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451.

If `(sin^(2)x-2cos^(2)x+1)/(sin^(2)x+2cos^(2)x-1)=4`, then the value of `2 tan^(2)x` isA. 3B. 4C. 5D. 6

Answer» Correct Answer - C
`(sin^(2)x-2soc^(2)x + 1)/(sin^(2)x+2cos^(2)x-1)=4`
`rArr sin^(2)x-2 cos^(2)x+1`
`=4 sin^(2)x +8 cos^(2)x-4`
`rArr 10 cos^(2)x+3 sin^(2)x-5=0`
`rArr 10+3tan^(2)x-5(1+tan^(2)x)=0`
`rArr 2 tan^(2)x=5`
`therefore tan^(2)x=(5)/(2)`
Discussion
452.

In `DeltaABC`, if `sin A + sin B + sin C= 1 + sqrt2 and cos A+cos B+cosC =sqrt2` then the triangle isA. equilateralB. isoscelesC. right angledD. right angle isosceles

Answer» Correct Answer - D
If the triangle is equilateral
`sin A+ sin B+sin C=(3sqrt(3))/(2)`
If the triangle isosceles, let
`A=30^(@),B=30^(@), C=120^(@)`. Then,
`sin A+ sin B+ sin C=1+(sqrt(3))/(2)`
If the triangle is right angled, let
`A=90^(@), B=30^(@),C=60^(@)`. Then
`sin A+ sin B + sin C=(3+sqrt(3))/(2)`
If the triangle is right angled isoceles, then one of the angles is `90^(@)` and the remaining two are `45^(@)` each, so that
`sin A+ sin B + sin C = 1+sqrt(2)`
and `cos A + cos B + cos C = sqrt(2)`
Discussion
453.

The value of `(tan^(2)20^(@)-sin^(2)20^(@))/(tan^(2)20^(@).sin^(2)20^(@))` isA. `1//2`B. 1C. 2D. none of these

Answer» Correct Answer - B
`tan^(2)-20^(@)-sin^(2)20^(@)=tan^(2)20^(@)(1-cos^(2)20^(@))=tan^(2)20^(@)sin^(2)20^(@)`.
Discussion
454.

In ` triangle ABC` prove that `(b^2-c^2)/(cos B+cos C) +(c^2-a^2)/(cos C+cos A) +(a^2-b^2)/(cos A + cos B)=0`

Answer» `a/sinA=b/SinB=c/sinC=k`
`a=KsinA,b=KSinB,c=KsinC`
`(b^2-c^2)/(cosB+cosC)=(k^2sin^2B-k^2sin^2c)/(cosB+cosC)=(k^2(1-cos^2B-(1-cos^2C)))/(cosC+cosB)`
`k^2(cos^2C-cos^2B)/(cosB+cosC)=k^2(cosC-CosB)`
LHS=`k^2(cosC-cosB)+k^2(cosA-cosC)+k^2(cosb-cosC)`
`K^2[0]=0=RHS`.`
Discussion
455.

If `y=(1+t a n A)(1-t a n B),`where `A-B=pi/4,`then `(y+1)^(y-1)`is equal to9 (b) 4(c) 27 (d) 81A. 9B. 4C. 27D. 81

Answer» Correct Answer - C
Discussion
456.

The values of x,`0

Answer» `81^(sin^2x) + 81^(1-sin^2x) = 30`
`81^(sin^2x) + 81 xx 81^(-sin^x)= 30`
`81^(sin^2x) + 81/(81^(sin^2x))= 30`
let `t= 81^(sin^2x)`
now, `t + 81/t = 30`
`t^2 -30t +81 = 0`
`t^2 - 27t - 3t + 81=0`
`(t-27)(t-3)= 0`
`t=27,3`
`81^(sin^2x) = 27`
`3^(4sin^2x) = 3^3`
so`4sin^2x = 3`
`sin^2x = 3/4`
`sinx = +- sqrt3/2`
`x= pi/3 or -pi/3 = n pi/3`
when `t=3`
`81^(sin^2x) = 3`
`3^(4sin^2x)= 3`
`4sin^2x = 1`
`sinx = +-1/2`
`x= pi/6 or -pi/6`
option A is correct
answer
Discussion
457.

In a triangle ABC `sin (A/2) sin (B/2) sin (C/2)

Answer» We know, arithmatic mean is always greater than or equal to geometric mean.
`:. (sin^2(A/2)sin^2(B/2)sin^2(C/2))^(1/3) le 1/3(sin^2(A/2)+sin^2(B/2)+sin^2(C/2))->(1)`
Also, we know, `sin^2(A/2)+sin^2(B/2)+sin^2(C/2) ge 3/4`
`:. (sin^2(A/2)sin^2(B/2)sin^2(C/2))^(1/3) le 1/3(3/4)`
`=>(sin^2(A/2)sin^2(B/2)sin^2(C/2))^(1/3) le 1/4`
`=>(sin^2(A/2)sin^2(B/2)sin^2(C/2)) le (1/4)^3`
`=>sin(A/2)sin(B/2)sin(C/2) le (1/64)^(1/2)`
`=>sin(A/2)sin(B/2)sin(C/2) le 1/8`
Discussion
458.

The value of `(1-tan^2 1 5^(@))/(1+tan^2 1 5^(@))` isA. 1B. `sqrt(3)`C. `(sqrt(3))/(2)`D. 2

Answer» Correct Answer - C
Given expression, `(1-tan^(2)15^(@))/(1+tan^(2)15^(@))`
Let `" "theta=15^(@)`
We know that, `" " cos2theta=(1-tan^(2)theta)/(1+tan^(2)theta)`
`" " cos2theta=(1-tan^(2)theta)/(1+tan^(2)theta)`
`rArr" "(1-tan^(2)15^(@))/(1+tan^(2)15^(@))=(sqrt(3))/(2)` `" "[becausecos30^(@)=(sqrt(3))/(2)]`
Discussion
459.

Which of the following is not correct ?A. `sintheta=-(1)/(5)`B. `costheta=1`C. `sectheta-(1)/(2)`D. `tantheta=20`

Answer» Correct Answer - C
We know that, the range of `sectheta` is `R-(-1, 1)`.
Hence, `sectheta` cannot be equal to `(1)/(2)`.
Discussion
460.

The value of `sin(45^(@)+theta)-cos(45^(@)-theta)` isA. `2costheta`B. `2sintheta`C. 1D. 0

Answer» Correct Answer - D
Given expression,
`sin(45^(@)+theta)-cos(45^(@)-theta)`
`" "=sin45^(@)*costheta+cos45^(@)*sintheta-cos45^(@)*costheta-sin45^(@)*sintheta`
`" "=(1)/(sqrt(2))*costheta+(1)/(sqrt(2))*sintheta-(1)/(sqrt(2))*costheta-(1)/(sqrt(2))sintheta`
= 0
Discussion
461.

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 in(a)A.P. (b)G.P. (c) H.P.(d) none of these

Answer» As `sin(y+z-x),sin(z+x-y) and sin(x+y-z)` are in A.P.
`:. sin(z+x-y)-sin(y+z-x) = sin(x+y-z)-sin(z+x-y)`
`=>2coszsin(x-y) = 2cosxsin(y-z)`
`=>coszsin(x-y) = cosxsin(y-z)`
`=>(coszsin(x-y))/(cosxcosycosz) = (cosxsin(y-z))/(cosxcosycosz)`
`=>(sin(x-y))/(cosxcosy) = (sin(y-z))/(cosycosz)`
Using `(sin(A-B))/(cosAcosB) = tanA-tanB`
`=>tanx-tany = tany - taz`
`=>tany-tanx = tanz-tany`
`:. tanx,tany and tanz` are in A.P.
Discussion
462.

If `(3pi)/4ltalphaltpi," then "sqrt(2cotalpha+1/sin^2alpha)`is equal toA. `1+cotalpha`B. `-1-cotalpha`C. `1-cotalpha`D. `-1+cotalpha`

Answer» Correct Answer - B
`sqrt(2cotalpha+1/sin^2alpha)=sqrt(2cot alpha+cosec^2alpha)`
`=sqrt(2cotalpha+1+cot^2alpha)`
`=abs(1+cotalpha)=-1-cotalpha`
[Since `aotalphalt-1" when "3pi//4ltalphaltpi`]
Discussion
463.

The value of `f(alpha)=sqrt(cos e c^2alpha-2cotalpha)+sqrt(cos e c^2alpha+2cotalpha)`can be`2cotalpha`(b) `-2cotalpha`(c) 2(d) `-2`A. `2cotalpha`B. `-2cotalpha`C. 2D. `-2`

Answer» Correct Answer - A::B::C
`f(alph)=sqrt(cosec^2alpha-2cotalpha)+sqrt(cosec^2alpha+2cotalpha)`
`=sqrt(1+cot^2alpha-2cotalpha)+sqrt(cosec^2lpha+2cotalpha)`
`=abs(cotalpha-1)=abs(cotalpha+1)`
Case I : `cotalphale-1`
`:. f(alpha)=-cotalpha=1-cotalpha-1=-2cotalpha`
Case II : `-1lecotalphale1`
`:. f(alpha)=-cotalpha+1+cotalpha+1=2`
Case III : `cotalphage 1`
`:. f(alpha)=(cotalpha-1)+(cotalpha+1)=2cotalpha`
Discussion
464.

Suppose the point with coordinates `(-12 ,5)`is on the terminal side of angle `theta`. Find the values of the six trigonometric functions of `thetadot`

Answer» We can draw the point `(-12,5)` at the XY-axis.
Please refer to see the graph.
It will become a right angle triangle with base = 12 and perpendicular = 5.
`:.` Hypotenuse `= 12^2+5^2 = 13`
As it is in second quadrant, only `sin theta` and `cosec theta` will be positive.
`:.sin theta = 5/13`
`cos theta = -12/13`
`tan theta = -5/12`
`cot theta = -12/5`
`cosec theta = 13/5`
`sec theta = -13/12`
Discussion
465.

The value of `cot((pi)/(4)+theta)cot((pi)/(4)-theta)` isA. -1B. 0C. 1D. Not defined

Answer» Correct Answer - C
Given expression, `" "cot((pi)/(4)+theta)-cot((pi)/(4)-theta)`
`" "=((cot""(pi)/(4)cottheta-1)/(cot""(pi)/(4)+cottheta))*((cot""(pi)/(4)cottheta+1)/(cottheta-cot""(pi)/(4)))`
` " " =((cottheta-1)/(cottheta+1))*((cottheta+1)/(cottheta-1))`
=1
Discussion
466.

If `(tanx)/2=(tany)/3=(tanz)/5,x+y+z=pi` and `tan^2x+tan^2y+tan^2z=(38)/K` then K=_________`

Answer» Aas `x+y+z = pi`,
`:. tanx+tany+tanz = tanxtanytanz->(1)`
Let `tanx/2 = tany/3 = tanz/5 = k`
`=>tanx = 2k, tany = 3k, tanz = 5k`
Putting these values in (1),
`2k+3k+5k = 2k(3k)(5k)`
`=>10k = 30k^3`
`=>k^2 = 1/3`
`=>k = 1/sqrt3`
Here, we will not take `k = -1/sqrt3` as it will make `x+y+z` negative.
Now, `tanx = 2/sqrt3 ,tany= 3/sqrt3, tanz = 5/sqrt3`
Now, `tan^2x+tan^2y+tan^2z = 38/K`
`=>4/3+9/3+25/3 = 38/K`
`=>38/3 = 38/K`
`=>K = 3`
Discussion
467.

If `(3pi)/4

Answer» `sqrt(2cotalpha+1/sin^2alpha)`
`=sqrt(2cotalpha+cosec^2alpha)`
`=sqrt(2cotalpha+cot^2alpha+1)`
`=sqrt((1+cotalpha)^2)`
`=|1+cotalpha|`
As `(3pi)/4 lt alpha lt pi`, `cotalpha` will be negative.
`:.|1+cotalpha| = -1-cot alpha`
So, optopon -`(b)` is the correct option.
Discussion
468.

Prove that `(cos(pi +theta)cos (-theta))/(cos(pi-theta) cos (pi/2+theta))=-cot theta`

Answer» LHS
`(cos(pi+theta)cos(-theta))/(cos(pi-theta)*cos(pi/2+theta))`
`(costheta*costheta)/(costheta*(-sintheta)`
`-costheta/sintheta=-cottheta=RHS`
Discussion
469.

Assuming the distance of the earth from the moon to be 38,400 km andthe angle subtended by the moon at the eye of a person on the earth to be31’, find the diameter of the moon.

Answer» Diameter of moon can be given by,
`l = r theta`
Here, `r` = distance between moon and earth `=38400 km`
`theta` = Angle in radians `= 31*1/60*pi/180`
`:. l = 31*1/60*pi/180**38400 = 364.32(approx)`
So, the diameter of the moon is approximately `364.32` km.
Discussion
470.

If `alpha=pi/3`, prove that `cosalphacos2alphacos3alphacos4alphacos5alphacos6alpha=-1/16`

Answer» `cosalphaxxcos2alphaxxcos3alphaxxcos4alphaxxcos5alphaxxcos6alpha`
`=cos. pi/3xxcos. (2pi)/3xxcos. pixxcos. (4pi)/3xxcos. (5pi)/3xxcos. (6pi)/3`
`=1/2xx(-1/2)(-1)(-1/2)(1/2)1`
`=-1/16`
Discussion
471.

Find the value of the expression `sec610^@cosec160^@-cot380^@tan470^@`

Answer» Correct Answer - -1
`sec610^@cosec160^@-cot380^@tan470^@`
`=sec(720^@-110^@)cosec(180^@-20^@)`
`-cot(360^@+20^@)tan(360^@+110^@)`
`=1/(cos110^@). 1/(sin20^@)-cot20^@tan110^@`
`= (-1)/sin^2 20^@+cot^2 20^@`
`=-[cosec^2 20^@-cot^2 20^@]`
`=-1`
Discussion
472.

If : `sin(x+(pi)/(3)) = cos (x-(pi)/(6)),"then" : tan x = `A)`1//sqrt3`B)`-1//sqrt3`C)`sqrt3`D)`-sqrt3`A. `1//sqrt3`B. `-1//sqrt3`C. `sqrt3`D. `-sqrt3`

Answer» Correct Answer - A
Discussion
473.

`(sin 2 theta)/(sin theta)-(cos 2 theta)/(cos theta)=`A)`sec theta` B)`tan theta` C)1 D)2A. `sec theta`B. `tan theta`C. 1D. 2

Answer» Correct Answer - A
Discussion
474.

Prove that: `t a n 720^0-cos 270^0-s in 150^0cos 120^0=1/4`

Answer» (a) `tan720^@-cos270^@-sin150^@cos120^@`
`=0-0-(sin30^@)(-cos60^@)=1/2xx1/2=1/4`
(b) `sin780^@sin480^@+cos120^@sin150^@`
`=sin(720^@+60^@)sin(360^@+120^@)`
`+cos(180^@-60^@)sin(180^@-30^@)`
`=sin(60^@)sin(120^@)-cos(60^@)sin(30^@)`
`sqrt3/2xxsqrt3/2-1/2xx1/2=3/4-1/4=1/2`
Discussion
475.

If : `sin (theta - 30^(@))+cos (theta - 60^(@))=k * sin theta, "then" : k=`A. `1//sqrt3`B. `-1//sqrt3`C. `sqrt3`D. `-sqrt3`

Answer» Correct Answer - C
Discussion
476.

`(sin 3 theta)/(sin theta)-(cos 3 theta)/(cos theta)=`A)` sin 2 theta`B)` cos 4 theta`C)1D)2A. ` sin 2 theta`B. `cos 4 theta`C. 1D. 2

Answer» Correct Answer - D
Discussion
477.

`" cos x + cos " (120^(@) -x ) + " cos "(120^(@) +x)=0`A. -1B. 0C. `sqrt3`D. `1//sqrt3`

Answer» Correct Answer - B
Discussion
478.

`(1-tan^(2)(45^(@)-A))/(1+tan^(2)(45^(@)-A))` is equal toA. sin 2 AB. cos 2 AC. tan 2 AD. cot 2 A

Answer» Correct Answer - A
Discussion
479.

`1+tan theta * tan 2 theta=`A)`2 csc theta`B)`sec 2 theta `C)`2 sec theta`D)` csc 2 theta`A. `2 csc theta`B. `sec 2 theta `C. `2 sec theta`D. ` csc 2 theta`

Answer» Correct Answer - B
Discussion
480.

`tan(45^(@)+A)tan(45^(@)-A)=1`A. -1B. 0C. 1D. `sqrt2`

Answer» Correct Answer - C
Discussion
481.

In a right angled triangle the hypotenuse is `2sqrt(2)`times the perpendicular drawn from the opposite vertex. Then the other acuteangles of the triangle are(a)`pi/3a n dpi/6`(b) `pi/8a n d(3pi)/8`(c) `pi/4a n dpi/4`(d) `pi/5a n d(3pi)/(10)`

Answer» We can create a right angle traigle `ABC` with the given details.
Please refer to video to see the diagram.
Let the length of the perpendicular drawn from `B` to `AC` is `p`.
Then, `AC = 2sqrt2p`
`BC = psec theta`
`AB = pcosec theta`
Now, `AB^2+BC^2 = AC^2`
`=> p^2cosec^2theta +p^2sec^2 theta = (2sqrt2p)^2`
`=>cosec^2 theta+sec^2theta = 8`
`=>1/sin^2theta+1/cos^2theta = 8`
`=>cos^2theta+sin^2theta = 8sin^2thetacos^2theta`
`=>1 = 8sin^2thetacos^2theta`
`=>1 = 2(2sinthetacostheta)^2`
`=>(sin2theta)^2 = 1/2`
`=>sin2theta = 1/sqrt2`
`=>2theta = pi/4`
`=>theta = pi/8`
`:. C = pi/8`
`:. A = pi/2-pi/8 = (3pi)/8`
So, the other two acute angles will be `pi/8` and `(3pi)/8.`
Discussion
482.

With usual notations, in triangle `A B C ,acos(B-C)+bcos(C-A)+c"cos"(A-B)`is equal to(a)`(a b c)/R^2`(b)`(a b c)/(4R^2)`(c)`(4a b c)/(R^2)`(d) `(a b c)/(2R^2)`

Answer» `acos(B-C)+bcos(C-A)+c cos(A-B)`
`=2RsinAcos(B-C)+2RsinBcos(C-A)+2RsinC cos(A-B)`
`=2Rsin(pi-(B+C))cos(B-C)+2Rsin(pi-(C+A))cos(C-A)+2Rsin(pi-(A+B)) cos(A-B)`
`=2Rsin(B+C)cos(B-C)+2Rsin(C+A)cos(C-A)+2Rsin(A+B) cos(A-B)`
`=R(sin2B+sin2C+sin2C+sin2A+sin2A+sin2B)`
`=2R(sin2A+sin2B+sin2C)`
`=2R(4sinAsinBsinC)`
`= 8RsinAsinBsinC`
`= 8R(a/(2R))(b/(2R))(c/(2R))`
`=(abc)/R^2`
`:. acos(B-C)+bcos(C-A)+c cos(A-B) = (abc)/R^2`
Discussion
483.

The number of values of `x`in the in interval `[0,5pi]`satisfying the equation `3sin^2x-7sinx+2=0`is0 (b)5 (c) 6(d) 10

Answer» `3sin^2x-7sinx+2=0`
`3sin^2x-6sinx-sinxx+2=0`
`3sinx(sinx-2)-1(sinx-2)=0`
`(sinx-2)(3sinx-1)=0`
`sinx=2`
No solution
`sinx=1/3`
Number of x=6
Option C is correct.
Discussion
484.

The nubmber of solutions of x in the interval `[0,5pi]` satisfying the equation `3sin^(2)x-7sinx+2=0` is

Answer» Correct Answer - C
Given ,`3sin^(2)x-7sinx+2=0`
`rArr3sin^(2)x-6sinx-sinx-+2=0`
`rArr3sinx(sinx-2)-1(sinx-2)=0`
`rArr(3sinx-1)(sinx-2)=0`
`rArrsinx=(1)/(3)or2rArrsinx=(1)/(3) " " [becausesinxne2]`
`rArrx=sin^(-1)((1)/(3))`
Let `"sin"^(-1)(1)/(3)=alpha,0ltalphalt(pi)/(2),alphain[-(pi)/(2),(pi)/(2)]`
Then `alpha,pi-alpha,2pi+alpha,3pi-alpha,4pi+alpha,5pi-alpha` are the solutions in `[0,5pi]`.
`therefore` Required number of solutions are 6.
Discussion
485.

If `u_n=sin^("n")theta+cos^ntheta,`then prove that `(u_5-u_7)/(u_3-u_5)=(u_3)/(u_1)`.

Answer» `(u_5-u_(7))/(u_3-u_(5))=((sin^5theta +cos^5theta)-(sin^7theta+cos^7theta))/((sin^3theta +cos^3theta)-(sin^5theta+cos^5theta))`
`(sin^5theta(1-sin^2theta)+cos^5theta(1-cos^2theta))/(sin^3theta(1-sin^2theta)+cos^3theta(1-cos^2theta))`
`=(sin^2thetacos^2theta[sin^3theta+cos^3theta])/(sin^2thetacos^2theta[sintheta+costheta])=u_3/u_1`
Discussion
486.

In `triangleABC, (c+a-b)tan((B)/(2))=`A. `(A(triangleABC))/(s)`B. `(2A(triangleABC))/(s)`C. `A(triangleABC)`D. `2A(triangleABC)`

Answer» Correct Answer - B
Discussion
487.

In `triangleABC, (a+b-c)tan((C)/(2))=`A. `(A(triangleABC))/(s)`B. `(2A(triangleABC))/(s)`C. `A(triangleABC)`D. `2A(triangleABC)`

Answer» Correct Answer - B
Discussion
488.

In `triangleABC, s(s-c)tan((C)/(2))=`A. `(A(triangleABC))/(2)`B. `(2A(triangleABC))/(4)`C. `A(triangleABC)`D. `2A(triangleABC)`

Answer» Correct Answer - C
Discussion
489.

In `triangleABC, s(s-a)tan((A)/(2))=`A. `(A(triangleABC))/(2)`B. `(2A(triangleABC))/(4)`C. `A(triangleABC)`D. `2A(triangleABC)`

Answer» Correct Answer - C
Discussion
490.

In `triangleABC, s(s-b)tan((B)/(2))=`A. `(A(triangleABC))/(2)`B. `(2A(triangleABC))/(4)`C. `A(triangleABC)`D. `2A(triangleABC)`

Answer» Correct Answer - C
Discussion
491.

In `DeltaABC`, if 2s=a+b+c , then the value of `(s(s-a))/(bc)-((s-b)(s-c))/(bc)` isA. `sinA`B. `cosA`C. `tanA`D. `cotA`

Answer» Correct Answer - B
Discussion
492.

In `triangleABC, tan((A)/(2))tan((B)/(2))=`A. `(c-(a+b))/(2(a+b+c))`B. `(c-(a+b))/(a+b+c)`C. `(a+b-c)/(2(a+b+c))`D. `(a+b-c)/(a+b+c)`

Answer» Correct Answer - D
Discussion
493.

In `triangleABC, (cos((B-C)/(2)))/(sin(A/2))=`A. `(b-c)/(a)`B. `(b+c)/(a)`C. `(a)/(b-c)`D. `(a)/(b+c)`

Answer» Correct Answer - B
Discussion
494.

`"In a "DeltaABC, if(cosA)/a=(cosB)/b,"show that the triangle is isosceles".`A. an equilateralB. an isoscelesC. a right angledD. a scalene

Answer» Correct Answer - B
Discussion
495.

In `triangleABC`, if `asinA=bsinB`, thenA. `a+b=c`B. `agtb`C. `altb`D. `a=b`

Answer» Correct Answer - D
Discussion
496.

In `DeltaABC`, with usual notations, if `cosA=(sinB)/(sinC)`, then the triangle isA. a scaleneB. a right angledC. an isoscelesD. an equilateral

Answer» Correct Answer - b
Discussion
497.

`tan 20^(@) + tan 25^(@) +tan 20^@ tan25^(@)=`A. `sin 45^(@)`B. `cos 45 ^(@)`C. `cot 45^(@)`D. `csc 45^(@)`

Answer» Correct Answer - C
Discussion
498.

Suppose ABCS (in order) is a quadrilateral inscribed in a circle. Whichof the following is/are always true?`secB=secD`(b) `cotA+cotC=0``cos e cA=cos e cC`(d) `tanB+tanD=0`A. `secB=secB`B. `cotA+cotC=0`C. `cosecA=cosecC`D. `tanB+tanD=0`

Answer» Correct Answer - B::C::D
Opposite angles of a cyclic quadrilateral are supplementary.
Discussion
499.

Which of the following is/are correct ?A. `(tanx)^("In"(sinx))gt(cotx)^("In"(sinx)), AA x in(0,pi//4)`B. `4^("In"cosecx)lt5^("In"cosecx), AA x in(0,pi//2)`C. `(1//2)^("In"(cosx))lt(1//3)^("In"(cosx)), AA x in(0,pi//2)`D. `2^("In"(tanx))lt2^("In"(tanx)), AA x in(0,pi//2)`

Answer» Correct Answer - A::B::C::D
(1) For ` x in (0,pi/4), tanxltcotx`
Also `In(sinx)lt0`
`rArr (tanx)^(in(sinx))gt(cotx)^(in(sinx))`
(2) For `x in (0,pi/2),cosecxge1`
`rArr In(cosecx)ge0`
`rArr 4^(In(cosecx))lt5^(in(cosecx))`
`(3) x in (0,pi/2)`
`rArr cosx in (0,1)`
`rArr In(cosx)lt0`
Also `1/2gt1/3`
`rArr (1/2)^(In(cosx))lt(1/3)^(In(cosx))`
(4) For `x in (0,pi/2)`
Since `sinxlttanx`,we get `In(sinx)ltIn(tanx)`
`rArr2^(In(sinx))lt2^(In(tanx))`
Discussion
500.

`(a+2)sinalpha(2a-1)cosalpha=(2a+1)if tanalpha` isA. `3//4`B. `4//3`C. `2a//(a^2+1)`D. `2a//(a^2-1)`

Answer» Correct Answer - B::D
Divide by `cosalpha` and square both sides abd let `tanalpha=t` so that ` sec^2alpha=1+t^2`
`rArr [(a+2)t+(2a-1)]^2=[(2a+1)^2(1+t^2)]`
`rArr t^3[(a+2)^2-(2a+1)^2]+2(a+2)(2a-1)t+[(2a-1)^2-(2a+1)^2]=0`
`or 3(1-a^2)t^2+2(2a^2+3a-2)t-4xx2a=0`
`or 3(1-a^2)t^2+2(2a^2+3a-2)t-4xx2a=0`
`or t(1-a^2)t^2-4(1-a^2)t-6at-8a=0`
`or (3t-4)[(1-a^2)t+2a]=0`
`or t-tanalpha=4/3or(2a)/(a^2-1)`
Discussion