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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.
| 351. |
Find in degrees the angle subtended at the centre of a circle ofdiameter 50cm by an arc of length 11cm. |
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Answer» `D=50cm` `r=D/2=25cm` Length of an arc=`(pirtheta)/180=11cm` `theta=396/(5pi)` `theta=79.2/pi`. |
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| 352. |
If any quadrilateral ABCD, prove that`"sin"(A+B)+sin(C+D)=0``"cos"(A+B)=cos(C+D)` |
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Answer» In quadrilateral `ABCD, A+B+C+D=2pi`. (a) `sin(A+B)+sin(C+D)=sin(A+B)+sin(2pi-(A+B))` `=sin(A+B)-sin(A+B)=0` (b) `cos(A+B)=cos(C+D)` `=cos(2pi-(C+D))=cos(C+D)` |
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| 353. |
In cyclic quadrilateral ABCD (none of these being `90^(@)`), which of the following is not true ?A. tan A cot C = -1B. sec B cos D = -1C. cosec B sin D = 1D. none of these |
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Answer» Correct Answer - D `angle A = pi - C` `rArr tan A = - tan C` `rArr A cot C=-1` `angle B=pi-D` `rArr sec B=-sec D` `rArr sec B cos D=-1` `rArr angle B=pi-D` `rArr cosec B sin D = 1` |
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| 354. |
The two legs of a right triangle are `sin theta +sin ((3pi)/2-theta)` and `cos theta -cos( (3pi)/2-theta)` The length of its hypotenuse is hypotenuse isA. 1B. 2C. `sqrt(2)`D. none of these |
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Answer» Correct Answer - C `a =sin theta + sin((3pi)/(2)-theta)` `=sin theta - cos theta` `b = cos theta - cos((3pi)/(2)-theta)` `= cos theta + sin theta` `a^(2)+b^(2)=(sin theta - cos theta)^(2)+(sin theta +cos theta)^(2)=2` `therefore` Hypotenuse `= c = sqrt(2)` |
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| 355. |
Prove that`(cos10^0+sin10^0)/(cos10^0-sin 10^0)=tan55^0` |
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Answer» `L.H.S. = (cos10^@ + sin10^@)/(cos10^@ - sin10^@)` `=(cos10^@(1+tan10^@))/(cos10^@(1-tan10^@))` `=(1+tan10^@)/(1-tan10^@)` `=(tan45^@+tan10^@)/(1-tan45^@tan10^@)` `=tan(45^@+10^@)` `=tan55^@ = R.H.S.` |
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| 356. |
The value of `2 cos10^(@)+sin 100^(@)+sin 1000^(@)+sin 10000^(@)` is |
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Answer» Correct Answer - C `2cos 10^(@)+sin 100^(@)+sin 1000^(@)+sin 10000^(@)` `=2cos 10^(@)+sin(90^(@)+10^(@))+sin(3xx60^(@)-80^(@))+sin(28xx360^(@)-80^(@))` `=2cos10^(@)+cos10^(@)-sin80^(@)-sin80^(@)` `=3 cos10^(@)-2cos 10^(@)` `=cos 10^(@)` |
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| 357. |
Let `f(x)=a sin x+c`, where a and c are real numbers and a>0. Then `f(x)lt0, AA x in R` ifA. `c lt -a`B. `c gt -a`C. `-a lt c lt a`D. `c lt a` |
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Answer» Correct Answer - A a `sin x + c lt 0, AA x in R` `rArr sin x lt - (c )/(a)` `rArr -(c )/(a)gt sin x, AA x in R` `rArr -(c )/(a)gt 1` `rArr -c gt a` `rArr c lt -a` |
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| 358. |
Find sum of maximum and minimum values of the function `f(x) = sin^2x + 8cosx - 7`A. `-4`B. `-5`C. 4D. 5 |
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Answer» Correct Answer - B `y = sin^(2)x + 8cos x-7` `1-cos^(2)x+8cos x-7` `=-[cos^(2)x-8 cos x+6]` `=-[(cos x-4)^(2)-10]` `=10-(4-cos x)^(2)` `therefore y_(min)=10-16=-6` `y_(max)=10-9=1` |
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| 359. |
In `triangleABC, (cos2A)/(a^(2))-(cos2B)/(b^(2))=`A. `b^(2)-a^(2)`B. `a^(2)-b^(2)`C. `(1)/(b^(2))-(1)/(a^(2))`D. `(1)/(a^(2))-(1)/(b^(2))` |
| Answer» Correct Answer - D | |
| 360. |
If `tantheta =-4/3," then " sintheta` isA. `-4/5" but not "4/5`B. `-4/5or4/5`C. `4/5" but not "-4/5`D. None of these |
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Answer» Correct Answer - B `tantheta=(-4)/3` `rArr theta in2nd` quadrant or 4th eqyadrant `rArr sintheta =pm4//5` If `theta in 2nd` equadrant, `sintheta=4//5` If `theta` in 4th quadrant, `sintheta=-4//5` |
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| 361. |
Find the general solution of : `2tanx-cotx+1=0`.A. `npi+(3pi)/(4), npi+tan^(-1)((1)/(2)), ninZ`B. `npi+(7pi)/(4), npi+tan^(-1)((1)/(2)), ninZ`C. `npi+(3pi)/(4), npi-tan^(-1)((1)/(2)), ninZ`D. `npi+(7pi)/(4), npi-tan^(-1)((1)/(2)), ninZ` |
| Answer» Correct Answer - A | |
| 362. |
The value of `3(sin^4t+cos^4t-1)/(sin^6t+cos^6t-1)`is equal to __________ |
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Answer» Correct Answer - 2 `sin^4t+cos^4t-1=(sin^2t=cos^2t)^2-2sin^2tcos^2t-1` `=-2sin^2tcos^2t` `sin^6t+cos^6t-1=(sin^2t+cos^2t)^3-3sin^2tcos^2t-1` `=-3sin^2tcos^2t` Hence, `3(sin^4t+cos^4t-1)/(sin^6t+cos^6t-1)`=2` |
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| 363. |
ABC is a triangle such that `sin(2A+B)=sin(C-A)=-sin(B+2C)=1/2`. If A,B, and C are in AP. then the value of A,B and C are.. |
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Answer» Here, `sin(B+2C) = -1/2` As it is a negative value, `(B+2C)` is greater than `180^@` .`:. B+2C = 180+30=> B+2C = 210^@->(1)` `sin(C-A) = 1/2=> C-A = 30^@->(2)` `sin(2A+B) = 1/2 => 2A+B = 180-30=>2A+B = 150^@->(3)` As, `A,B and C` are in AP. `:. A+C = 2B->(4)` From (1), `2B+4C = 420^@` `=>A+5C = 420^@->(5)` Adding (2) and (5), `C-A+A+5C = 420+30=> 6C = 450=> C = 75^@` `A = 75-30 = 45^@` `B = (45+75)/2 = 60^@` |
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| 364. |
If `sintheta-costheta=1`, then the value of `sin^3theta-cos^3theta` is __________ . |
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Answer» Correct Answer - 1 `sintheta-costheta=1` `orsin^2theta+cos^2theta-2sinthetacostheta=1` `or sinthetacostheta=0` Now `sin^3theta-cos^3theta=1(sin^2thta+cos^2theta+sinthetacostheta)` `=(sintheta-costheta)^2+3sinthetacostheta` `=1+3sinthetacostheta=1` |
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| 365. |
The number of real solution of equation `Sin(e^x) = 5^x + 5^(-x)` is : |
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Answer» `sin(e^x) = 5^x+5^(-x)` `sin(e^x) :` Maximum value of `sin(e^x)` can be `1` and minimum value can be `-1`. `5^x+5^-x = 5^x+1/5^x` Now, `5^x` is always greater than `0` which means, which means `5^x+1/5^x` will always be greater than or equall to `2`. so, minimum value of `5^x+5^-x` will be `2`. So, we can see that in given equation, maximum value of left side is `1` and minimum value of right side is `2`. It means there is no solution possible that will satisfy the given equation. |
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| 366. |
Prove that :`sintheta/cos[3theta]+sin[3theta]/cos[9theta]+sin[9theta]/cos[27theta]=1/2[tan[27theta]-tan[theta]]` |
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Answer» `tan3theta=tantheta=(sin3theta)/cos3theta-sintheta/costheta` `=(sin3thetatantheta-sinthetacos3theta)/(cos3theta*costheta)` `=sin(3theta-theta)/(cos3theta*costheta)` `=(2sintheta)/(cos3theta)` `sintheta/cos3theta=1/2(tan3theta-tantheta)-(1)` `(sin3theta)/(cos9theta)=1/2(tan9theta-tan3theta)-(2)` `(sin9theta)/(cos27theta)=1/2(tan27theta-tan9theta)-(3)` adding equation 1,2 and3 `sintheta/(cos3theta)+(sin3theta)/(cos9theta)+(sin9theta)/(cos27theta)=1/2(tan27theta-tantheta)`. |
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| 367. |
If`(sin^3theta-cos^3theta)/(sin theta-cos theta)-cos theta/sqrt(1+cot^2theta)-2tan theta.cot theta=-1` |
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Answer» `L.H.S. = (sin^3theta-cos^3theta)/(sintheta-costheta) - costheta/(sqrt(1+cot^2theta)) - 2tanthetacot theta` `=((sintheta-costheta)(sin^2theta+cos^2theta+sinthetacostheta))/(sintheta-costheta) - costheta/(sqrt(cosec^2theta)) - 2tantheta(1/tantheta)` `=(1+sinthetacostheta) -costheta/(cosec theta) -2` `=1+sinthetacostheta - costhetasintheta -2 ` `=1-2 =-1 = R.H.S.` |
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| 368. |
If `S_n=cot^-1(3)+cot^-1(7)+cot^-1(13)+cot^-1(21)+.....`, `n` terms, then |
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Answer» `S_n = cot^-1(3)+ cot^-1(7) + cot^-1(13) + cot^-1(21) .....` n terms `T_n = cot^-1(n^2 +n +1)` `S_n = sum_(n=1)^n cot^-1(n^2+n+1)` `= sum_(n=1)^n tan^-1[ 1/(n^2+n+1)]` `= sum_(n=1)^n tan^-1((n+1-n)/(1+n(n+1)))` `sum_(n=1)^n tan^-1 (n+1) - tan^-1(n)` `= tan^-1 2 - tan^-1 1 + tan^-13 ..... tan^-12+ ....+ tan^-1 (n+1) ` `= tan^-1(n+1) - tan^-1(1)` `S_10 = tan^-1(11) - tan^-1(1)` `= tan^-1((11-1)/(1+11)) = tan^-1(10/12) = tan^-1(5/6)` `S_oo = n-> oo { tan^-1 (n+1) - tan^1(1)}` `pi2 - pi/4 = pi/4` `S_6 = tan^-1 7 - tan^-1 1= tan^-1((7-1)/(1+7)) = tan^-1(6/8) = tan^-1(3/4)` `S_20 = tan^-1 21 - tan^-1= tan^-1((21-1)/(1+21)) = tan^-1(20/22)` option `a, b,d` are correct answer |
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| 369. |
The upper `3/4`th portion of a vertical pole subtends an angle `theta`such that tan `theta=3/5`at a point in the horizontal plane through its foot and at a distance 40mfrom the foot. Find the possible height of the vertical pole. |
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Answer» Let `h` is the height of the vertical pole. It is given that , `tan theta = 3/5` We can draw a diagram wuth the given details. Please refer to video for the diagram. From the diagram, `beta = alpha+theta` `=>tan beta = tan(alpha+theta)` `=>h/(CD) = (tanalpha+tantheta)/(1-tanalphatantheta)` `=>h/40 = ((h/4)/40 +3/5)/(1-(h/4)/40(3/5)` `=>h/40(1-h/160(3/5)) = h/160+3/5` `=>h/8(800-3h) = 800(h/32+3)` `=>3h^2-600h+19200 = 0` `=>h^2-200h+6400 = 0` `=>(h-40)(h-160) = 0` `=>h = 40m or h = 160m` |
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| 370. |
Find the general solution of the equation `sin2x+sin4x+sin6x=0`.A. `(npi)/(2), npipm(2pi)/(3), ninZ`B. `(npi)/(2), npipm(pi)/(3), ninZ`C. `(npi)/(4), npipm(2pi)/(3), ninZ`D. `(npi)/(4), npipm(pi)/(3), ninZ` |
| Answer» Correct Answer - D | |
| 371. |
General solution of `cos(x+(pi)/(10))=0` isA. `(2n+1)(pi)/(2)-(pi)/(10), ninZ`B. `(2n+1)(pi)/(2)+(pi)/(10), ninZ`C. `(2n+1)(pi)/(2)-(pi)/(5), ninZ`D. `(2n+1)(pi)/(2)+(pi)/(5), ninZ` |
| Answer» Correct Answer - A | |
| 372. |
General solution of `sinx=(sqrt(3))/(2)` isA. `npi+(-1)^(n)(pi)/(3), ninZ`B. `2npi+(-1)^(n)(pi)/(3), ninZ`C. `npi+(-1)^(n-1)(pi)/(3), ninZ`D. `npi+(-1)^(n+1)(pi)/(3), ninZ` |
| Answer» Correct Answer - A | |
| 373. |
General solutions of `sin3x=0` isA. `npi, ninZ`B. `2npi, ninZ`C. `(npi)/(3), ninZ`D. `(2npi)/(3), ninZ` |
| Answer» Correct Answer - C | |
| 374. |
General solution of `cos2x=0` isA. `(2n-1)(pi)/(2), ninZ`B. `(2n-1)(pi)/(4), ninZ`C. `(2n+1)(pi)/(2), ninZ`D. `(2n+1)(pi)/(4), ninZ` |
| Answer» Correct Answer - D | |
| 375. |
General solution of `sin((3x)/(2))=0` isA. `(npi)/(3), ninZ`B. `(2npi)/(3), ninZ`C. `npi, ninZ`D. `(3npi)/(2), ninZ` |
| Answer» Correct Answer - C | |
| 376. |
Number of ordered pairs (a, x) satisfying the equation `sec^2(a+2)x+a^2-1=0;-pi < x< pi` isA. 2B. 1C. 3D. infinite |
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Answer» Correct Answer - C We have `sec^(2)(a+2)x=1-a^(2)` Now `sec^(2)x ge 1` `rArr 1-alpha^(2)ge 1` `rArr a = 0` So, `sec^(2)(a+2)x = 1` `rArr sec^(2)2x = 1` `rArr x=-(pi)/(2), 0,(pi)/(2)` |
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| 377. |
If `1+sinx+sin^2x+sin^3x+oo`is equal to `4+2sqrt(3),0ltxltpi,` then `x`is equal to |
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Answer» `1+sinx+sin^2x+sin^3x+...oo = 4+2sqrt3` `=>1/(1-sinx) = 4+2sqrt3` `=>1-sinx = 1/(4+2sqrt3)**(4-2sqrt3)/(4-2sqrt3)` `=>1-sinx = (4-2sqrt3)/(16-12) = (4-2sqrt3)/4` `=>1-sinx = 1-sqrt3/2` `=>sinx = sqrt3/2` `:. x = pi/3 or (2pi)/3.` |
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| 378. |
The number of ordered pairs which satisfy the equation `x^2+2xsin(x y)+1=0`are (where `y in [0,2pi]`)1 (b)2 (c) 3(d) 0 |
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Answer» `x^2+2xsin(xy) +1 = 0` `=>(x+sin(xy))^2 - sin^2(xy)+1 = 0` `=>(x+sin(xy))^2 + cos^2(xy) = 0` `=>x+sin(xy) = 0 and cos(xy) = 0` When, `cos(xy) = 0, sin(xy) = sqrt(1-cos^2(xy)) = +-1` When `sin(xy) = 1` `x+sin(xy) = 0=>x+1 = 0 => x = -1` `=>sin(-y) = 1 =>siny = -1 => y = (3pi)/2` When `sin(xy) = -1` `x+sin(xy) = 0=>x-1 = 0 => x = 1` `=>sin(y) = -1 => y = (3pi)/2` So, there are two ordered pairs, `(1,(3pi)/2)` and `(-1,(3pi)/2)` that satisfy the given equation. |
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| 379. |
Prove that: `"sec"((3pi)/2-theta)"sec"(theta-(5pi)/2)+t a n((5pi)/2+theta)"tan"(theta-(3pi)/2)=-1.` |
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Answer» `L.H.S. = sec(((3pi)/2) - theta)sec(theta - ((5pi)/2))+ tan (((5pi)/2)+theta)tan(theta-((3pi)/2))` `= sec((3pi)/2 - theta)sec(-((5pi)/2- theta))+ tan ((5pi)/2+theta)tan(-((3pi)/2-theta))` As, `sec(-theta) = sec theta and tan (-theta) = -tan theta` So, our expression becomes, `= sec((3pi)/2 - theta)sec((5pi)/2- theta)- tan ((5pi)/2+theta)tan((3pi)/2-theta)` `=-cosecthetasec(2pi+(pi/2-theta))-tan(2pi+(pi/2+theta))tantheta` `= -cossecthetacosectheta+cotthetacottheta` `=cot^2theta - cosec^2theta` `=-(cosec^2theta - cot^2theta)` `=-1 = R.H.S.` |
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| 380. |
General solution of `cot4x=-1` isA. `(npi)/(4)+(3pi)/(4), ninZ`B. `(npi)/(4)+(3pi)/(16), ninZ`C. `npi+(3pi)/(4), ninZ`D. `npi+(3pi)/(16), ninZ` |
| Answer» Correct Answer - B | |
| 381. |
General solution of `tan5theta=cot2theta` isA. `(npi)/(7)+(pi)/(2), ninZ`B. `(npi)/(7)+(pi)/(14), ninZ`C. `(npi)/(7)+(pi)/(7), ninZ`D. `(npi)/(5)+(pi)/(10), ninZ` |
| Answer» Correct Answer - B | |
| 382. |
General solution of `cotx+cosecx=sqrt(3)` isA. `2npi+(pi)/(6), ninZ`B. `2npi+(pi)/(3), ninZ`C. `2npi+(pi)/(4), ninZ`D. `2npi+(pi)/(2), ninZ` |
| Answer» Correct Answer - B | |
| 383. |
Prove that `sec((3pi)/2-theta)sec(theta-(5pi)/2)+tan((5pi)/2+theta)tan(theta-(3pi)/2)=-1` |
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Answer» `sec((3pi)/2-theta)sec(theta-(5pi)/2)+tan((5pi)/2+0)tan(theta-(3pi)/2)` `=-cosecthetacosectheta+cotthetacot theta` `=-(cosec^2theta-cot^2theta)=-1` |
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| 384. |
Prove that`(cos3x)/(sin2xsin4x)+(cos5x)/(sin4xsin6x)+(cos7x)/(sin6xsin8x)+(cos9x)/(sin8xsin10 x)`= `1/2(cos e cx)[cos e c2x-cos e c10x]` |
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Answer» `L.H.S. = (cos3x)/(sin2xsin4x)+(cos5x)/(sin4xsin6x)+(cos7x)/(sin6xsin8x)+(cos9x)/(sin8xsin10x)` `=1/(2sinx)[ (2sinxcos3x)/(sin2xsin4x)+(2sinxcos5x)/(sin4xsin6x)+(2sinxcos7x)/(sin6xsin8x)+(2sinxcos9x)/(sin8xsin10x)]` `=1/(2sinx)[ (2sinxcos3x)/(sin2xsin4x)+(2sinxcos5x)/(sin4xsin6x)+(2sinxcos7x)/(sin6xsin8x)+(2sinxcos9x)/(sin8xsin10x)]` Using `2cosAsinB = sin(A+B) - sin(A-B)` `=1/(2sinx)[ (sin4x-sin2x)/(sin2xsin4x)+(sin6x-sin4x)/(sin4xsin6x)+(sin8x-sin6x)/(sin6xsin8x)+(sin10x-sin8x)/(sin8xsin10x)]` `=1/(2sinx)[ 1/(sin2x) - 1/(sin4x)+1/(sin4x) - 1/(sin6x)+1/(sin6x) - 1/(sin8x)+1/(sin8x) - 1/(sin10x)]` `=1/(2sinx)[1/(sin2x) - 1/(sin10x)]` `=1/2cosecx[cosec2x-cosec10x] = R.H.S.` |
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| 385. |
Find the number of pairs of integer `(x , y)`that satisfy the following two equations:`{cos(x y)=x ` and `tan(x y)=y`(a)`1` (b)` 2`(c) `4 `(d)` 6` |
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Answer» `cos(xy) = x and tan(xy) = y` `=>sin(xy)/cos(xy) = y` `=>sin(xy)/x = y` `=>sin(xy) = xy` Only possible value of `theta` when `sintheta = theta` is `0`. `:. xy = 0` `=>x = 0 or y = 0` When `x = 0`, `cos(0*y) = 0` `=>cos0 = 0`, which is not possible. `:. x = 0` is not possible. When `y = 0`, `cos(x*0) = x` `=>cos0 = x` `=>x = 1` `:. x = 1 and y = 0` is the solution for the given equation. `:.(1,0)` is the only pair that satisfy the given equation. |
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| 386. |
If `x, y in [0,2pi]` then find the total number of order pair `(x,y)` satisfying the equation `sinx .cos y = 1` |
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Answer» `sinxcosy = 1` `=> cosy = 1/sinx` `=> cosy = cosec x` Now, there are only two cases when `cos theta and cosec theta ` are equal.When both of them are `1` or `-1`. Case 1: When `cos y = cosecx = 1` `=>cos y = 1 , cosecx =1` `=>y = 0,2pi and x = pi/2` So, ordered pairs in this case are `(pi/2,0) and (pi/2,2pi)`. Case 2: When `cos y = cosecx = -1` `=>cos y = -1 , cosecx = -1` `=>y = pi and x = (3pi)/2` So, ordered pair in this case is`((3pi)/2,pi)`. So, there are `3` ordered pairs that satify the given equation. |
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| 387. |
`sin ((C-A)/2) =(c-a)/b cos(B/2)` |
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Answer» `a/sinA=b/sinB=c/sinA=k` `a=ksinA` `b=ksinB` `c=ksinC` RHS `=(ksinC-ksinA)/(ksinB)*cos(B/2)` `=(sinC-sinA)/sinB*cos(B/2)` `=(sin((C-A)/2)cos((C+A)/2))/(sin(B/2)cos(B/2))*cos(B/2)` `=sin((C-A)/2)[cos((C+A)/2)]/sin(B/2)` `=sin((C-A)/2)` LHS. |
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| 388. |
One of the general solutions of `4sinthetasin2thetasin4theta=sin3theta`is`(3n+-1)pi/(12),AAn in Z``(4n+-1)pi/9,AAn in Z``(3n+-1)pi/(12),AAn in Z``(3n+-1)pi/3,AAn in Z` |
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Answer» `4sinthetasin2thetasin4theta=sin3theta` `4sintheta*sin2theta*sin4theta-(3sintheta-4sin^3theta)=0` `sintheta(4sin2theta*sin4theta-3+4sin^2theta)=0` `2(2sinthetasin2theta)-3+2*(2sin^2theta)=0` `2(cos2theta-cos6theta)-3+2(1-cos2theta)=0` `2cos2theta-2cos6theta-3+2-2cos2theta=0` `-2cos6theta-1=0` `cos6theta=-1/2` `cos6theta=cos(2/3pi)` `6theta=2npipm2/3pi` `theta=(npi)/3pmpmpi/9` `=(3npm1)/pi/9` Option D is correct. |
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| 389. |
If `3tan(theta-15^0)=tan(theta+15^0),`then `theta`is equal to `n in Z)`n`pi+pi/4`(b) `npi+pi/8``npi+pi/3`(d) none of these |
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Answer» `3tan(theta-15^@) = tan(theta+15^@)` `=>(tan(theta+15^@))/(tan(theta-15^@)) = 3/1` Using componendo and dividendo, `=>(tan(theta+15^@)+tan(theta-15^@))/(tan(theta+15^@)-tan(theta-15^@)) = (3+1)/(3-1)` `=>(sin(theta+15^@)/cos(theta+15^@)+sin(theta-15^@)/cos(theta-15^@))/(sin(theta+15^@)/cos(theta+15^@)-sin(theta-15^@)/cos(theta-15^@)) = 2` `=>(sin(theta+15^@)cos(theta-15^@)+sin(theta-15^@)cos(theta+15^@))/(sin(theta+15^@)cos(theta-15^@)-sin(theta-15^@)cos(theta+15^@)) = 2` `=>(sin(theta+15^@+theta-15^@))/(sin(theta+15^@-theta+15^@)) = 2` `=>(sin2theta)/(sin30^@) = 2` `=>(sin2theta)/(1/2) = 2` `=>sin2theta = 1` `=>2theta =2npi+ pi/2` `=>theta = npi +pi/4 ` So, option-`(a)` is the correct option. |
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| 390. |
Solve `2sin^(2)x-5sinxcosx-8cos^2x=-2` |
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Answer» `2sin^2x-5sinxcosx-8cos^2x = -2` Dividing this equation by `cos^2x`, `=>2tan^2x - 5tanx - 8 = -2sec^2x` `=>2tan^2x - 5tanx - 8 = -2(1+tan^2x)` `=>2tan^2x - 5tanx - 8 = -2-2tan^2x)` `=>4tan^2x - 5tanx - 6 = 0` `=>4tan^2x - 8tanx +3tanx - 6 = -0` `=>(4tanx+3)(tanx-2) = 0` `=>tanx = -3/4 and tanx = 2` `=> x = npi+tan^-1(-3/4) and x = npi+tan^-1(2).` |
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| 391. |
Prove that :`sintheta/(cos(3theta)) + (sin3theta ) /(cos9theta) + (sin 9theta) /(cos27theta) =1/2( tan27theta-tantheta)` |
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Answer» `tan3theta - tan theta =( sin3theta)/(cos3 theta) - sin theta/cos theta` `= (sin3thetacostheta - sinthetacos3theta)/(cos3thetacos theta)` `=sin(3theta-theta)/(cos3thetacos theta)` `=sin(2theta)/(cos3thetacos theta)` `=(2sinthetacostheta)/(cos3thetacos theta) = (2sintheta)/(cos3theta)` `:. 1/2(tan3theta - tan theta) = (sintheta)/(cos3theta)` Similarly, `1/2(tan9theta - tan 3theta) = (sin3theta)/(cos9theta)` Similarly, `1/2(tan27theta - tan 9theta) = (sin9theta)/(cos27theta)` `:. (sintheta)/(cos3theta)+ (sin3theta)/(cos9theta) +(sin3theta)/(cos9theta) = 1/2(tan3theta - tan theta+tan9theta - tan 3theta+tan27theta - tan 9theta)` `=> (sintheta)/(cos3theta)+ (sin3theta)/(cos9theta) +(sin3theta)/(cos9theta) = 1/2(tan27 theta - tantheta)` |
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| 392. |
Prove that `sintheta+sin3theta+sin5theta+.....+sin(2n-1)theta=(sin^2ntheta)/(sintheta)dot` |
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Answer» Here, we will use, `sinalpha+sin(alpha+beta)+sin(alpha+2beta)+...+sin(alpha+(n-1)beta) = (sin((nbeta)/2))/(sin(beta/2))**sin(alpha+((n-1)beta)/2)` Now, `L.H.S. = sintheta+sin3theta+sin5theta+...+sin(2n-1)theta` `=sintheta+sin(theta+2theta)+sin(theta+2*2theta)+...sin(theta+(n-1)2theta)` `=(sin((2ntheta)/2))/(sin((2theta)/2))**sin(theta+((n-1)2theta)/2)` `=(sin(ntheta)/(sintheta))*sin(theta+ntheta-theta)` `=sin^2(ntheta)/sintheta = R.H.S.` |
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| 393. |
If `theta=pi/(4n)` then the value of `tanthetatan(2theta)tan(3theta)....tan((2n-1)theta)` is |
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Answer» Correct Answer - 1 `2n theta=pi//2` `:. theta,(2n-1)theta=(pi//2)-theta` `or 2theta,(2n-2)theta=(pi//2)-2theta,...` They form complementary angles A and B so that `tanA tanB=tanAcotA=1` for each pair. Also, `tanntheta=tan. Pi/4=1` Hence, the value of product is 1 |
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| 394. |
If A,B,C are angles of a triangle, then `2sin(A/2)cosec (B/2)sin(C/2)-sinAcot(B/2)-cosA` is(a)independent of A,B,C(b) function of A,B(c)function of C (d) none of these |
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Answer» `2sin(A/2)cosec(B/2)sin(C/2) - sinAcot(B/2) - cosA` `=2sin(A/2)cosec(B/2)sin((pi-(A+B))/2) - 2sin(A/2)cos(A/2)cos(B/2)/sin(B/2) - cosA` `=2sin(A/2)cosec(B/2)cos((A+B)/2) - 2sin(A/2)cosec(B/2)cos(A/2)cos(B/2) - cosA` `=2sin(A/2)cosec(B/2)[cos((A+B)/2) - cos(A/2)cos(B/2)] - cosA` `=2sin(A/2)cosec(B/2)[cos(A/2)cos(B/2)-sin(A/2)sin(B/2) - cos(A/2)cos(B/2)] - cosA` `=-2sin^2(A/2) - cosA` `=-(1-cosA) - cosA` `=-1` `:. 2sin(A/2)cosec(B/2)sin(C/2) - sinAcot(B/2) - cosA = -1` So, it is independent of `A,B and C`. |
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| 395. |
If in a triangle ABC, `sin^4A+sin^4B+sin^4C=sin^2B sin^2C+2 sin^2C sin^2A+2sin^2A sin^2B`, show that, one of the angles of the triangle is `30^@` or `150^@` |
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Answer» `sin^4A+sin^4B-2sin^2Asin^2B=sin^2Bsin^2C+2sin^2Csin^2Csin^2A-sin^4c` `(sin^2A-sin^2B)^2=sin^2c(sin^2B+2sin^2A-sin^2(A+B))` `sin^2(A-B)sin^2C=sin^2C(sin^2B+2sin^2A-sin^2(A+B)` `sin^2Acos^2B+sin^2Bcos^2A-2sinAcosBsinBcosA= sin^2B+2sin^2A-sin^2Acos^2b-sin^2Bcos^2A-2sinacosBsinBcosA` `sin^2A[cos^2B+cos^2B-2]=sin^2B[1-cos^2A-cos^2A]` `2sin^2A[cos^2B-1]=sin^2B[1-2cos^2A]` `-2sin^2A=1-2cos^2A` `2(cos^2A-sin2A)=1` `2A=pi/3,5pi/3` `A=pi/6,5/6pi` `A=30^0,150^0`. |
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| 396. |
In triangle `A B C ,`if `cosA+cosB+cosC=7/4, t h e n R/r`is equal to`3/4`(b) `4/3`(c) `2/3`(d) `3/2` |
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Answer» Here, we will use, `cosA+cosB+cosC = 1+4sin(A/2)sin(B/2)sin(C/2)` `cosA+cosB+cosC = 7/4` `:. 1+4sin(A/2)sin(B/2)sin(C/2) = 7/4` `=> 4sin(A/2)sin(B/2)sin(C/2) = 7/4-1` `=> 4sin(A/2)sin(B/2)sin(C/2) = 3/4->(1)` Now, `r = 4Rsin(A/2)sin(B/2)sin(C/2)` `:. r/R = 4sin(A/2)sin(B/2)sin(C/2)->(2)` From (1) and (2), `r/R = 3/4` `:. R/r = 4/3.` |
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| 397. |
Find the range of `f(x)=sin(cosx).` |
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Answer» Correct Answer - `[-sin1, sin 1]` `f(x)=sin(cosx)` For `AA x in R, theta=cosx in [-1,1]` Since `sintheta`is increasing from `-pi//2" to"pi//2`, the maximum value occurs at `thet=1` and the minimum value occurs at ` theta=-1`. Hence, rangs is `[-sin1, sin1]`. |
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| 398. |
Find the range of `12sintheta-9sin^2theta` |
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Answer» Correct Answer - [-21, 4] `f(x)=12sintheta-9sin^2theta` `=-(9sin^2theta-12sintheta)` `=-(9sin^2theta-12sintheta+4-4)` `=-((3sintheta-2)^2-4)=4-(3sintheta-2)^2` Minimum value of `f(theta)` occurs when `(3sintheta-2)^2` is minimum, which is 0. Hence, range of `f(theta)` is `[4-25,4]-=[-21,4]`. |
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| 399. |
Range of `f(theta)=cos^2theta(cos^2theta+1)+2 sin^2theta` isA. `[3//4,1]`B. `[3//16,1]`C. `[3//4,7//4]`D. `[7//4,2]` |
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Answer» Correct Answer - D `f(theta)=cos^2theta(cos^2theta+1)+2sin^2theta` `=cos^4theta+cos^2theta+sin^2theta+sin^2theta` `=cos^4theta+1+1-cos^2theta` `=(cos^2theta-1/2)^2+2-1/4` `=(cos^2theta-1/2)^2+7/4` `f_("min")=7/4` `f_("max")=1/4+7/4=2` |
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| 400. |
If `secxcos5x+1=0 `, where `0ltxle(pi)/(2)`, then find the value of x. |
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Answer» Given that, `" "secxcos5x+1=0` `" "(cos5x)/(cosx)+1=0rArrcos5x+cosx=0` `rArr2cos((5x+x)/(2))*cos((5x-x)/(2))=0` `" "[because cosx+cosy=2cos""(x+y)/(2)*cos""(x-y)/(2)]` `rArr" "2cos3x*cos2x=0` `rArr" "cos3x=0orcos2x=0` `rArr " "cos3x=cos""(pi)/(2) or cos2x=cos""(pi) /( 2)` `therefore" "3x=(pi)/(2)rArr2x=(pi)/(2)` and `" "x=(pi)/(6)rArrx=(pi)/(4)` Hence, the solutions are `(pi)/(2), (pi)/(4) and (pi)/(6)`. |
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