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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.
| 1. |
In triangle `A B C ,`if `sinAcosB=1/4 and 3t a n A=t a n B ,t h e ncot^2A`is equal toA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - B `3 sin A cos B=sin B cos A` `rArr cos A sin B =(3)/(4)` `rArr C=(pi)/(2).B=(pi)/(2)-A` `rARr 3tan A=tan ((pi)/(2)-A)` `rArr 3=cot^(2)A`. |
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| 2. |
The value of `(cos25^(@))/(sin70^(@)sin85^(@))+(cos70^(@))/(sin 25^(@)sin85^(@))+(cos85^(@))/(sin25^(@)sin70^(@))` isA. `1//2`B. 1C. 2D. `3//2` |
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Answer» Correct Answer - C Given expression `=(sin 50^(@)+sin 140^(@)+sin 170^(@))/(2sin 25^(@)sin 70^(@)sin 85^(@))` `=(4)/(2)` (Using sin 2A + sin 2B + sin 2C =4 sin A sin B sin C, where `A+B+c=180^(@)`) = 2 |
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| 3. |
If `tantheta=a/b` , show that `(asintheta-bcostheta)/(asintheta+bcostheta)=(a^2-b^2)/(a^2+b^2)` |
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Answer» `((asin theta- b cos theta ))/(( a sin theta+b cos theta ))=((atan theta-b))/((a tan theta +b))" " ["dividing num , and denom , by cos "theta]` `=((axx(a)/(b)-b))/((axx(a)/(b)+b))=((a^(2)-b^(2)))/((a^(2)+b^(2))).` |
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| 4. |
In a triangle `tan A+ tan B + tan C=6` and `tan A tan B= 2,` then the values of `tan A, tan B` and `tan C` areA. 1,2,3B. `3,2//3,7//3`C. `4,1//2,3//2`D. none of these. |
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Answer» Correct Answer - A In triangle. `tanA+tanB+tanC=tanA tan B tan C` `rARr 6=2tanC` or `tanC=3` Also `an A+tanB=6-3=3` By Eqs. I and ii, `tanA` and `tanB` are roots of `x^(2)-3x+2=0`. thus, `tanA,tanB=2,1` or `2,2` and `tanC=3`. |
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| 5. |
If `cos x+cos y-cos(x+y)=(3)/(2)`, thenA. `x+y=0`B. x=2yC. x=yD. 2x=y |
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Answer» Correct Answer - B `cos x+cosy-cos(x+y)=(3)/(2)` or `2cos((x+y)/(2))cos((x-y)/(2))-2cos^(2)((x+y)/(2))+1=(3)/(2)` or `2cos^(2)((x+y)/(2))-2cos((x+y)/(2))cos((x-y)/(2))+(1)/(2)=0`. Now `cos((x+y)/(2))` is always real, then discriminant `ge0`. Thus, `4cos^(2)((x-y)/(2))-4ge0` or `cos^(2)((x-y)/(2))ge1` or `cos^(2)((x-y)/(2))=1` or `(x-y)/(2))=0` or x=y. |
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| 6. |
If `alpha and beta` are non-zero real number such that `2(cos beta-cos alpha)+cos alpha cos beta=1.` Then which of the following is treu?A. `tan ((alpha )/(2)) + sqrt3 tan ((beta)/(2)) =0`B. `sqrt3 tan ((alpha)/(2)) + tan ((beta)/(2)) =0`C. `tan ((alpha )/(2)) - sqrt3 tan ((beta)/(2)) =0`D. `sqrt3 tan (( alpha )/(2)) - tan ((beta)/(2))= 0` |
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Answer» Correct Answer - A::C `cos beta ( 2+ cos alpha ) = 1+ 2 cos alpha ` `rArr (cos beta)/(1) = (1+2 cos alpha )/(2 + cos alpha )` `rArr (1-cos beta)/(1+cos beta) = (1-cos alpha )/(3(1+cos alpha ))` `rArr tan ^(2) ""(alpha )/(2) = 3 tan ^(2) ""(beta)/(2)` `rArr tan ""(alpha )/(2) = pm sqrt3 tan ""(beta)/(2)` |
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| 7. |
In a `DeltaPQR`. if `3 sinP +4 cosQ=6` and `4sinQ+3cosP=1`, then the angle `R` is equal to:A. `( 5pi)/(6)`B. `(pi)/(6)`C. `(pi)/(4)`D. `(3pi)/(4)` |
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Answer» Correct Answer - B `3 sin P + 4 cos Q =6" " `(i) `4 sin Q + 3 cos P =1" " `(ii) Squaring and adding (i) and (ii) we get ` sin (P+Q) = (1)/(2)` `rArr P + Q = (pi)/(6) or (5pi)/(6)` `rArr R =(5pi)/(6) or (pi)/(6)` If `R = (5pi)` then `0 lt P, Q lt (pi)/(6)` `rArr cos Q lt 1 and sin P lt (1)/(2)` `rArr 3 sin P + 4 cos Q lt (11)/(2)`, which is not possible So `R= (pi)/(6)` |
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| 8. |
If `A = sin^2x + cos^4 x`, then for all real x :A. `(3)/(4) le A le (13)/(16)`B. `(3)/(4) le A le 1`C. `(13)/(16) le A le 1`D. `1 le A le 2` |
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Answer» Correct Answer - B `A = sin ^(2) x + cos ^(4) x ` `= 1- cos ^(2) x + cos ^(4)x` `= 1 - cos ^(2) x (1- cos ^(2)x)` ` = 1- cos^(2)x sin ^(2) x ` ` = 1- ( sin ^(2) 2x)/( 4)` Now, `0 le sin ^(2) 2 x le 1` `rArr - (1)/(4) le - ( sin^(2) 2x )/(4) le 0` `rArr (3)/(4) le 1 - ( sin ^(2)2x)/(4x) le 1 ` `rArr 3//4 le A le 1` |
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| 9. |
Let `A and B` denote the statements `A : cos alpha + cos beta + cos gamma =0` `B : sin alpha + siin beta + sin gamma = 0` If `cos(beta - gamma) + cos (gamma -alpha) + cos (alpha -beta) = - (3)/(2)`, thenA. `A` is true and `B` is false.B. `A` is false and `B` is true.C. Both `A and B` are true.D. Both `A and B` are false. |
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Answer» Correct Answer - C ` cos (beta -gamma) +cos (gamma - alpha ) + cos (alphea - beta) =-(3)/(2)` `rArr 2 [cos (beta -gamma) +cos (gamma -alpha ) + cos (alpha - beta)] + 3 =0` `rArr 2 [cos (beta -gamma ) + cos (gamma -alpha) + cos (alpha -beta)] + sin ^(2) alpha + cos ^(2) alpha + sin ^(2) beta + cos ^(2) beta + sin ^(2) gamma + cos ^(2) gamma =0` `rArr (sin alpha + sin beta + sin gamma)^(2) + ( cos alpha + cos beta + cos gamma)^(2) =0` `rArr cos alpha + cos beta + cos gamma =0` and `sin alpha + sin beta + sin gamma =0` |
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| 10. |
If `alpha,beta,gamma,delta` are the four solutions of the equation `tan(theta+pi/4)=3 tan 3theta.` No two of which have equal tangents, then the value of `tan alpha+tan beta+tan gamma+tan delta=`A. `1//3`B. `8//3`C. `-8//3`D. 0 |
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Answer» Correct Answer - D We have `tan( theta + (pi)/(4))= 3tan 3 theta ` or `(1+ tan theta)/(1-tan theta) = 3 xx (3tan theta - tan^(3) theta)/(1-3tan^(2) theta)` `rArr (1+t)/(1-t) = 3 ((3t-t^(3))/(1-3t^(2)))" "` (putting `t = tan theta`) `or 3t^(4) - 6t^(2) + 8t- 1 =0` Hence, `S_1 `= sum of roots `=t_1 + t_2 + t_3 + t_4 =0` `S_2 ` = sum of product of roots taken two at a time = -2 `S_3` = sum of product of roots taken three at time `= -8//3` `S_4` = product of all roots `=-1//3` `(1)/(t_1) + (1)/(t_2) + (1)/(t_3) + (1)/(t_4) = (sum t_1 t_2 t_3)/(t_2t_2t_3t_4) = (-8//3)/(-1//3) =8` |
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| 11. |
If `alpha, beta,gamma` are the solutions of the equation `tan (theta+(pi)/(4))=3tan 3theta`, no two of which have equal tangents. The value of `(1)/(tan alpha)+(1)/(tan beta)+(1)/(tan gamma)+(1)/(tan delta)` isA. `-8`B. `8`C. `2//3`D. `1//3` |
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Answer» Correct Answer - B We have `tan( theta + (pi)/(4))= 3tan 3 theta ` or `(1+ tan theta)/(1-tan theta) = 3 xx (3tan theta - tan^(3) theta)/(1-3tan^(2) theta)` `rArr (1+t)/(1-t) = 3 ((3t-t^(3))/(1-3t^(2)))" "` (putting `t = tan theta`) `or 3t^(4) - 6t^(2) + 8t- 1 =0` Hence, `S_1 `= sum of roots `=t_1 + t_2 + t_3 + t_4 =0` `S_2 ` = sum of product of roots taken two at a time = -2 `S_3` = sum of product of roots taken three at time `= -8//3` `S_4` = product of all roots `=-1//3` `(1)/(t_1) + (1)/(t_2) + (1)/(t_3) + (1)/(t_4) = (sum t_1 t_2 t_3)/(t_2t_2t_3t_4) = (-8//3)/(-1//3) =8` |
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| 12. |
If `alpha,beta,gamma,delta` are the four solutions of the equation `tan(theta+pi/4)=3 tan 3theta.` No two of which have equal tangents, then the value of `tan alpha+tan beta+tan gamma+tan delta=`A. `-1//3`B. `-2`C. 0D. none of these |
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Answer» Correct Answer - A We have `tan( theta + (pi)/(4))= 3tan 3 theta ` or `(1+ tan theta)/(1-tan theta) = 3 xx (3tan theta - tan^(3) theta)/(1-3tan^(2) theta)` `rArr (1+t)/(1-t) = 3 ((3t-t^(3))/(1-3t^(2)))" "` (putting `t = tan theta`) `or 3t^(4) - 6t^(2) + 8t- 1 =0` Hence, `S_1 `= sum of roots `=t_1 + t_2 + t_3 + t_4 =0` `S_2 ` = sum of product of roots taken two at a time = -2 `S_3` = sum of product of roots taken three at time `= -8//3` `S_4` = product of all roots `=-1//3` `(1)/(t_1) + (1)/(t_2) + (1)/(t_3) + (1)/(t_4) = (sum t_1 t_2 t_3)/(t_2t_2t_3t_4) = (-8//3)/(-1//3) =8` |
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| 13. |
Let ` cos (alpha+beta) = 4/5` and `sin(alpha-beta)=5/13 ` where `0 |
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Answer» Correct Answer - C `cos (alpha + beta )=(4)/(5)` `rArr tan (alpha + beta) = (3)/(4)` `" " sin (alpha - beta) = (5)/(13)` `rArr tan (alpha - beta) = (5)/(12)` ` therefore tan 2 alpha = tan (alpha + beta + alpha - beta)` ` " "= ((3)/(4)= (5)/(12))/(1-(3)/(4)(5)/(12)) = (56)/(33)` |
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| 14. |
If `x , y , z`are in A.P., then `(sinx-sinz)/(cosz-cosx)`is equal to`tany`(b) `coty`(c) `siny`(d) `coty`A. `tan y`B. `cot y`C. `sin y`D. `cos y` |
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Answer» Correct Answer - B `(sinx-sinz)/(cosz-cosx)=(2cos((x+z)/(2))sin((x-z)/(z)))/(2sin((x+z)/(z))sin((x-z)/(z)))` `=cot((x+z)/(2))=cot(y)` |
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| 15. |
If `cosx=(2cosy-1)/(2-cosy),`where `x , y in (0,pi)`then `tanx/2coty/2`is equal to`sqrt(2)`(b) `sqrt(3)`(c) `1/(sqrt(2))`(d) `1/(sqrt(3))`A. `sqrt(2)`B. `sqrt(3)`C. `(1)/sqrt(2)`D. `(1)/sqrt(3)` |
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Answer» Correct Answer - B `cos x=(2cos y-1)/(2-cosy)` `rArr(1-tan^(2)((x)/(2)))/(1+tan^(2)((X)/(2)))=((2(1-tan^(2)y//2))/(1+tan^(2)y//2))/(2-(1-tan^(2)y//2)/(1+tan^(2)y//2))=((1-3tan^(2)((y)/(2)))/(1+3tan^(2)((y)/(2))))` `rArr tan^(2)((x)/(2))=3tan^(2)((y)/(2))` or `tan((x)/(2))cot((y)/(2))=sqrt(3)`. |
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| 16. |
Given that `(1+sqrt(1+x))tany=1+sqrt(1-x)`. Then `sin4y`is equal to`4x`(b) `2x`(c) `x`(d) none of theseA. 4xB. 2xC. xD. none of these. |
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Answer» Correct Answer - C `tan y=(1+sqrt(1-x))/(1+sqrt(1+x))` Let `x=cos theta,` then `sqrt(1-x)=sqrt(2)sin(theta//2),sqrt(1+x)=sqrt(2)cos(theta//2)` `rArr tany=(sqrt(2)[(1)/sqrt(2)+sin((theta)/(2))])/(sqrt(2)[(1)/sqrt(2)+cos((theta)/(2))])``=(sin((pi)/(4))+sin((theta)/(2)))/(cos((pi)/(4))+cos((theta)/(2)))` `=(2sin((pi)/(8)+(theta)/(4))cos((pi)/(8)-(theta)/(4)))/(2cos((pi)/(8)+(theta)/(2))cos((pi)/(8)-(theta)/(4)))` `=tan((pi)/(8)+(theta)/(4))` |
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| 17. |
If `(tan(alpha+beta+gamma))/("tan"(alpha-beta-gamma))=(tangamma)/(tanbeta),(beta!=gamma)`then `sin2alpha+s in2beta+s in2gamma=`0 (b) 1(c) 2 (d) |
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Answer» Correct Answer - A `(tan (alpha+beta-gamma))/(tan(alpha-beta+gamma))=(tan gamma)/(tan beta)` `rArr(sin(alpha+beta-gamma)cos(alpha-beta+gamma))/(sin(alpha-beta+gamma)cos(alpha+beta-gamma))=(sin gamma cos beta)/(sin beta cos gamma)` Applying componendo and dividendo, we get `(sin 2alpha)/(sin 2(beta-gamma))=(sin(gamma+beta))/(sin(gamma-beta))` `rArr sin2(beta-gamma)sin(beta+gamma)+sin 2alpha sin(beta-gamma)=0` `rArr sin(beta-gamma)(sin 2alpha+sin2beta+sin2gamma)=0` `rArr sin2alpha+sin 2beta+sin 2y=0` (as `beta ne gamma`) |
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| 18. |
If `cos (x-y),, cosx and cos(x+y)` are in H.P., are in H.P., then `cosx*sec(y/2)`=A. `-sqrt(3)`B. `-sqrt(2)`C. `sqrt(2)`D. `sqrt(3)` |
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Answer» Correct Answer - B::C We have, `" " (2)/(cosx ) = (1)/(cos(x-y)) + (1)/(cos(x+y)) = ( 2cosx*cosy)/( cos^(2)x - sin^(2)y)` `rArr cos^(2) x - sin^(2)y = cos^(2) x * cos y` `rArr cos^(2)x ( 1-cosy) = sin^(2)y` `rArr cos^(2) x* 2 sin^(2)""(y)/(2) = 4 sin^(2) ""(y)/(2)cos^(2)""(y)/(2)` `rArr cos ^(2)x= 2 cos^(2)""(y)/(2)` `rArr cos^(2)x sec^(2) ""(y)/(2)= 2` `rArr cosx * sec""(y)/(2) = pm sqrt2` |
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| 19. |
If `sin(y+z-x),sin(z+x-y),"sin"(x+y-z)`are in A.P., then `t a n x ,tany ,tanz`are inA.P. (b)G.P. (c) H.P.(d) none of theseA. APB. GPC. HPD. none of these. |
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Answer» Correct Answer - A Applying `b-a=c-b` for AP we get `2cos sin(x-y)=2cos x sin(y-z)` Dividing by `2 cos x cos y cos z`, etc, we get `tan x-tan y=tan y-tan z`. |
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| 20. |
If `(sinx)/(siny)=1/2,(cosx)/(cosy)=3/2,` where `x,y in (0,pi/2),` then the value of `tan(x+y)` is equal toA. `sqrt(13)`B. `sqrt(14)`C. `sqrt(17)`D. `sqrt(15)` |
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Answer» Correct Answer - D `(sinx)/(siny)=(1)/(2),(cos x)/(cosy)=(3)/(2)` `rArr (tan x)/(tany)=(1)/(3)` `rArr tan(x+y)=(tan x+tany)/(1-tan xtany)=(4tan x)/(1-3tan^(2)x)` Also `siny=2sinx, cos y=(2)/(3)cosx` or `sin^(2)y+cos^(2)y=4sin^(2)x+(4cos^(2))/(9)=1` or `36tan^(2)x+4=9sec^(2)x=9(1+tan^(2)x)` or `27tan^(2)x=5` or `tan x=(sqrt(5))/(3sqrt(3))` `rArr tan(x+y)=((4sqrt(5))/(3sqrt(3)))/(1-(15)/(27))=(4sqrt(5)xx27)/(12xx3sqrt(3))=sqrt(15)` |
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| 21. |
The value of `sum_(k=1)^(13) (1)/(sin((pi)/(4) + ((k-1)pi)/(6)) sin ((pi)/(4)+ (kpi)/(6)))` is equal toA. ` 3-sqrt3`B. `2(3-sqrt3)`C. `2(sqrt3-1)`D. `2(2+sqrt3)` |
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Answer» Correct Answer - C `2 overset(13) underset(k=1) (sum) (sin((pi)/(6)))/(sin ((pi/(4) + ((k-1)pi)/(6)) sin ((pi)/(4) +(kpi)/(6))))` `" " = 2 sum (sin{((pi)/(4) + (kpi)/(6))- ((pi)/(4) + ((k-1)pi)/(6))})/(sin ((pi)/(4) + ((k-1)pi)/(6))* sin ((pi)/(4)+ (kpi)/(6)))` `" " = 2 overset(13)underset(k=1) (sum)(cot((pi)/(4)+ ((k-1)pi)/(6)) - cot ((pi)/(4)+ (k pi)/(6)))` `= 2[ cot ((pi)/(4)) - cot ((pi)/(4) + (13pi)/(6))]` `= 2[1-(2-sqrt3)]` `= 2 (sqrt3-1)` |
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| 22. |
If `cos^3xsin2x=sum_(r=0)^n a_xsin(r x),AAx in R` thenA. `n=5,a_(1)=1//2`B. `n=5,alpha_(1)=1//4`C. `n=5,a_(2)=1//8`D. `n=5,a_(2)=1//4` |
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Answer» Correct Answer - B `cos^(3)xsin2x=cos^(2)x cos x sin 2x` `=((1+cos 2x)/(2))(2sin2xcosx)/(2))` `=(1)/(4)(1+cos 2x)(sin3x+sinx)` `=(3)/(4)[sin3x+sinx+(1)/(2)(2sin 3xcos2x)` `=(1)/(2)(2cos 2xsin x]` `=(1)/(4)[sin3x+sinx+(1)/(2)(sin5x+sinx)+(1)/(2)(sin3x-sinx)]` `=(!)/(4)[sinx+((3)/(2))sin3x+((1)/(2))sin5x]` `rARr a_(1)=1//4,a_(3)=3//8,n=5` |
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| 23. |
If `x_(1)` and `x_(2)` are two distind roots of the equation `a cos x+b sinx=c`, then `tan"" (x_(1)+x_(2))/(2)` is equal toA. `(a)/(b)`B. `(b)/(a)`C. `(c)/(a)`D. `(a)/(c)` |
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Answer» Correct Answer - B `a cos x+b sin x=c` `rArr (a(1-tan^(2)(x)/(x)))/(1+tan^(2)(x)/(2))+(2btan((x)/(2)))/(1+tan^(2)((x)/(2)))=c` `rArr(c+a)tan^(2)((x)/(2))-2btan((x)/(2))+c-a=0` `rARr tan((x_(1))/(2))+tan ((x^(2))/(2))=(2b)/(c+a)` and `tan((x_(1))/(2))tan((x_(2))/(2))=(c-a)/(c+a)` `rArr tan((x_(1)+x_(2))/(2))=((2b)/(c+a))/(1-(c-a)/(c+a))=(2b)/(2a)=(b)/(a)` |
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| 24. |
The roots of the equation `4x^2-2sqrt5 x +1=0` are .A. `sin36^(@),sin18^(@)`B. `sin18^(@),cos36^(@)`C. `sin36^(@),cos18^(@)`D. `cos 18^(@),cos36^(@)` |
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Answer» Correct Answer - B `4x^(2)-2sqrt(5)x+1=0` `therefore x=(2sqrt(5)+-2)/(8)=(sqrt(5)+-1)/(4)` `therefore` Roots are `sin 18^(@)` and `cos36^(@)` |
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| 25. |
Difference between maximum and minimum values of `(60sin alpha+p cos alpha)` is 122 then p can beA. 61B. 11C. `-61`D. `-11` |
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Answer» Correct Answer - B::D `E= 60 sin alpha + pcosalpha ` Maximum value of `E = sqrt(3600 + p^(2))` Minimum value of E `=-sqrt(3600 + p^(2))` `sqrt(3600 + p^(2)) + sqrt(3600 + p^(2)) = 122" "` (given) `therefore 2 sqrt ( 3600 + p^(2)) = 122` `rArr p^(2) = 121` `rArr p = pm 11` |
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| 26. |
In ` A B C`Prove that `cos^2A/2+cos^2B/2+cos^2C/2lt=9/4dot`In `cos^2A/2+cos^2B/2+cos^2C/2=y(x^2+1/(x^2))`then find the maximum value of `ydot` |
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Answer» Correct Answer - `9//8` (a) In `triangle ABC`, we know that `cosA+cosB+cos Cle(3)/(2)` `cos^(2)""(A)/(2)+cos^(2)""(B)/(2)+cos^(2)""(C)/(2)` `=(1+cosA)/(2)+(1+cosB)/(2)+(1+cosC)/(2)` `=(3)/(2)+(cos A+cosB+cosC)/(2)le(3)/(2)+(3)/(4)` (using eq. i) `therefore cos^(2)""(A)/(2)+cos^(2)""(B)/(2)+cos^(2)""(C)/(2)le(9)/(4)`. (b) `cos^(2)""(A)/(2)+cos^(2)""(B)/(2)+cos^(2)""(C)/(2)=y(x^(2)+(1)/(x^(2)))` `therefore y(x^(2)+(1)/(x^(2)))le(9)/(4)`. `therefore yle(9)/(4(x^(2)+(1)/(x^(2)))` Now `x^(2)+(1)/(x^(2))ge2` `therefore yle(9)/(8)`. thus, maximum value of `y` is `(9)/(8)`. |
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| 27. |
The value of `sum_(r=0)^10cos^3((rpi)/3)` isA. `1//4`B. `1//8`C. `-1//4`D. `-1//8` |
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Answer» Correct Answer - D `sum_(r=0)^(10)cos^(3)((rpi)/(3))=(1)/(4)sum_(r=0)^(10)(3cos((rpi)/(3))+cos rpi)` `=(1)/(4)[3(cos0+cos((pi)/(3))+.......+cos((10pi)/(3)))+(1-1+.....-1+1])` `=(3)/(4)((cos((10pi)/(6))sin((11pi)/(6)))/(sin((pi)/(6))))+(1)/(4)=(1)/(8)` |
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| 28. |
In a triangle `A B C ,/_C=pi/2dot`If `tan(A/2)a n dtan(B/2)`are the roots of the equation `a x^2+b x+c=0,(a!=0),`then the value of `(a+b)/c`(where `a , b , c ,`are sides of ``opposite to angles `A , B , C ,`respectively) is |
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Answer» Correct Answer - 1 `tan ((A)/(2)) + tan ((B)/(2)) = - (b)/(a)` `tan ((A)/(2)) xx tan ((B)/(2)) = (c ) /(a)` `A + B = 90^(@) or (A+B)/(2) = 45^(@)` `rArr tan ((A+B)/(2)) =1= (-(b)/(a))/(1-(c )/(a))` or `1- (c )/(a) = - (b)/(a)` or ` a +b = c` or `(a+b)/(c ) =1`. |
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| 29. |
The maximum value of `y=1/(sin^6x+cos^6x)`is ______ |
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Answer» Correct Answer - 4 `sin ^(6)x + cos ^(6)x = ( sin ^(2)x + cos ^(2)x ) ( sin ^(4)x + cos ^(4)x - sin ^(2)x cos ^(2)x)` ` " " = 1-3 sin^(2)x cos ^(2) x =1 - ( 3(sin 2 x )^(2))/(4)` `rArr y = (4)/(4-3( sin 2x)^(2))` `rArr y_(max) =m (4)/(4-3(1)) = 4`. |
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| 30. |
Let `0lt=a , b , c ,dlt=pi,`where `ba n dc`are not complementary, such that `2cosa+6cosb+7cosc+9cosd=0a n d2sina-6sinb+7sinc-9a n dd=0,`then the value of `3(cos(a+d))/("cos"(b+c))`is_____ |
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Answer» Correct Answer - 7 From the given equations, we have ` " " (2 cos a + 9 cos d)^(2) = (6 cos b + 7 cos c)^(2) ` `and ( 2 sin a - 9 sin d)^(2) = (6 sin b - 7 sin c )^(2)` Adding, we have `36 cos (a+d) = 84 cos (b+c)` or `(cos(a+d))/(cos (b+c)) = (7)/(3)` |
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| 31. |
If `x , y in R`satisfies `(x+5)^2+(y-12)^2=(14)^2,`then the minimum value of `sqrt(x^2=y^2)`is__________ |
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Answer» Correct Answer - 1 Let `x = 5 = 14 cos theta and y -12 = 14 sin theta` `therefore x ^(2) +y^(2) = (14 cos theta - 5)^(2) + (14 sin theta + 12)^(2)` `" " = 196 + 25 + 144 + 28 ( 12 sin theta - 5 cos theta)` ` " "= 365 + 28(12 sin theta - 5 cos theta)` ` therefore sqrt(x^(2) + y^(2)) :|_(min)= sqrt(365- 28 xx 13)` `" " =sqrt(365 - 364) = 1` |
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| 32. |
If the angles `alpha, beta, gamma` of a triangle satisfy the relation, `sin((alpha-beta)/(2)) + sin ((alpha -gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)`, then Triangle isA. acute angledB. right angled but not isoscelesC. isoscelesD. isosceles right angled |
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Answer» Correct Answer - C We have `sin((alpha - beta)/(2)) + sin ((alpha - gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)` `therefore sin (((pi-(beta+gamma))-beta)/(2))+ sin (((pi-(gamma +beta))-gamma)/(2)) + sin((3pi-3(beta+gamma))/(2)) = (3)/(2)` `therefore cos((2beta + gamma)/(2)) + cos((2gamma + beta)/(2)) - cos((3(beta+gamma))/(2)) = (3)/(2)` `therefore 2cos ((3)/(4)(beta + gamma)) cos ((beta-gamma)/(4)) +1 - 2cos^(2)((3)/(4)(beta +gamma)) = (3)/(2)` `therefore 4 cos^(2) ((3)/(4) (beta + gamma)) - 4cos((3)/(4) (beta + gamma)) cos ((beta-gamma)/(4)) +1 =0`...(1) Above equation is quadratic in `cos((3)/(4)(beta +gamma))` Since `cos((3)/(4)(beta+gamma))` is a real number, Discriminant `D ge 0` `therefore 16 cos^(2)((beta-gamma)/(4)) -16 ge 0` `rArr cos ^(2)((beta -gamma)/(4)) ge 1` `rArr cos^(2)((beta -gamma)/(4)) =1 ` `rArr beta = gamma ` From equation (1) for `beta = gamma`, we get `[ 2 cos((3)/(4)(beta+gamma)) -1]^(2) =0` `rArr 2cos""(3)/(4) (beta + gamma) = 1` `rArr cos""(3)/(4) (beta +gamma) = (1)/(2)` `rArr (3)/(4) (beta +gamma) = 60^(@)` `therefore beta +gamma = 80^(@)` `therefore alpha = 100^(@)` `therefore beta = 40^(@), gamma = 40^(@)` |
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| 33. |
`alphaa n dbeta`are the positive acute angles and satisfying equation `5sin2beta=3s in2alphaa n dtanbeta=3tanalpha`simultaneously. Then the value of `tanalpha+tanbeta`is _________ |
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Answer» Correct Answer - 4 `5 xx (2 tanbeta)/( 1+ tan^(2) beta ) = 3xx ( 2tan alpha)/(1+tan^(2) alpha )` or ` ( 5tan beta)/( 1+ tan^(2) beta ) = (3 tan alpha )/( 1+ tan^(2) alpha)` Substituting `tan beta = 3 tan alpha`, we have `" " ( 5xx 3 tan alpha ) /( 1+9 tan^(2) alpha ) = ( 3tan alpha )/( 1+ tan^(2) alpha)` or `5 +5 tan^(2) alpha = 1+ 9 tan ^(2) alpha ` or `4 tan ^(2) alpha = 4` or `tan alpha = 1`, i.e., `tan beta = 3` `therefore tan alpha + tan beta = 4` |
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| 34. |
`sin alpha+ sinbeta=(1)/(4)` and `cos alpha+cos beta=(1)/(3)` The value of `tan (alpha+beta)` isA. `(25)/(7)`B. `(25)/(12)`C. `(25)/(13)`D. `(24)/(7)` |
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Answer» Correct Answer - D `sin alpha + sin beta = (1)/(4)" "`...(i) `rArr 2 sin((alpha + beta)/(2)) cos ((alpha -beta)/(2))= (1)/(4)" "`...(ii) `rArr cosalpha + cos beta = (1)/(3)" " `...(iii) `rArr 2 cos ((alpha + beta )/(2) ) cos((alpha -beta) /(2)) = (1)/(3)" "` (iv) Dividing Eq. (ii) by Eq. (iv), we have `" " tan((alpha + beta) /(2)) = (3)/(4)` `rArr sin(alpha + beta) = (2tan((alpha + beta)/(2)))/(1+tan^(2)((alpha + beta)/(2)))` `" " = (2xx (3)/(4))/(1+((3)/(4))^(2)) = (24)/(25)` `therfore cos (alpha + beta) = (7)/(25) and tan(alpha + beta) = (24)/(7)` |
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| 35. |
If `f(x)=2(7cos x +24 sin x) (7 sinx -24 cos x)`, for even `x in R` then maximum value of f(x) is __________ |
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Answer» Correct Answer - 625 `f(x) = 2(7 cos x + 24 sin x ) (7 sinx - 24 cos x )` Let `r cos theta = 7 , r sin theta = 24` `therefore r ^(2) = 625 , tan theta = (24)/(7)` `therefore f(x) = 2rcos (x- theta) xx r sin (x- theta)` `= r^(2) ( sin 2 (x - theta))` `f(x)_(max)= 25^(2)` |
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| 36. |
If the angles `alpha, beta, gamma` of a triangle satisfy the relation, `sin((alpha-beta)/(2)) + sin ((alpha -gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)`, then The measure of the smallest angle of the triangle isA. `30^(@)`B. `40^(@)`C. `45^(@)`D. `50^(@)` |
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Answer» Correct Answer - B We have `sin((alpha - beta)/(2)) + sin ((alpha - gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)` `therefore sin (((pi-(beta+gamma))-beta)/(2))+ sin (((pi-(gamma +beta))-gamma)/(2)) + sin((3pi-3(beta+gamma))/(2)) = (3)/(2)` `therefore cos((2beta + gamma)/(2)) + cos((2gamma + beta)/(2)) - cos((3(beta+gamma))/(2)) = (3)/(2)` `therefore 2cos ((3)/(4)(beta + gamma)) cos ((beta-gamma)/(4)) +1 - 2cos^(2)((3)/(4)(beta +gamma)) = (3)/(2)` `therefore 4 cos^(2) ((3)/(4) (beta + gamma)) - 4cos((3)/(4) (beta + gamma)) cos ((beta-gamma)/(4)) +1 =0`...(1) Above equation is quadratic in `cos((3)/(4)(beta +gamma))` Since `cos((3)/(4)(beta+gamma))` is a real number, Discriminant `D ge 0` `therefore 16 cos^(2)((beta-gamma)/(4)) -16 ge 0` `rArr cos ^(2)((beta -gamma)/(4)) ge 1` `rArr cos^(2)((beta -gamma)/(4)) =1 ` `rArr beta = gamma ` From equation (1) for `beta = gamma`, we get `[ 2 cos((3)/(4)(beta+gamma)) -1]^(2) =0` `rArr 2cos""(3)/(4) (beta + gamma) = 1` `rArr cos""(3)/(4) (beta +gamma) = (1)/(2)` `rArr (3)/(4) (beta +gamma) = 60^(@)` `therefore beta +gamma = 80^(@)` `therefore alpha = 100^(@)` `therefore beta = 40^(@), gamma = 40^(@)` |
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| 37. |
In a `Delta ABC`, if `cosA cos B cos C= (sqrt3-1)/(8) and sin A sin B sin C= (3+ sqrt3)/(8)`, then The value of `tan A tan B tanC ` isA. `5-4sqrt3`B. `5+4sqrt3`C. `6+sqrt3`D. `6-sqrt3` |
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Answer» Correct Answer - B `because cos A cos B cos C = (sqrt3 -1)/(8)` `sin A sin B sin C = (3+sqrt3)/(8)` `therefore tanA tan B tan C = (3+ sqrt3)/(sqrt3-1)" " `...(1) `therefore tan A+ tanB + tan C = tan A tan B tan C ` `" " = (3 + sqrt3)/(sqrt3 -1) " "` ...(2) Now `A + B +C= pi` `therefore cos(A+B +C) = -1 ` `therefore cos A cos B cos C [1 -sum tan A tan B] =-1` `therefore sum tanA tan B = 5 + 4sqrt3" "` ...(3) Form (1), (2) and (3), we get `tanA, tan B, tanC` are roots of `x^(3)- ((3+sqrt3)/( sqrt3-1)) x^(2) + (5+4sqrt3)x - ((3+sqrt3))/(sqrt3-1)=0` or `x^(3) - (2+ sqrt3)sqrt3 x^(2) + (5+4sqrt3)x - (2+ sqrt3) sqrt3 =0` or ` x^(3) - (3 + 2 sqrt3)x^(2) + (5+ 4sqrt3)x - (3+ 2sqrt3)=0` or `(x-1)(x-sqrt3)(x-(2+sqrt3)) =0` `therefore tanA =1, tan B = sqrt3, tan C = 2 + sqrt3` |
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| 38. |
To find the sum `sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7)`, we follow the following method. Put `7theta = 2npi`, where `n ` is any integer. Then `" " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta` This means that `sin theta` takes the values `0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7)`. From Eq. (i), we now get `" " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta ` or `4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3)` Rejecting the value `sin theta =0`, we get `" " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3` or ` 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2)` or `16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta)` `" " = 16 sin ^(4) theta - 24 sin ^(2) theta +9` or `" " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0` This is cubic in `sin^(2) theta` with the roots `sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7)`. The sum of these roots is `" " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4)`. The value of `tan^(2)""(pi)/(7)tan ^(2)""(2pi)/(7) tan ^(2)""(3pi)/(7)` isA. `-3`B. `7`C. `-5`D. none of these |
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Answer» Correct Answer - B Let `theta = (npi)/(7)` (so that `7theta = npi`), `n in Z` `rArr 4theta + 3theta = npi` `or 4theta = tan (npi - 3theta ) = -tan 3theta` or ` (4tan theta - 4tan^(3)theta)/(1-6tan^(2) theta + tan^(4)theta) =- (3tan theta - tan^(3) theta)/(1-3tan^(2) theta )` or ` " " ( 4z - 4z^(3))/(1-6z^(2)+ z^(4)) = - (3z-z^(3))/(1-3z^(2))" "` [where `tan theta =z` (say)] or `(4-4z^(2)) (1-3z^(2))= -(3-z^(2)) (1-6z^(2)+z^(4))` or ` z^(6) - 21 z^(4) = 35 z^(2) - 7 =0" "` ... (i) This is a cubic equation in `z^(2)`, i.e., in `tan^(2)theta`. The roots of this equation are therefore `" " tan ^(2)pi//7, tan^(2) 2pi//7, and tan^(2) 3pi//7`. From Eq. (i), `" "` Sum of the roots `= (-(-21))/(1) = 21` `rArr tan^(2)""(pi)/(7) + tan^(2) ""(2pi)/(7) + tan^(2)"" (3pi)/(7) = 21" " `...(ii) Putting `1//y` in place of z in Eq. (i), we get `" " -7y^(6) + 35y^(4)- 21y^(2) +1 =0` `or 7y^(6)- 35 y^(4)+ 21y^(2) -1=0" "`...(iii) This is a cubic equation in `y^(2)`, i.e., in `cot^(2) theta`. The roots of this equation are therefore `cot^(2)pi//7, cot^(2) 2pi//7, and cot^(2)3pi//7`. Sum of the roots of Eq. (iii) = `(35)/(7) =5` `rArr cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) +cot ^(2)""(3pi)/(7)=5" "` ...(iv) By multiplying Eqs. (ii) and (iv), we get `(tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2) ""(3pi)/(7))(cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7))` `= 21xx5 = 105` |
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| 39. |
In a `Delta ABC`, if `cosA cos B cos C= (sqrt3-1)/(8) and sin A sin B sin C= (3+ sqrt3)/(8)`, then The value of `tan A + tan B + tan C` isA. `(3+ sqrt3)/(sqrt3-1)`B. `(sqrt3 +4)/(sqrt3-1)`C. `(6-sqrt3)/(sqrt3-1)`D. `(sqrt3+ sqrt2)/(sqrt3-1)` |
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Answer» Correct Answer - A `because cos A cos B cos C = (sqrt3 -1)/(8)` `sin A sin B sin C = (3+sqrt3)/(8)` `therefore tanA tan B tan C = (3+ sqrt3)/(sqrt3-1)" " `...(1) `therefore tan A+ tanB + tan C = tan A tan B tan C ` `" " = (3 + sqrt3)/(sqrt3 -1) " "` ...(2) Now `A + B +C= pi` `therefore cos(A+B +C) = -1 ` `therefore cos A cos B cos C [1 -sum tan A tan B] =-1` `therefore sum tanA tan B = 5 + 4sqrt3" "` ...(3) Form (1), (2) and (3), we get `tanA, tan B, tanC` are roots of `x^(3)- ((3+sqrt3)/( sqrt3-1)) x^(2) + (5+4sqrt3)x - ((3+sqrt3))/(sqrt3-1)=0` or `x^(3) - (2+ sqrt3)sqrt3 x^(2) + (5+4sqrt3)x - (2+ sqrt3) sqrt3 =0` or ` x^(3) - (3 + 2 sqrt3)x^(2) + (5+ 4sqrt3)x - (3+ 2sqrt3)=0` or `(x-1)(x-sqrt3)(x-(2+sqrt3)) =0` `therefore tanA =1, tan B = sqrt3, tan C = 2 + sqrt3` |
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| 40. |
In a `Delta ABC`, if `cosA cos B cos C= (sqrt3-1)/(8) and sin A sin B sin C= (3+ sqrt3)/(8)`, then the respective values of `tan A, tan B and tanC` areA. `1, sqrt3, sqrt2`B. `1, sqrt3, 2`C. `1, 2, sqrt3`D. `1, sqrt3, 2+sqrt3` |
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Answer» Correct Answer - D `because cos A cos B cos C = (sqrt3 -1)/(8)` `sin A sin B sin C = (3+sqrt3)/(8)` `therefore tanA tan B tan C = (3+ sqrt3)/(sqrt3-1)" " `...(1) `therefore tan A+ tanB + tan C = tan A tan B tan C ` `" " = (3 + sqrt3)/(sqrt3 -1) " "` ...(2) Now `A + B +C= pi` `therefore cos(A+B +C) = -1 ` `therefore cos A cos B cos C [1 -sum tan A tan B] =-1` `therefore sum tanA tan B = 5 + 4sqrt3" "` ...(3) Form (1), (2) and (3), we get `tanA, tan B, tanC` are roots of `x^(3)- ((3+sqrt3)/( sqrt3-1)) x^(2) + (5+4sqrt3)x - ((3+sqrt3))/(sqrt3-1)=0` or `x^(3) - (2+ sqrt3)sqrt3 x^(2) + (5+4sqrt3)x - (2+ sqrt3) sqrt3 =0` or ` x^(3) - (3 + 2 sqrt3)x^(2) + (5+ 4sqrt3)x - (3+ 2sqrt3)=0` or `(x-1)(x-sqrt3)(x-(2+sqrt3)) =0` `therefore tanA =1, tan B = sqrt3, tan C = 2 + sqrt3` |
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| 41. |
If `sin^(-1)a+sin^(-1)b+sin^(-1)c=pi,`then `asqrt(1-a^2)+bsqrt(1-b^2)+csqrt(1-c^2)`is equal to`a+b+c`(b) `a^2b^2c^2``2a b c`(d) `4a b c`A. a+b+cB. `a^(2)b^(2)c^(2)`C. 2abcD. 4abc |
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Answer» Correct Answer - C Let `A=sin^(-1)a,B=sin^(-1)b` and `C=sin^(-1)C` we have `A+B+C=pi`. `asqrt(1-a^(2))+bsqrt(1-b^(2))+csqrt(1-c^(2))` `=(1)/(2)(sin 2A+sin2B+sin2C)` `=(1)/(2)(4sin A sin B sin C)=2abc`. |
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| 42. |
The value of `cos""(pi)/(11)+cos""(3pi)/(11)+cos""(5pi)/(11)+cos""(7pi)/(11)+cos""(9pi)/(11),` is |
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Answer» Correct Answer - `1//2` We have `cos""(pi)/(11)+cos""(3pi)/(11)+cos""(5pi)/(11)+cos""(7pi)/(11)+cos""(9pi)/(11)` `=(cos((5pi)/(11)).sin((pi)/(11)+(9pi)/(1))/(2))/(sin((pi)/(11)))` `=(cos""(5pi)/(11)sin""(5pi)/(11))/(sin""(pi)/(11))=(1)/(2)((10pi)/(11))/(sin((pi)/(1)))=(1)/(2)` |
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| 43. |
In any triangle `A B C ,sin^2A-sin^2B+sin^2C`is always equal to`2sinAsinBcosC`(b) `2sinAcosBsinC``2sinAcosBcosC`(d) `2sinAsinBsinC`A. `2sin A sin B cosC`B. `2 sin A cos B sinC`C. `2 sin A cos B cos C`D. `2 sin A sin B sinC` |
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Answer» Correct Answer - B `sin^(2)A-sin^(2)sin^(2)C` `=sin(A+B)sin(A-B)+sin^(2)C` `=sinC(sin(A-B)+sinC)` `=sinC(sin(A-B)+sin(A+B))` `=2sin A cos B sin C`. |
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| 44. |
If `A+B+C=3pi/2`. Then `cos 2A +cos 2B+cos2C` is equal toA. `1-4cos A cos B cos C`B. `4 sin A sin B sinC`C. `1+2cos A cos B cos C`D. `1-4sin A sin B sinC` |
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Answer» Correct Answer - D `cos 2A+cos2B+cos 2C` `=2cos(A+B)cos(A-B)+cos2C` `=2cos(A+B)cos(A-B)+cos2C` `=2cos((3pi)/(2)-C)cos(A-B)+cos2C` `=2cos((3pi)/(2)-C)cos(A-B)+cos2C` `=-2sin C cos (A-B)+1-2sin^(2)C` `=1-2sinC cos(A-B)+sinC)` `=1-2sinC(cos(A-B)+sin[3pi//2-(A+B)]]` `=1-2sinC[cos(A-B)-cos(A+B)]` `=1-4sinA sin B sin C`. |
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| 45. |
If `cosx-sinalphacotbetasinx=cosa ,`then the value of `tan(x/2)`is`-tan(alpha/2)cot(beta/2)`(b) `tan(alpha/2)tan(beta/2)``-cot((alphabeta)/2)tan(beta/2)`(d) `cot(alpha/2)cot(beta/2)`A. `-tan (alpha2)cot(beta2)`B. `tan (alpha//2)tan(beta//2)`C. `-cot(alpha//2)tan(beta//2)`D. `cot(alpha//2)cot(beta//2)` |
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Answer» Correct Answer - A::B `cos x - sin alpha cos beta sinx = cos alpha ` `rArr (1-tan^(2)(x//2))/(1+tan^(2)(x//2)) - sin alpha cos beta (2tan(x//2))/(1+tan^(2)(x//2)) = cos alpha` `rArr tan^(2) ""(x)/(2) (1+ cosalpha ) + 2 sin alpha cos beta tan ""(x)/(2) - (1- cos alpha) =0` `rArr tan ^(2)""(x)/(2) + (2sinalpha cos beta)/(1+ cos alpha) tan""(x)/(2) - (1-cos alpha)/( 1+cos alpha) =0` `rArr tan^(2)""(x)/(2) + 2 tan""(alpha)/(2) cos beta tan ""(x)/(2) - tan^(2) ""(alpha)/(2)=0` `rArr tan^(2""(x)/(2) + 2 tan""(alpha)/(2)*(1)/(2)(cot""( beta)/(2) - tan ""(beta)/(2)) tan ""(x)/(2) - tan^(2) (alpha)/(2) =0` `rArr (tan ""(x)/(2) + cot""(beta)/(2) tan""(alpha)/(2)) (tan""(x)/(2) - tan""(beta)/(2) tan""(alpha)/(2)) =0` `rArr tan((x)/(2)) = - tan((alpha)/(2)) cot((beta)/(2))` or `" " tan ((x)/(2)) = tan((alpha)/(2) tan((beta)/(2))` |
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| 46. |
If sin A + sin 2A = x and cos A + cos 2A = y, then `(x^(2)+y^(2)) (x^(2)+y^(2)-1)=`A. 2yB. yC. 3yD. none of these |
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Answer» Correct Answer - A Squaring and adding, we get `x^(2)+y^(2)=1+1+2 cos (2A-A)` `therefore (x^(2)+y^(2)-2)/(2)=cos A` ….(1) Also `cos A + 2 cos^(2)A-1=y` or `(cos A+1)(2cos A-1)=y` Put value of cos A from (1) and and get the answer. |
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| 47. |
If the value of `cot(11 1/4^@)+tan(112 1/2^@)-cot(112 1/2^@)-tan(11 1/4^@)=sqrtn` where `n in N` then find the value of `n` |
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Answer» Correct Answer - `2sqrt(2)` `cot(11(1^(@))/(4))+tan(112(1^(@))/(2))-cot(112(1^(@))/(2))-tan(11(1^(@))/(4))` `=(cot11(1^(@))/(4)-tan11(1^(@))/(4))+tan(90^(@)+22(1^(@))/(2))-cot(90^(@)+22(1^(@))/(2))` `=(2(1-tan^(2)11(1^(@))/(4)))/(2tan11(1^(@))/(4))-cot22(1^(@))/(2)+tan22(1^(@))/(2)` `=2cot22(1^(@))/(2)-cot22(1^(@))/(2)+tan22(1^(@))/(2)` `cot 22(1^(@))/(2)+tan22(1^(@))/(2)` `=(sqrt(2)+1)+(sqrt(2)-1)` `=2sqrt(2)`. |
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| 48. |
`sin(9pi)/14sin(11pi)/14sin(13pi)/14` is equal to |
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Answer» Correct Answer - `1//8` `sin""(9pi)/(14)sin""(11pi)/(14)sin""(13pi)/(14)` `=sin""(5pi)/(14)sin""(3pi)/(14)sin""(pi)/(14)` `=cos""(pi)/(8)cos""(2pi)/(7)cos""(3pi)/(7)` `=-cos""(pi)/(7)cos""(2pi)/(7)cos ""(4pi)/(7)` `-(sin""(8pi)/(7))/(8sin""(pi)/(7))=(1)/(8)` |
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| 49. |
If `x =cos alpha+cos beta-cos(alpha+beta)` and `y=4 sin.(alpha)/(2)sin.(beta)/(2)cos.((alpha+beta)/(2))`, then (x-y) equals |
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Answer» Correct Answer - A `y=4 sin.(alpha)/(2)sin.(beta)/(2)cos((alpha+beta)/(2))` `=2[cos.(alpha-beta)/(2)-cos.(alpha+beta)/(2)]cos((alpha+beta)/(2))` `= cos alpha + cos beta - (1+cos (alpha + beta))=x-1` `rArr x-y=1` `cos 5^(@)cos 20^(@)+cos 35^(@)cos 50^(@)` |
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| 50. |
Find the value of `(tan 9^(@)+cot9^(@))/(tan27^(@)+cot27^(@))`. |
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Answer» Correct Answer - `(3+sqrt(5))/(2)` `(tan9^(@)+cot9^(@))/(tan27^(@)+cot27^(@))=(tan9^(@)+(1)/(tan9^(@)))/(tan27^(@)+(1)/(tan27^(@)))` `=(Tan^(2)9^(@)+1)/(2 tan 9^(@))(2tan 27^(@))/((tan^(2)27^(@)+1)` `=(sin 54^(@))/(sin18^(@))=(cos 36^(@))/(sin18^(@))` `=(sqrt(5)+1)/(4).(4)/(sqrt(5)-1)` `=(sqrt(5)+1)/(sqrt(5)-1)xx(sqrt(5)+1)/(sqrt(5)+1)=(3+sqrt(5))/(2)`. |
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