InterviewSolution
Saved Bookmarks
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.
| 1. |
In an acute angled `triangle ABC`, the minimum value of `tanA tanB tanC` is |
| Answer» As, `tanA*tanB*tanC= tanA+ tanB + tanC`Since, `A.M ge G.M``(tanA+ tanB + tanC)/3 ge root(3)(tanA*tanB*tanC)``root(2/3)(tanA+ tanB + tanC) ge 3`So, `(tanA + tanB+ tanC) ge 3^(3/2)`Hence,` min(tanA + tanB+ tanC)= 3sqrt3` | |
| 2. |
Consider the system of linear equations in `x , ya n dz :``""(sin3theta)x-y+z=0(cos2theta)x+4y+3z=0``3x+7y+7z=0`Which of the following can be the value of `theta`for which the system has a non-trivial solution`npi+(-1)^npi/6,AAn in Z``npi+(-1)^npi/3,AAn in Z``npi+(-1)^npi/9,AAn in Z`none of these |
|
Answer» As given equations has a non-trivial solution, `:. |[sin3theta,-1,1],[cos2theta,4,3],[2,7,7]| = 0` `=>sin3theta(28-21)-cos2theta(-7-7) +2(-3-4) = 0` `=>7sin3theta + 14cos2theta -14 =0` `=>sin3theta +2cos2theta -2 = 0` `=>3sintheta - 4sin^3theta +2- 4sin^2theta -2 =0` `=>-sintheta(4sin^2theta + 4sintheta -3) = 0` `=>sintheta(4sin^2theta + 6sintheta - 2sintheta -3) = 0` `=>sintheta(2sintheta(2sintheta+3) - 1(2sintheta +3)) = 0` `=>sintheta(2sintheta- 1)(2sintheta +3) = 0` `=>sintheta = 0 or sin theta = 1/2 or sin theta = -3/2` `sin theta` can not be less than `-1`, so `sin theta != -3/2`. `:. sintheta = 0 or sin theta = 1/2` `theta = npi or theta = npi+(-1)^n(pi/6).` |
|
| 3. |
In `(0, 4pi)`, the number of solutions of `2sin^(2)theta=cos2theta` areA. `4`B. `2`C. `8`D. `6` |
| Answer» Correct Answer - C | |
| 4. |
If `f(theta)=(1-sin2theta+cos2theta)/(2cos2theta)`, then value of `8f(11^0)*f(34^0)`is ____ |
|
Answer» `f(theta) = (1-sin2theta+cos2theta)/(2cos2theta)` `=((cos^2theta+sin^2theta - 2sinthetacostheta) +(cos^2theta-sin^2theta) )/(2(cos^2theta - sin^2theta))` `= ((costheta - sintheta)^2+(costheta+sintheta)(costheta - sintheta))/(2(costheta+sintheta)(costheta - sintheta))` `=(costheta-sintheta+costheta+sintheta)/(2(costheta+sintheta))` `=(2costheta)/(2(costheta+sintheta))` `=1/(1+tantheta)` `:. f(theta) = 1/(1+tantheta)` `:. 8f(11^@)f(34^@) = 8(1/(1+tan11^@))(1/(1+tan34^@))->(1)` Now, `tan 34^@ = tan(45^@-11^@) = (tan45^@-tan11^@)/(1+tan45^@tan11^@) = (1-tan11^@)/(1+tan11^@)` `:. 1+tan34^@ = 1+ (1-tan11^@)/(1+tan11^@) = 2/(1+tan11^@)` Putting value of `1+tan34^@` in (1), `8f(11^@)f(34^@) = 8(1/(1+tan11^@))*((1+tan11^@)/2)` `=>8f(11^@)f(34^@) = 4.` |
|
| 5. |
If A = `4 sin theta + cos^(2) theta` , which of the following is not true ?(A) Maximum value of A is 5 .(B) Minimum value of A is `-4`(C) Maximum value of A occurs when `sin theta = 1//2`(D) Minimum value of A occurs when `sin theta = 1`.A. Maximum value of A is 5 .B. Minimum value of A is `-4`C. Maximum value of A occurs when `sin theta = 1//2`D. Minimum value of A occurs when `sin theta = 1`. |
|
Answer» (a) , ( c) , (d) `f(theta) = 4 sin theta cos^(2) theta = 4 sin theta + 1 - sin^(2) theta` `= 5 -(4-4sin theta + sin^(2) theta) = 5 - (sin theta -2)^(2)` Now maximum value of f`(theta)` occurs when `(sin theta -2)^(2)` is minimum . Minimum value of `(sin theta -2)^(2)` occurs when `sin theta =1` , then maximum value of `f(theta)" is " 5-(-1-2)^2=4`. Also minimum value of `f(theta)` occurs when `(sintheta-2)^2` is maximum. Mximum value of `(sintheta-2)^2` occurs when `sintheta=-1`, then minimum value of `f(theta)" is " 5-(-1-2)^2=-4`. |
|
| 6. |
Show that the equation `sintheta=x+1/x`is not possible if `x`is real. |
|
Answer» `x+1/x = (sqrtx)^2+(1/sqrtx)^2` `=(sqrtx-1/sqrtx)^2 +2*sqrtx*1/sqrtx` `=>x+1/x = (sqrtx-1/sqrtx)^2 +2` `:. x+1/x ge 2.` But, `-1 le sin theta le 1` , so `sin theta` can not be greater than `1`. Thus, `sintheta = x+1/x` is not possible if `x` is real. |
|
| 7. |
In `(0, 2pi)`, the number of solutions of `cos2theta=sintheta` areA. `1`B. `2`C. `3`D. `4` |
| Answer» Correct Answer - C | |
| 8. |
Find the values of x for which `3costheta=x^2-8x+19` holds good. |
|
Answer» `3cos theta=x^2-8x+19` `rArr3costheta=(x-4)^2+3` Now, `L.H.S. =3costhetale3` or L.H.S. has greatest value 3 But `R.H.S.(x-4)^2+3ge3` or R.H.S.has least value 3 Hence, L.H.S.=R.H.S. when `3costheta=(x-4)^2+3=3` `rArr costheta=1" and "x-4=0` `rArr theta=2n pi " and " x=4," where "in Z`. |
|
| 9. |
Find the number of solution of `theta in [0,2pi]`satisfying the equation `((log)_(sqrt(3))tantheta(sqrt((log)_(tantheta)3+(log)_(sqrt(3))3sqrt(3))=-1` |
|
Answer» `log_(sqrt3)tantheta[sqrt(log_tantheta 3+log_sqrt3 3sqrt3)]=-1` `log_(sqrt3) tantheta[sqrt(2/log_sqrt3 tantheta)+3)]=-1` `ysqrt(2/y+3)=-1` `y^2(2/y+3)=1` `2y+3y^2=1` `3y^2+2y-1=0` `3y(y+1)-1(y+1)=0` `(y+1)(3y-1)=0` `y=-1` `log_sqrt3 tantheta=-1` `tantheta=(sqrt3)^(-1)` `tantheta=1/sqrt3` `theta=pi/6,7/6pi` `y=1/3` `log_sqrt3 tantheta=1/3` `tantheta=(sqrt3)^(1/3)`. |
|
| 10. |
If `sin alpha + sin beta = a and cos alpha + cos beta =b`, show that `sin(alpha+beta)=(2ab)/(alpha^2+beta^2)` |
|
Answer» `Sin(alpha) + Sin(beta)=a``Cos(alpha) + cos(beta)=b``2Sin((alpha +beta)/2)Cos((alpha-beta)/2)=a``2cos((alpha+beta)/2)cos((alpha-beta)/2)=b`then, `tan((alpha+beta)/2)= a/b` By using angles of triangle, `Sin((alpha+beta)/2)=a/sqrt(a^2+b^2)` `cos((alpha+beta)/2)= b/sqrt(a^2+b^2)` Hence, `Sin(alpha+beta)=2Sin((alpha+beta)/2)cos((alpha+beta)/2)``= (2ab)/(a^2+b^2)` |
|
| 11. |
The total number of solution of `sin{x}=cos{x}`(where `{}`denotes the fractional part) in `[0,2pi]`is equal to5 (b)6 (c) 8(d) none of theseA. 5B. 6C. 8D. None of these |
|
Answer» Correct Answer - option 2 |
|
| 12. |
Show that the equation `sintheta=x+1/x` is not possible if x is real. |
|
Answer» Given, `sintheta=x+1/x` `:. Sin^2theta=x^2+1/x^2+2=(x-1/x)^2+4ge4` Which is not possible since `sintheta^2le1`. |
|
| 13. |
if `sin^2theta=(x^2+y^2+1)/(2x)` then `x` must beA. -3B. -2C. 1D. None of these |
|
Answer» Correct Answer - C `sin^2thetale1` `rArr (x^2+y^2+1)/(2x)le1` `orx^2+y^2-2x+1le0 (asxgt0)` `or (x-1)^2+y^2le0` It is possible if x - 1 = 0 and y = 0. |
|
| 14. |
If `alpha, beta` are different values of `theta` satisfying the equation `5 cos theta-12 sin theta=11`. If the value of `sin (alpha + beta) =-(5k)/(169)` , then find the value of k. |
|
Answer» given that, `5cos theta - 12sintheta= 11` `(5/13)sin theta - (12/13)sin theta = 11/13` `sin (phi- theta) = 11/13` let `sin x = sin a` we can say that `x=a or pi - a` `phi- theta = alpha- phi or pi-(alpha- phi) ` `pi - (alpha- theta) = (beta- phi)` `pi- alpha+ phi= beta - phi` `pi+ phi + phi = alpha + beta` `sin(pi + 2phi) = sin(alpha+ beta)` `sin (alpha+beta) = - sin 2 phi = -2sinphicos phi` `= -2*5/13* 12/13` `sin (alpha+ beta)= -120/169` `(-5k)/cancel(169) = -120/cancel(169)` we get, `:. k = 24` |
|
| 15. |
If : `(csc alpha - cot alpha)(csc beta -cot beta)(csc gamma-cot gamma)=(csc alpha +cot alpha)(csc beta + cot beta)(csc gamma + cot gamma),` then the value of each side isA. 1B. 2C. 3D. none of these |
| Answer» Correct Answer - A | |
| 16. |
Given that:`(1 + cos alpha) (1 + cos beta) (1 + cos gamma) = (1 - cos alpha) (1 - cos beta) (1 - cos gamma),` Show that one of the values of each member of this equality is `sin alpha sin beta sin gamma.`A. `cos alpha* cos beta* cos gamma`B. `sin alpha* sin beta* sin gamma`C. `sec alpha* sec beta* sec gamma`D. `tan alpha* tan beta* tan gamma` |
| Answer» Correct Answer - B | |
| 17. |
If `sin^2theta_1+sin^2theta_2+sin^2theta_3=0,`then which of the following is not the possible value of `costheta_1+costheta_2+costheta_3?`3 (b) `-3`(c) `-1`(d) `-2`A. 3B. -3C. -1D. -2 |
|
Answer» `sin^2theta_1+sin^2theta_2+sin^2theta_3=0` `rArr sin^2theta_1=sin^2theta_2=sin^2theta_3=0` `rArr cos^2theta_1,cos^2theta_2,cos^2theta_3=1` `rArr costheta_1,costheta_2,costheta_3=pm1` `cos^2theta_1+cos^2theta_2+cos^2theta_3"can be-3 (when all are -I)"` or 3 (when all are +I) or -1 (when any two are -1 and one is +1) or 1 (when any two are +11 and one is -1) but-2 is not a possible value. |
|
| 18. |
If `sin^2theta_1+sin^2theta_2+...+sin^2theta_n=0`, then find the minimum value of `costheta_1+costheta_2+...+costheta_n`. |
|
Answer» Correct Answer - `-n` `sin^2theta_1+sin^2theta_2+...+sin^2theta_n=0` `rArr sintheta_1=sintheta_2=...sintheta_n=0` `rArr costheta_1,costheta_2...,costheta_n=pm1` Therefore, the minimum value of `costheta_1+costheta_2+…+costheta_n` `=(-1)+(-1)+(-1)+…n" times "=-n` |
|
| 19. |
`sec^2 x=(4xy)/(x+y)^2` is true if and only ifA. `x+yne0`B. `x=y,xne0`C. `x=y`D. `xne0,yne0` |
|
Answer» Correct Answer - B Given, `sec^2theta=(4xy)/((x+y)^2ge1` Now `sec^2thetage1rArr(4xy)/((x+y)^2ge1` `or (x+y)^2le4xy` `or (x+y)^2-4xyle0` `or (x-y)^2le0` But for real values of x and y, `(x-y)^2ge0or(x-y)^2=0` `:. x=y` Also `x+yne0rArrxne0.yne0` |
|
| 20. |
If `x+y+z=pi/2,`then prove that`|[sinx,siny,sinz],[cosx,cosy,cosz],[cos^3x,cos^3 y,cos^3z]|=0` |
|
Answer» `L.H.S. = |[sinx,siny,sinz],[cosx,cosy,cosz],[cos^3x,cos^3y,cos^3z]|` By expending the determinant along `R_3`, `=sum cos^3x[sinycosz - cosysinz]` `=sum cos^3xsin(y-z)` As, `x+y+z = pi/2`, so the determinant becomes, `=sum cos^3(pi/2-(y+z))sin(y-z)` `=sum sin^3(y+z)sin(y-z)` `=sum sin^2(y+z) sin(y+z)sin(y-z)` `=sum sin^2(pi/2-x) (sin^2y-sin^2z)` `=sum cos^2x (sin^2y-sin^2z)` `=sum (1-sin^2x)(sin^2y-sin^2z)` Let `sin^2x = a, sin^2y = b, sin^2z = c`, then, our determinant becomes, `=sum(1-a)(b-c)` `=(1-a)(b-c)+(1-b)(c-a)+(1-c)(a-b)` `=(b-c+c-a+a-b)-ab+ac-bc+ab-ac+bc` `=0 = R.H.S.` |
|
| 21. |
If `sintheta_1+sintheta_2+sintheta_3=3," then "costheta_1+costheta_2+costheta_3` is equal toA. 3B. 2C. 1D. 0 |
|
Answer» Correct Answer - D The given relation is satisfied only when `sintheta_1=sintheta_2=sintheta_3=1` `rArrcostheta_1=costheta_2=costheta_3=0` `rArrcostheta_1+costheta_2+costheta_3=0` |
|
| 22. |
The minimum value of the expression `sin alpha + sin beta+ sin gamma`, where `alpha,beta,gamma` are real numbers satisfying `alpha+beta+gamma=pi` isA. positiveB. zeroC. negativeD. -3 |
|
Answer» Correct Answer - C For `alpha=-pi//2,beta=pi//2andgamma=2pi` `sinalpha+sinbeta=singamma=-2` Hence, the minimum value of the expression is negative. |
|
| 23. |
If `sinx+siny+sinz+sinw=-4" then the value of "sin^400x+sin^300y+sin^200z+sin^100w` isA. `sin^400x.sin^300y.sin^200z.sin^100w`B. `sinx.siny.sinz.sinw`C. 4D. 3 |
|
Answer» Correct Answer - C `sinx+siny+sinz+sinw=-4` `rArrsinx=siny=sinz=sinw` So,`sin^400x+sin^300y+sin^200x+sin^100w` `=-(-1)^400+(-1)^300+(-1)^200+(-)^100=4` |
|
| 24. |
Prove that:`(cos e c(90^0+theta)+cot(450^0+theta))/(cos e c(90^0-theta)+t a n(180^0-theta))+(t a n(180^0+theta)+s e c(180^0-theta))/(t a n(360^0+theta)-s e c(-theta))`=2 |
|
Answer» `L.H.S. = (cosec(90^@+theta)+cot(450^@+theta))/(cosec(90^@-theta)+tan(180^@-theta))+(tan(180^@+theta)+sec(180^@-theta))/(tan(360^@+theta)-sec(-theta)` `=(cosec(90^@+theta)+cot(360^@+90^@+theta))/(cosec(90^@-theta)+tan(180^@-theta))+(tan(180^@+theta)+sec(180^@-theta))/(tan(360^@+theta)-sec(theta)` `=(cosec(90^@+theta)+cot(90^@+theta))/(cosec(90^@-theta)+tan(180^@-theta))+(tan(180^@+theta)+sec(180^@-theta))/(tan(360^@+theta)-sec(theta)` `=(sec(theta)-tan(theta))/(sec(theta)-tan(theta))+(tan(theta)-sec(theta))/(tan(theta)-sec(theta)` `1+1 = 2= R.H.S.` |
|
| 25. |
If : `sin^(2)theta+sin^(4)theta=1, "then": cot^(2)theta+cot^(4)theta=`A. -1B. 1C. 2D. 4 |
| Answer» Correct Answer - B | |
| 26. |
If : `tan theta+cot theta=2, "then" : tan^(2)theta-tan^(3)theta=`A. `cot^(3)theta-cot^(2)theta`B. 1C. `cot^(2)theta+cot^(3)theta`D. 2 |
| Answer» Correct Answer - A | |
| 27. |
Prove that`(cos(90^0+theta)sec(-theta)"tan"(180^0-theta))/(sec(360^0-theta)sin(180^0+theta)cot(90^0-theta))=-1` |
|
Answer» `L.H.S.=(cos(90^@+theta)sec(-theta)tan(180^@-theta))/(sec(360^@-theta)sin(180^@+theta)cot(90^@-theta))` `=((-sintheta)(sectheta)(-tantheta))/((sectheta)(-sintheta)(tantheta))=-1=R.H.S.` |
|
| 28. |
If : `sectheta+costheta=2 "then" : sec^(2)theta-sec^(4)theta=`A. 1B. `cos^(2)theta+cos^(4)theta`C. -1D. `cos^(4)theta-cos^(2)theta` |
| Answer» Correct Answer - D | |
| 29. |
Show that `tan^(-1)[(cosx+sinx)/(cosx-sinx)]=(pi)/(4)+x`.A. `-x`B. `x`C. `(pi)/(4)-x`D. `(pi)/(4)+x` |
| Answer» Correct Answer - D | |
| 30. |
Prove that:`(sin(180^0+theta)cos(90^0+theta)t a n(270^0-theta)cot(360^0-theta))/(sin(360^0-theta)cos(360^0+theta)cos e c(-theta)"sin"(270^0+theta))=1` |
|
Answer» `L.H.S. = (sin(180^@+theta)cos(90^@+theta)tan(270^@-theta)cot(360^@-theta))/(sin(360^@-theta)cos(360^@+theta)cosec(-theta)sin(270^@+theta))` `= (sin(theta)(-sin(theta))cot(theta)(-cot(theta)))/((-sin(theta))(cos(theta)(-cosec(theta)(-cos(theta))` `=(sin^2thetacot^2theta)/(sinthetacosecthetacos^2theta)` `=(sin^2theta(cos^2theta/sin^2theta))/(sintheta(1/sintheta)cos^2theta)` `=cos^2theta/cos^2theta` `=1 = R.H.S.` |
|
| 31. |
Solve the equation `(sqrt(3))/2sinx-cosx=cos^2x` |
|
Answer» `sqrt3/2sinx=cos^2x+cosx` `sqrt3sinx=2(cos^2x+cosx)` `3sin^2x=4(cos^2x+cosx)^2` `3(1-cos^2x)=4(cos^4x+2cos^3x+cos^2x)` `4cos^4x+8cos^3x+7cos^2x-3=0` `(cosx+1)(2cosx-1)(2cos^2x+3cosx+3)=0` `cosx=-1,1/2` `x=(2n+1)pi,2npi+pi/3`. |
|
| 32. |
If `cos(theta - alpha) , costheta , cos(theta + alpha)` are in H.P. then `costheta.sec(alpha)/2 = ` |
|
Answer» If three numbers `a,b and c` are in H.P., then, `b = (2ac)/(a+c)` As, `cos(theta-alpha),costheta,cos(theta+alpha)` are in H.P., `:. costheta = (2cos(theta-alpha)cos(theta+alpha))/(cos(theta-alpha)+cos(theta+alpha))` `=>costheta = (2(cos^2thetacos^2alpha - sin^2thetasin^2alpha))/(2costhetacosalpha)` `=>costheta = (cos^2thetacos^2alpha - sin^2thetasin^2alpha)/(costhetacosalpha)` `=>costheta = (cos^2thetacos^2alpha - (1-cos^2theta)sin^2alpha)/(costhetacosalpha)` `=>costheta = (cos^2thetacos^2alpha - sin^2alpha+sin^2alphacos^2theta)/(costhetacosalpha)` `=>costheta = (cos^2theta(cos^2alpha + sin^2alpha)-sin^2alpha)/(costhetacosalpha)` `=>cos^2thetacosalpha = cos^2theta-sin^2alpha` `=>sin^2alpha = cos^2theta(1-cosalpha)` `=>(2sin(alpha/2)cos(alpha/2))^2 = cos^2theta(2sin^2(alpha/2))` `=>4sin^2(alpha/2)cos^2(alpha/2) = 2cos^2thetasin^2(alpha/2)` `=>2cos^2(alpha/2) = cos^2theta` `=>2 = cos^2thetasec^2(alpha/2)` `=>costhetasec(alpha/2) = sqrt2` |
|
| 33. |
If `cosx = -sqrt15/4` and `pi/2 < x < pi`, find the value of `sinx` |
|
Answer» `sin^x+cos^2x=1` `sin^2x=1-cos^2x` `=1-(sqrt15/4)^2` `=1-15/16=1/16` `sinx=pm1/4` `sinx=1/4`. |
|
| 34. |
Prove that:`("cos"(2pi+theta)"c o s e c"(2pi+theta)"tan"(pi//2+theta))/("sec"(pi//2+theta)costheta"cot"(pi+theta))=1` |
|
Answer» `L.H.S. = (cos(2pi+theta)cosec(2pi+theta)tan(pi/2+theta))/(sec(pi/2+theta)costhetacot(pi+theta))` `= (cos(theta)cosec(theta)(-cottheta))/((-cosectheta)costhetacot(theta))` `= (-costhetacosecthetacottheta)/(-costhetacosecthetacottheta)` `=1 = R.H.S.` |
|
| 35. |
If `xsina+ysin2a+zsin3a=sin4a``xsinb+ysin2b+zsin3b=sin4b`,`xsinc+ysin2c+zsin3c=sin4c`,then the roots of the equation `t^3-(z/2)t^2-((y+2)/4)t+((z-x)/8)=0,a , b , c ,!=npi,`are(a)`sina ,sinb ,sinc`(b) `cosa ,cosb ,cosc`(b)`sin2a ,sin2b ,sin2c`(d) `cos2a ,cos2bcos2c` |
|
Answer» `xsina+ysin2a+2sin3a=sin4a` `xsina+y(2sinacosa)+2(3sina-4sin3a)=2sin2acos2a` `xsina+2ysinacosa+sina2(3-4sin^2a)=4sinacosacos2a` `a+2ycosa+2(3-4sin^2a)=4cosacos2a` `8cos^3a-4zcos^2a-(2y+4)cosa+(z-x)=0` `cos^3a-(z/2)cos^a-((y+2)/4)cosa+((z-x)/8)=0` `t=cosa` `t^3-(z/2)t^2-((y+z)/4)t+((z-x)/8)=0` Cosa is a root. cosb and cosc are roots. |
|
| 36. |
Prove that: `sin^2pi/(18)+sin^2pi/9+sin^2(7pi)/(18)+sin^2(4pi)/9=2` |
|
Answer» `L.H.S. = sin^2(pi/18)+sin^2((pi)/9)+sin^2((7pi)/18)+sin^2((4pi)/9)` `=sin^2 10^@+ sin^2 20^@+ sin^2 70^@+sin^2 80^@` `=sin^2 10^@+ sin^2 80^@+ sin^2 70^@+sin^2 20^@` `=sin^2 10^@+ sin^2 (90-10)^@+ sin^2 (90-20)^@+sin^2 20^@` `=(sin^2 10^@+ cos^2 10^@)+ (cos^2 20^@+sin^2 20^@)` `=1+1= 2 = R.H.S.` |
|
| 37. |
`cos 55^(@) + cos 65^(@) + cos 175^(@)=`A)0B)1C)`sin 18^(@)`D)`cos 36 ^(@)` |
| Answer» Correct Answer - A | |
| 38. |
The equation `tan^4x-2sec^2x+a=0`will have at least one solution if |
|
Answer» `tan^4x-2sec^2x +a = 0` `=>tan^4x-2(1+tan^2x) +a = 0` `=>tan^4x-2tan^2x -2 +a = 0` `=>tan^4x-2tan^2x +1-3 +a = 0` `=>(tan^2x - 1)^2 = 3-a` To have at least one solution for this equation, `3-a ge 0` `=>a le 3` So, option `c` is the correct option. |
|
| 39. |
Which of the following is the least?sin 3 (b) sin 2 (c) sin 1 (d) sin 7 |
|
Answer» `r` radian `r*180/pi` deegree `3->3*180/pi=171.96^@` `2->2*180/pi=114.64^@` `1->57.33^@` `7->401.27^@` `sin3=sin(171.96^@)=0.139` `sin2=sin(114.64^@)=0.9089` `sin1=sin(57.32^@)=0.8416` `sin7=sin(401.27^@)=0.6592` Sin3 is least. |
|
| 40. |
If `("tan"(theta+alpha))/a=""("tan"(theta+beta))/b=""("tan"(theta+gamma))/c``(a+b)/(a-b)s in^2(alpha-beta)+(b+c)/(b-c)s in^2(beta-gamma)+(c+a)/(c-a)s in^2(gamma-alpha)=0` |
|
Answer» `x/y=(tan(theta+alpha))/(tan(theta+beta))` `(x+y)/(x-y)=(tan(theta+alpha)+tan(theta+beta))/(tan(theta+alpha)-tan(theta+beta))` `(x+y)/(x-y)=(sin(2theta+alpha+beta))/(sin(alpha-beta))` `(x+y)/(x-y)sin^2(alpha-beta)=sin(2theta+alpha+beta)sin(alpha-beta)-(1)` `=1/2(cos2(theta+beta)-cos2(theta+alpha))` `(y+z)/(y-z)sin^2(beta-gamma)=1/2[cos2(theta+gamma)-cos2(theta+beta))-(2)` `(z+x)/(z-x)sin^2(gamma-alpha)=1/2[cos2(theta+alpha)-cos2(theta+gamma)]-(3)` adding equation 1,2and3 `sum(x+y)/(x-y)sin^2(alpha-beta)=0`. |
|
| 41. |
The general solution of the equation `sinx-3sin2x+sin3x=cosx-3cos2x+cos3x`is `(n in Z)``npi+pi/8`(b) `(npi)/2+pi/8``(-1)^n(npi)/2+pi/8`(d) `2npi+cos^(-1)2/3` |
|
Answer» `(sinx+sin3x) - 3sin2x = (cosx+cos3x) - 3cos2x` `=>(2sin2xcosx) - 3sin2x = (2cos2xcosx) - 3cos2x` `=>sin2x(2cosx-3) = cosx(2cosx-3)` `=>sin2x = cos2x` `=>tan2x = 1` `=>2x = npi+pi/4` `=>x = (npi)/2+pi/8` So, option - `(b)` is the correct option. |
|
| 42. |
The value of `sin50^(@)-sin70^(@)+sin10^(@)` isA. 1B. 0C. `(1)/(2)`D. 2 |
|
Answer» Correct Answer - B Given expression, `sin50^(@)-sin70^(@)+sin10^(@)` `" "=2cos((50^(@)+70^(@))/(2))*sin((50^(@)-70^(@))/(2))+sin10^(@)` `" "=-2cos60^(@)sin10^(@)+sin10^(@)` `" "=-2*(1)/(2)sin10^(@)+sin10^(@)=0` |
|
| 43. |
If H is the orthocentre of `Delta ABC,` then : `cos(angle AHB)=`A. `- cos A`B. `- cos B`C. `- cos C`D. `-cos D` |
| Answer» Correct Answer - C | |
| 44. |
If : `(1+sintheta)/(1+costheta)+(1-sintheta)/(1-costheta)=2u.(csctheta-costheta),"then" : u =`A. `sintheta`B. `costheta`C. `sectheta`D. `csctheta` |
| Answer» Correct Answer - D | |
| 45. |
The minimum of ` 3cosx +4sin x+8` isA. 5B. 9C. 7D. 3 |
|
Answer» Correct Answer - D Given expression, `3cosx+4sinx+8 ` Let `" "` ` y=3cosx+4sinx+8` `rArr" "y-8=3cosx+4sinx` `therefore" "` Minimum value of `y-8=-sqrt(9+16)` `rArr" "y-8=-5rArry=-5+8` `therefore" "y=3` Hence, the minimum value of `3cosx+4sinx+8` is 3. |
|
| 46. |
`tan^(3)x-3tanx=0`A. `npi, npipm(2pi)/(3), ninZ`B. `2npi, 2npipm(2pi)/(3), ninZ`C. `npi, npipm(pi)/(3), ninZ`D. `npi, 2npipm(pi)/(3), ninZ` |
| Answer» Correct Answer - C | |
| 47. |
The principal solutions of `2sin^(2)x=3cosx` areA. `(pi)/(3), (5pi)/(3)`B. `(pi)/(3), (2pi)/(3)`C. `(2pi)/(3), (4pi)/(3)`D. `(4pi)/(3), (5pi)/(3)` |
| Answer» Correct Answer - A | |
| 48. |
In ` A B C ,a , ca n dA`are given and `b_1,b_2`are two values of the third side `b`such that `b_2=2b_1dot`Then prove that `sinA=sqrt((9a^2-c^2)/(8c^2))` |
|
Answer» `cosA=(b^2+c^2-a^2)/(2bc)` `b^2-2bc*cosA+(c^2-a^2)=0` `b_1+b_2=2cosA` `b_1b_2=c^2-a^2` `3b_1=2cosa,2b_1^2=c^2-a_2` `2(2/3c cosA)^2=c^2-a^2` `8c^2(1-sin^2A)=9c^2-9a^2` `sinA=sqrt((9a^2-c^2)/(8c^2))`. |
|
| 49. |
If `A,B,C` are in A.P then `(sinA-sinC)/(cosC-cosA)=`A. tan AB. cot AC. tan BD. cot B |
| Answer» Correct Answer - D | |
| 50. |
Let ABC be an acute angled triangle whose orthocentre is at H. Ifaltitude from A is produced to meet the circumcircle of triangle ABC at `D`, then prove `H D=4RcosBcosC` |
|
Answer» `/_BHN` and `/_BDN` are congurent `HN=ND=2RcosBcosC` `HD=2HN` `2*2RcosBcosC` `4RcosBcosC`. |
|