` |
| Answer» Correct Answer - B | |
| 567. |
If `tan^(-1)x+cos^(-1)((y)/(sqrt(1+y^(2))))=sin^(-1)((3)/(sqrt(10)))`, thenA. `x=2, y=1`B. `x=2, y=3`C. `x=3, y=2`D. `x=1, y=2` |
| Answer» Correct Answer - D | |
| 568. |
In `triangleABC`, If the angles are in A.P., and `b:c=sqrt(3):sqrt(2)`, then `angleA, angleB, angleC` areA. `30^(@), 60^(@), 90^(@)`B. `50^(@), 60^(@), 70^(@)`C. `75^(@), 60^(@), 45^(@)`D. `15^(@), 60^(@), 105^(@)` |
| Answer» Correct Answer - C | |
| 569. |
In `triangleABC`, if `a=72, angleB=108^(@), angleA=25^(@)`, then c=A. `162.04`B. `162.14`C. `124.61`D. `124.16` |
| Answer» Correct Answer - C | |
| 570. |
In `triangleABC`, if `a^(2)+c^(2)-b^(2)=ac`, then `angleB=`A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - C | |
| 571. |
In `triangleABC,` If b=20, c=21 and sinA=(3)/(5)`, then a=A. `13`B. `14`C. `15`D. `16` |
| Answer» Correct Answer - A | |
| 572. |
For any triangle ABC, prove that`(b^2-c^2)/(a^2)sin2A+(c^2-a^2)/(b^2)sin2B+(a^2-b^2)/(c^2)sin2C=0` |
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Answer» For any triangle ABC, we have,`sinA/a = sinB/b = sinC/c = k`->(1) `cos A = (b^2+c^2-a^2)/(2bc), cosB = (c^2+a^2-b^2)/(2ca), cos C = (a^2+b^2-c^2)/(2ab)`->(2) Now, we will solve our equation. `L.H.S = (b^2-c^2)/a^2 *2sinAcosA+(c^2-a^2)/b^2*2sinBcosB+(a^2-b^2)/c^2*2sinCcosC` From (1), `=(b^2-c^2)/a^2*2kacosA+(c^2-a^2)/b^2*2kbcosB+(a^2-b^2)/c^2*2kccosC` `=2k((b^2-c^2)/a*(b^2+c^2-a^2)/(2bc)+(c^2-a^2)/b*(c^2+a^2-b^2)/(2ca)+(a^2-b^2)/c*(a^2+b^2-c^2)/(2ab))` `=k/(abc)(b^4-c^4-a^2(b^2-c^2)+c^4-a^4-b^2(c^2-a^2)+a^4-b^4-c^2(a^2-b^2))` `=k/(abc)(0) = 0= R.H.S.` |
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| 573. |
`cos^(-1)sqrt(1-x)+sin^(-1)sqrt(1-x)=`A. `(pi)/(4)`B. `1`C. `pi`D. `(pi)/(2)` |
| Answer» Correct Answer - D | |
| 574. |
In `triangleABC`, If `angleA=25^(@), angleB=85^(@) and c=3,4`, then b=A. `3.0604`B. `3.604`C. `1.0529`D. `1.529` |
| Answer» Correct Answer - B | |
| 575. |
prove that: ` 4 cot^2(45^ @) - sec^2(60^ @)+ sin^2(30^ @) = 1/8`A. `(1//2)`B. `(1//2)^(2)`C. `(1//2)^(3)`D. `(1//2)^(4)` |
| Answer» Correct Answer - C | |
| 576. |
`cos^(-1)((3)/(5))+cos^(-1)((4)/(5))=`A. `(3pi)/(2)`B. `(pi)/(2)`C. `(-pi)/(2)`D. `0` |
| Answer» Correct Answer - B | |
| 577. |
In `triangleABC`, if `a=3, b=4, c=5,` then `sin2B=`A. `(3)/(5)`B. `(4)/(5)`C. `(24)/(25)`D. `(12)/(25)` |
| Answer» Correct Answer - C | |
| 578. |
Find the acute angle A and B such that `sec A * tan B -sec A -2 tan B+2=0.`A. `(60^(@),45^(@))`B. `(45^(@),30^(@))`C. `(30^(@),45^(@))`D. `(30^(@),60^(@))` |
| Answer» Correct Answer - A | |
| 579. |
prove that: `sin^2 3 0^(@)+sin^2 4 5^(@)+sin^2 6 0^(@)=3/2`A. `2//3`B. `3//2`C. 13D. `3//4` |
| Answer» Correct Answer - B | |
| 580. |
If `tan^(2) theta-4sqrt3 * tan theta+3=0,"then" : tan theta=`A. `sqrt2`B. `1//sqrt2`C. `1//sqrt3`D. 2 |
| Answer» Correct Answer - C | |
| 581. |
Find the angle between the minute hand and the hour hand of a clockwhen the time is 7:20 AM. |
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Answer» Angle made by minute hand in `20` minutes `= 360/60**20 = 120^@` Now, hour hand makes an angle `360^@` in `720` minutes. Total number of minutes at `7:20` AM `= 7**60+20 = 440` So, angle made by hour hand at `7:20` AM `= 360/720**440 = 220^@` `:.` Angle made between the minute hand and the hour hand at `7:20` AM `=220-120 = 100^@` |
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| 582. |
If `A+B+C=pi,`prove that `tan^2A/2+tan^2B/2+tan^2C/2geq1.` |
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Answer» `(tan (A/2) - tan(B/2))^2 + (tan(B/2) - tan(C/2))^2 +(tan(C/2) - tan(A/2))^2 ge 0` `=>2(tan^2(A/2)+tan^2(B/2)+tan^2(C/2)) - 2(tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)) ge 0->(1)` Now, `tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)` `=tan(A/2)[tan(B/2)+tan(C/2)]+tan(B/2)tan(C/2)` `=tan(A/2)[tan((B/2)+(C/2))(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)` `=tan(A/2)[tan((B+C)/2)(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)` `=tan(A/2)[tan((pi-A)/2)(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)` `=tan(A/2)[tan(pi/2-A/2)(1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)` `=tan(A/2)cot(A/2)(1-tan(B/2)tan(C/2))+tan(B/2)tan(C/2)` `=1-tan(B/2)tan(C/2)+tan(B/2)tan(C/2)` `=1` `=>tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2) = 1->(2)` From (1) and (2), `=>2(tan^2(A/2)+tan^2(B/2)+tan^2(C/2)) - 2 ge 0` `=>2(tan^2(A/2)+tan^2(B/2)+tan^2(C/2)) ge 2` `=>tan^2(A/2)+tan^2(B/2)+tan^2(C/2) ge 1` |
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| 583. |
A circular ring of radius 3cm hangs horizontally form a point 4cm vertically above the centre by 4 strings attached at equal intervals to its circumference. If the angle between two consecutive strings be `theta` , then `costheta` is equal to(A)`4/5`(B) `4/(25)`(C) `(16)/(25)`(D) none of these |
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Answer» OP=4cm OA=OB=OC=OD=3cm `(OA)^2+(OP)^2=(AP)^2` `AP^2=3^3+4^2=25` `AP=5cm` AP=BP=DP=CP=5cm `AB^2+BC^2=AC^2` `2AB^2=6^2` `AB=6/sqrt2=3sqrt2` `AB=BC=CD=AD=3sqrt2cm` `sin(theta/2)=1/2*(BC)/(BP)=(3sqrt2)/5=3/(5sqrt2)` `cos2A=1-2sin^2A` `costheta=1-2(9/(25*2))` `costheta=16/25` Option D is correct. |
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| 584. |
`tan 75^(@)+ tan 15^(@)=`A)1B)2C)3D)4A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - D | |
| 585. |
The value of `cos 1^@ cos 2^@ cos 3^@... cos 179^@` isA. `(1)/(sqrt(2))`B. 0C. 1D. -1 |
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Answer» Correct Answer - B Given expression, `cos1^(@) cos2^(@) cos3^(@)...cos179^(@)` `" "=cos1^(@)cos2^(@)...cos90^(@)...cos179^(@)" "[becausecos90^(@)=0]` ltBrgt =0 |
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| 586. |
`Tan 75 - Cot 75=`A. `2sqrt(3)`B. `2+sqrt(3)`C. `2-sqrt(3)`D. 1 |
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Answer» Correct Answer - A Given expression, `tan75^(@)-cot75^(@)` `" "=(sin75^(@))/(cos75^(@))-(cos75^(@))/(sin75^(@))` `" "=(sin^(2)75^(@)-cos^(2)75^(@))/(sin75^(@)*cos75^(@))` `" "=(-2cos150^(@))/(sin150^(@))` `" "=(-2cos(90^(@)+60^(@)))/(sin(90^(@)+60^(@)))` ltBrgt `" "=(+2sin60^(@))/(cos60^(@))` `" "=(2*(sqrt(3))/(2))/((1)/(2))=2sqrt(3)` |
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| 587. |
PQ is a vertical tower having P as the foot. A,B,Care three points in the horizontal plane through P. The angles of elevationof Q from A,B,C are equal and each is equal to `theta`. The sides of the triangle ABC are a,b,c, and area of the triangle ABCis ``. Then prove that the height of the tower is (abc) `tantheta/(4)dot` |
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Answer» Let `h` is the height of the tower. Then, `AP = BP = CP = hcot theta` Please refer to video to see the diagram. Here, `P` is the circumcenter of `Delta APC`. `:. AP = BP = CP = R = (abc)/(4Delta)` `=>(abc)/(4Delta) = hcottheta` `=>h = 1/cottheta(abc)/(4Delta)` `=>h = (abc)tantheta/(4Delta).` So, height of the tower is ` (abc)tantheta/(4Delta).` |
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| 588. |
In a `DeltaP Q R`, if `3""s in""P""+""4""cos""Q""=""6`and `4""s in""Q""+""3""cos""P""=""1`, then the angle R is equal to(1) `(5pi)/6`(2) `pi/6`(3) `pi/4`(4) `(3pi)/4` |
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Answer» `P + Q+R = pi` `3 sin P + 4 cos Q =6 (i) ; 4 sin Q + 3 cos P= 1 (ii)` `R < pi/2` eqn `(i)^2 + (ii)^2:` `9sin^2 P + 24sinPcos Q + 16cos^2 Q+ 16sin^2 Q+ 24sin QcosP + 9cos^2 P= 36 +1 ` `9 + 16 + 24 (sin Pcos Q + cos P sin Q) = 37` `25 + 24 (sin(P+Q))= 37` `sin(P +Q) = (37-25)/24= 1/2` `(P+Q) = pi/6 or 5 pi/6 ` `R = pi - (P+Q) = pi/6` as `5pi/6` not possible option 1 is correct |
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| 589. |
If `tantheta=3` and `theta` lies in third quadrant then `sintheta=`A. `(1)/(sqrt(10))`B. `-(1)/(sqrt(10))`C. `(-3)/(sqrt(10))`D. `(3)/(sqrt(10))` |
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Answer» Correct Answer - C Given that, `" "tantheta=3` `rArr" "sec^(2)theta=1+tan^(2)theta` `rArr" "sectheta=sqrt(1+9)=pmsqrt(10)` `rArr" "sectheta=-sqrt(10)` ` rArr" "costheta=-(1)/(sqrt(10))` `rArr" "sintheta=pmsqrt(1-(1)/(10))=pmsqrt((9)/(10))=pm(3)/(sqrt(10))" "` [since, `theta` lines in third quadrant] ` therefore" "sintheta=-(3)/(sqrt(10))` |
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| 590. |
The period of `cosx^2` is |
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Answer» A function is periodic if `f(x) = f(x+T)`. Let us assume that `cos(x^2)` has a period of T. In that case: `cos(x+T)^2 = cosx^2` `=>(x+T)^2 = 2npi+x^2` `=>x^2+T^2+2xT= 2npi+x^2` `=>T^2+2xT-2npi = 0` As we can see, `T` is dependent on the value of `x` and hence, is not a constant. So `cos(x^2)` is not periodic. So, we can say that period of `cosx^2` does not exist. |
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| 591. |
The angle of elevation of to be top point `P`of the vertical tower PQ of height `h`from poin A is `45^0`and from a point B, the angle of elevation is `60^0,`where B is point at a distance `d`from the point A measured along the line `A B`which makes an angle `30^0`with AQ. Prove that `d=(sqrt(3)-1)hdot` |
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Answer» `cosc=(a^2+b^2-c^2)/(2ab)` `=(7^2+8^2-9^2)/(2*7*8)` `=2/7` `BM^2=BC^2+CM^2-2BC*CM*cosc` `=7^2+4^2-2*7*4*2/7` `=49` `BM=7` `In/_BMP` `h/(BM)=tan15^o` `h=BMtan15^o` `=7tan15^o-(1)` `tan30^o=(2tan15^o)/(1-tan^215^o` `x=tan15^o` `1/sqrt3=(2x)/(1-x^2)` `1-x^2=2/3x` `x=(-2sqrt3pmsqrt((2sqrt3)^2+4))/2` `x=(-2sqrt3pm4)/2` `2-sqrt3` putting this value in equation 1 `=7(2-sqrt3)m` |
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| 592. |
A vertical pole with height more than 100 m consists of two parts, the lower being one-third of the whole. At a point on a horizontal plane through the foot and 40 m from it,the upper part subtends an angle whose tangent is `1/2` What is the height of the pole |
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Answer» `tanphi=x/(3*40)` `tan(theta+phi)=x/40` `tantheta=1/2` `(tantheta+tanphi)/(1-tantheta*tanphi)=x/40` `((1/2)+(x/120))/(1-(1/2*x/120))=x/40` `(cotx/120)/((240-x)/240)=x/40` `(2(60+x))/(140-x)=x/40` `80(60+x)=x(240-x)` `x^2-160x+4800=0` `(x-120)(x-40)=0` `x=40,120`. |
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| 593. |
ABis a vertical pole with B at the ground level and A at the top. A man findsthat the angle of elevation of the point A from a certain point C on theground is `60o`.He moves away from the pole along the line BC to a point D such that `C D""=""7""m`.From D the angle of elevation of the point A is `45o`.Then the height of the pole is(1) `(7sqrt(3))/2dot1/(sqrt(3)-1)m`(2) `(7sqrt(3))/2dot(sqrt(3)+1)m`(3) `(7sqrt(3))/2dot(sqrt(3)-1)m`(4) `(7sqrt(3))/2dot1/(sqrt(3)+1)m` |
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Answer» `BC=x` `AB=y` `tan60^@=AB/BC=y/x` `sqrt3=y/x` `x=y/sqrt3` `tan45^@=(AB)/(BD)=(AB)/(DC+CB)` `=y/(7+x)` `=1` `y=7+x` `y=7+y/sqrt3` `y-y/sqrt3=7` `y(sqrt3-1)=7sqrt3` `y=(7sqrt3)/(sqrt3-1)xx(sqrt3+1)/(sqrt3+1)` `=(7sqrt3)/(3-1)xx(sqrt3+1)` `=(7sqrt3)/(2)xx(sqrt3+1)` option`2` |
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| 594. |
if `3sec^4theta + 8 = 10sec^2theta` , then find the value of `tantheta` |
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Answer» `sec^2theta-tan^2theta=1` `sec^2theta=1+tan^2theta` `3(sec^2theta)^2+8=10sec^2theta` `3(1+tan^2theta)^2+8=10(1+tan^2theta)` `Let tan^2theta=t` `3(1+t)^2+8=10(1+t)` `3(1+t^2+2t)+8=10+10t` `3+3t^2+6t+8=10+10t` `3t^2+(6t-10t)+(8+3-10)=0` `3t^2-4t+1=0` `3t^2-3t-t+1=0` `3t(t-1)-1(t-1)=0` `(t-1)(3t-1)=0` `t=1,1/3` `tan^2theta=1,1/3` `tantheta=pm1,pm1/sqrt3` `tantheta=1,-1,1/sqrt3,-1/sqrt3`. |
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| 595. |
`3sec^4theta+8=10sec^2theta` then find the value of `tantheta` |
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Answer» `3sec^4theta +8 = 10sec^2theta` `=>3(1+tan^2theta)^2 +8 -10(1+tan^2theta) = 0` Let `tan^2 theta = x`, then, `=>3(1+x)^2+8 -10(1+x)= 0` `=>3(1+x^2+2x)-10-10x +8 = 0` `=>3x^2-4x +1 = 0` `=>3x^2-3x -x +1 = 0` `=>3x(x-1)-1(x-1) = 0` `=>(3x-1)(x-1) = 0` `=>x = 1/3 or x = 1` `=>tan^2theta = 1/3 or tan theta = 1` `=>tantheta = +-1/sqrt3 or tan theta = +-1.` |
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| 596. |
If `sec^4theta+sec^2theta=10+tan^4theta+tan^2theta", then "sin^2theta=`A. `2/3`B. `3/4`C. `4/5`D. `5/6` |
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Answer» Correct Answer - C `sec^4theta+sec^2theta=10+tan^4theta+tan^2theta` `rArrsec^4theta-tan^4theta+sec^4theta-tan^2theta=10` `rArr sec^2theta+tan^2theta+1=10` `rArr2sec^2theta=10` `rArrcos^2theta=1/5` `rArr sin^2theta=4/5` |
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| 597. |
There exists a value of `theta`between 0 and `2pi`that satisfies the equation `sin^4theta-2sin^2theta-1=0` |
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Answer» `sin^4theta-2sin^2theta-1=0` Let `sin^2theta=y` `y^2-2y-1=0` `y=(-(-2)pmsqrt(4-4(1)(-1)))/2` `y=(2pmsqrt8)/2` `y=(2pm2sqrt2)/2` `sin^2theta=1pmsqrt2` `sin^2theta=1+sqrt2` `sin^2theta=1+1.414` `sin^2theta=2.414` `sintheta=pmsqrt2.414` `sintheta=pm1.55` There is no value of `theta`. |
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| 598. |
Prove that `(a^2sin(B-C))/(sinb+sinC)+(b^2"sin"(C-A))/(sinC+sinA)+(c^2"sin"(A-B))/(sinA+sinB)=0` |
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Answer» `(a^2sin(B-C))/(sinB+sinC) = ((2RsinA)^2 sin(B-C))/(sinB+sinC)` `=(4R^2sin^2Asin(B-C))/(sinB+sinC)` `=(4R^2sinA*sinAsin(B-C))/(sinB+sinC)` `=(4R^2sinA*sin(pi-(B+C)sin(B-C))/(sinB+sinC)` `=(4R^2sinA*sin(B+C)sin(B-C))/(sinB+sinC)` `=(4R^2sinA*(sin^2B-sin^2C))/(sinB+sinC)` `=(4R^2sinA*(sinB+sinC)(sinB-sinC))/(sinB+sinC)` `=4R^2sinA(sinB-sinC)` Similarly, `(b^2sin(C-A))/(sinC+sinA) =4R^2sinB(sinC-sinA)` `(c^2sin(A-B))/(sinA+sinB) =4R^2sinC(sinA-sinB)` `:. L.H.S. = (a^2sin(B-C))/(sinB+sinC) +(b^2sin(C-A))/(sinC+sinA)+(c^2sin(A-B))/(sinA+sinB) = 4R^2[sinAsinB -sinAsinC+sinBsinC-sinBsinA+sinCsinA-sinCsinB]` `=4R^2[0] = 0 = R.H.S.` |
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| 599. |
Given `A=sin^2theta+cos^4theta,`then for all real `theta,`(a)`1lt=Alt=2`(b) `3/4lt=Alt=1`(c)`(13)/(16)lt=Alt=1`(d) `3/4lt=Alt=(13)/(16)` |
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Answer» `A = sin^2theta+cos^4theta` `=>A = 1-cos^2theta+cos^4theta` `=>A= 1-cos^2theta(1-cos^2theta)` `=>A= 1-cos^2theta(sin^2theta)` `=>A= 1-1/4(2sinthetacostheta)^2` `=>A= 1-1/4sin^2 2theta` Now, we know, `0 le sin^2 2theta le 1` `=>0 le 1/4sin^2 2theta le 1/4` `=>3/4 le 1- 1/4sin^2 2theta le 1` `=>3/4 le A le 1` So, option - `(b)` is the correct option. |
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| 600. |
Solve : `5cos2theta+2cos^2(theta/2)+1=0,-pi lt theta lt pi` |
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Answer» `5cos2theta+2cos^2(theta/2)+1=0` `5(2cos^2theta-1)+(1+costheta+1=0` `10cos^2theta-5+1+costheta+1=0` `10cos^2theta+costheta-3=0` `10cos^2theta+6costheta-5costheta-3=0` |
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