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301.

The principal solution `tanx=sqrt(3)` isA. `(pi)/(3)`B. `(2pi)/(3)`C. `(-pi)/(3)`D. `(-4pi)/(3)`

Answer» Correct Answer - A
Discussion
302.

The principal solution `cosx=(1)/(2)` isA. `(2pi)/(3)`B. `-(2pi)/(3)`C. `(pi)/(3)`D. `(7pi)/(4)`

Answer» Correct Answer - C
Discussion
303.

The principal solution `tanx=1` isA. `(7pi)/(4)`B. `(3pi)/(4)`C. `(-5pi)/(4)`D. `(5pi)/(4)`

Answer» Correct Answer - D
Discussion
304.

The principal solution `tanx=-1` isA. `npi+(3pi)/(4), ninZ`B. `npi-(3pi)/(4), ninZ`C. `npi+(pi)/(4), ninZ`D. `npi-(pi)/(4), ninZ`

Answer» Correct Answer - A
Discussion
305.

The principal solution `tanx=-1` isA. `-(3pi)/(4)`B. `(5pi)/(4)`C. `(pi)/(4)`D. `(7pi)/(4)`

Answer» Correct Answer - D
Discussion
306.

The principal solution `tanx=-1` isA. `(5pi)/(4)`B. `(pi)/(4)`C. `(3pi)/(4)`D. `-(3pi)/(4)`

Answer» Correct Answer - C
Discussion
307.

The principal solution `tanx=1` isA. `(3pi)/(4)`B. `(pi)/(4)`C. `(7pi)/(4)`D. `(-pi)/(4)`

Answer» Correct Answer - B
Discussion
308.

The principal solution `tanx=(1)/(sqrt(3))` isA. `(7pi)/(6)`B. `(5pi)/(6)`C. `(11pi)/(6)`D. `(pi)/(6)`

Answer» Correct Answer - A
Discussion
309.

The principal solution `tanx=(1)/(sqrt(3))` isA. `(5pi)/(6)`B. `(11pi)/(6)`C. `(-pi)/(6)`D. `(pi)/(6)`

Answer» Correct Answer - D
Discussion
310.

निम्नलिखित के मान ज्ञात कीजिए : `cos^(-1)(cos""(3pi)/(6))`A. `(-pi)/(6)`B. `(11pi)/(6)`C. `(pi)/(6)`D. `(13pi)/(6)`

Answer» Correct Answer - C
Discussion
311.

निम्नलिखित के मान ज्ञात कीजिए : `tan^(-1)(tan""(3pi)/(6))`A. `(7pi)/(6)`B. `(pi)/(6)`C. `(-pi)/(6)`D. `(-5pi)/(6)`

Answer» Correct Answer - B
Discussion
312.

`cos^(-1)(cos""(7pi)/(6))` का मान बराबर हैA. `(7pi)/(6)`B. `(5pi)/(6)`C. `(pi)/(6)`D. `(11pi)/(6)`

Answer» Correct Answer - B
Discussion
313.

The principal solution `cosx=-(1)/(sqrt(2))` isA. `(pi)/(4)`B. `(7pi)/(6)`C. `(5pi)/(4)`D. `-(pi)/(4)`

Answer» Correct Answer - C
Discussion
314.

The principal solution `cosx=-(1)/(2)` isA. `(2pi)/(3)`B. `(pi)/(3)`C. `(5pi)/(4)`D. `-(pi)/(3)`

Answer» Correct Answer - A
Discussion
315.

The principal solution `cosx=-(1)/(2)` isA. `(5pi)/(4)`B. `(pi)/(3)`C. `(4pi)/(3)`D. `-(pi)/(3)`

Answer» Correct Answer - C
Discussion
316.

The principal solution `cosx=(1)/(sqrt(2))` isA. `(3pi)/(4)`B. `-(3pi)/(4)`C. `(5pi)/(4)`D. `(7pi)/(4)`

Answer» Correct Answer - D
Discussion
317.

The principal solution of `cos^(-1)(cos((9pi)/4))` isA. `(7pi)/(4)`B. `(-pi)/(4)`C. `(9pi)/(4)`D. `(pi)/(4)`

Answer» Correct Answer - D
Discussion
318.

The number of principal solution of `tan2theta=1` isA. oneB. twoC. threeD. four

Answer» Correct Answer - D
Discussion
319.

The principal solution `cosx=-(1)/(sqrt(2))` isA. `(pi)/(4)`B. `(3pi)/(4)`C. `(7pi)/(4)`D. `-(pi)/(4)`

Answer» Correct Answer - B
Discussion
320.

The principal solution `cosx=(1)/(2)` isA. `(7pi)/(3)`B. `-(2pi)/(3)`C. `(2pi)/(3)`D. `(5pi)/(3)`

Answer» Correct Answer - D
Discussion
321.

The principal solution `cosx=(1)/(sqrt(2))` isA. `(pi)/(4)`B. `(3pi)/(4)`C. `(5pi)/(4)`D. `-(3pi)/(4)`

Answer» Correct Answer - A
Discussion
322.

The principal value of `sin^(-1)(1/2)` isA. `(pi)/(6)`B. `(-pi)/(6)`C. `(pi)/(3)`D. `(2pi)/(3)`

Answer» Correct Answer - A
Discussion
323.

The principal solution `cosx=(-sqrt(3))/(2)` isA. `(5pi)/(6)`B. `(pi)/(6)`C. `(pi)/(3)`D. `(-pi)/(6)`

Answer» Correct Answer - A
Discussion
324.

The principal solution `cosx=(sqrt(3))/(2)` isA. `(7pi)/(6)`B. `-(5pi)/(6)`C. `(5pi)/(6)`D. `(11pi)/(6)`

Answer» Correct Answer - D
Discussion
325.

The principal solution `cosx=(-sqrt(3))/(2)` isA. `(pi)/(6)`B. `(7pi)/(6)`C. `(-pi)/(6)`D. `(pi)/(3)`

Answer» Correct Answer - B
Discussion
326.

Find the principal solution of the equation `tanx=-1/(sqrt(3))`.

Answer» `0 <= x <= 2 pi`
`tan x = - tan(pi/6)`
`= tan(pi - pi/6)`
`= tan((5pi)/6)`
also, `tan x = tan(pi + (5pi)/6)`
`tan x = tan((11 pi)/6)`
answer
Discussion
327.

Find the principal value of `sin^(-1)``(1/(sqrt(2)))`A. `(-pi)/(4)`B. `(3pi)/(4)`C. `(pi)/(4)`D. `(-5pi)/(4)`

Answer» Correct Answer - C
Discussion
328.

The principal solution of `sinx=(-1)/(sqrt(2))` isA. `(pi)/(4)`B. `(3pi)/(4)`C. `-(7pi)/(4)`D. `(7pi)/(4)`

Answer» Correct Answer - D
Discussion
329.

The principal solution of `sinx=(-1)/(2)` isA. `(7pi)/(6)`B. `-(7pi)/(6)`C. `(pi)/(6)`D. `(5pi)/(6)`

Answer» Correct Answer - A
Discussion
330.

The principal solution `cosx=(-sqrt(3))/(2)` isA. `(5pi)/(6)`B. `(pi)/(6)`C. `(7pi)/(6)`D. `-(5pi)/(6)`

Answer» Correct Answer - B
Discussion
331.

The principal solution of `sinx=(-1)/(2)` isA. `(pi)/(6)`B. `(-pi)/(6)`C. `(5pi)/(6)`D. `(7pi)/(6)`

Answer» Correct Answer - B
Discussion
332.

The principal solution of `sinx=-(sqrt(3))/(2)` isA. `(pi)/(3)`B. `(4pi)/(3)`C. `(2pi)/(3)`D. `-(4pi)/(3)`

Answer» Correct Answer - B
Discussion
333.

The principal solution of `sinx=-(sqrt(3))/(2)` isA. `(4pi)/(3)`B. `(-pi)/(3)`C. `(pi)/(3)`D. `(-2pi)/(3)`

Answer» Correct Answer - B
Discussion
334.

The principal solution of `sinx=-(sqrt(3))/(2)` isA. `(pi)/(3)`B. `(-5pi)/(3)`C. `(5pi)/(3)`D. `(2pi)/(3)`

Answer» Correct Answer - C
Discussion
335.

The principal solution of `sinx=(-1)/(sqrt(2))` isA. `(11pi)/(6)`B. `-(11pi)/(6)`C. `(pi)/(6)`D. `(5pi)/(6)`

Answer» Correct Answer - B
Discussion
336.

The principal solution of `sinx=(-1)/(sqrt(2))` isA. `(pi)/(4)`B. `(3pi)/(4)`C. `(5pi)/(4)`D. `(-5pi)/(4)`

Answer» Correct Answer - D
Discussion
337.

Find the principal solution of the equation `sinx=(sqrt(3))/2`.

Answer» `0<= x <= 2 pi`
as we know that, `sqrt3/2 = sin (pi/3)`
sin x = sin (pi/3)
so, `x=pi/3 ` principal solution
also, `pi-x = pi/3 `
`x= pi- pi/3 = (2pi)/3` principal solution
Discussion
338.

Prove that `3sin``pi/6``sec``pi/3``-4sin``(5pi)/6``cot``pi/4``=1`

Answer» `L.H.S = 3sin(pi/6)sec(pi/3)-4sin(5pi/6)cot(pi/4)`
`=3(1/2)(2)-4sin(pi-(pi/6))(1)`
As we know, `sin(pi -x) = sinx`
Our equation becomes,
`=3(1/2)(2)-4sin(pi/6)(1)`
`=3-4(1/2) = 1 = R.H.S.`
Discussion
339.

Prove that `(sin(x+y))/(sin(x-y))=(tanx+tany)/(tanx-tany)`

Answer» `L.H.S. = (sin(x+y))/(sin(x-y))`
`=(sinxcosy+sinycosx)/(sinxcosy-cosxsiny)`
Dividing numerator and denominator by `cosxcosy`
`=((sinx/cosx)+(siny/cosy))/((sinx/cosx)-(siny/cosy))`
`=(tanx+tany)/(tanx-tany)=R.H.S`
Discussion
340.

Show that `tan 3x tan 2x tan x = tan 3x - tan 2x - tan x`.

Answer» `3x = 2x + x`
`tan 3x = tan(2x+x)`
`tan 3x = (tan 2x + tan x)/(1- tan2x*tanx)`
`tan3x (1- tan2x*tanx) = tan2x + tanx)`
`tan3x - tan3x*tan2x*tanx = tan2x + tanx`
`tan3x - tan2x - tanx = tan3x tan2x tanx`
hence proved
Discussion
341.

Prove that `(cos7x+cos5x)/(sin7x-sin5x)=cotx`.

Answer» Here, we will use the following formulas:
`sin A - sin B = 2sin((A-B)/2)cos((A+B)/2)`
`cos A + cos B = 2cos((A-B)/2)cos((A+B)/2)`

`L.H.S. = (cos7x+cos5x)/(sin7x-sin5x)`
`=(2cos((7x+5x)/2)cos((7x-5x)/2))/(2cos((7x+5x)/2)sin((7x-5x)/2))`
`=cos(x)/sin(x)=cotx=R.H.S.`
Discussion
342.

Given x`gt` 0, then value of `f(x)=-3cossqrt(3+x+x^(2))` lie in the interval ….. .

Answer» Given function `f(x)=-3cossqrt(3+x+x^(2))`
We know that, `" "-1lecosxle1`
`rArr" "-3le3cosxle3`
`rArr" "3ge-3cosxge-3`
`rArr" "-3le-3cosxle3`
So, the value of `f(x)` lies in [-3, 3].
Discussion
343.

The value of `3(sinx-cosx)^(4)+6(sinx+cosx)^(2)+4(sin^(6)xcos^(6)x)` is

Answer» Given expression, `3(sinx-cos)^(4)+6(sinx+cosx)^(2)+4(sin^(6)x+cos^(6)x)`
`" "=3[sin^(2)x+cos^(2)x-2sinxcosx]^(2)+6[sin^(2)x+cos^(2)x+2*sinx*cosx]+4[(sin^(2)x)^(3)+(cos^(2)x)^(3)]`
`" "=3(1-sin2x)^(2)+6(1+sin2x)+4[(sin^(2)+cos^(2)x)(sin^(4)x-sinxcos^(2)x+cos^(4)x)`
`" "=3(1-sin^(2)23x-2sin2x)+6+6sin2x+4[(sin^(2)x+cos^(2)x)^(2)3sincos^(2)x]`
`" "=3+3sin^(2)2x-6sin2x+6+6sin2x`
`" "=4-3sin^(2)2x=13`
Discussion
344.

Which of the following is correct?A. `sin1^@gt sin1`B. `sin1^@ltsin1`C. `sin1^@=sin1`D. `sin1^@=pi/180sin1`

Answer» Correct Answer - B
Since f(x) =sin x is an invreasing function for `0 ltx ltpi//2` and 1 rad is approximately `57^@` , we have
`1^@lt1^RrArr sin1^@,sin1`
Discussion
345.

Evaluate `cosacos2acos3a cos 999 a ,`where `a=(2pi)/(1999)dot`

Answer» Let `P = cosacos2acos3a...cos999a`
Let `Q = sinasin2asin3a...sin999a`
Then, `2^999PQ = (2sinacosa)(2sin2acos2a)...(2sin999acos999a)`
`= sin2asin4a...sin1996asin1998a`
`=(sin2asin4a...sin998a)((-sin(2pi-1000a))(-sin(2pi-1002a))...(-sin(2pi-1998a)))`...[As `-sin(2pi-theta) = sintheta`]
`=(sin2asin4a...sin998a)((-sin(2pi-1000((2pi)/1999)))(-sin(2pi-1002((2pi)/1999))))...(-sin(2pi-1998((2pi)/1999)))))`...[As `a = (2pi)/1999`]
`=(sin2asin4a...sin998a)((-sin((2pi)/1999(999)))(-sin((2pi)/1999(997)))...(-sin((2pi)/1999)))`
`=(sin2asin4a...sin998a)(sin999asin997a...sin3asina)`
`=>2^999PQ = sinasin2asin3a...sin999a`
`=>2^999PQ = = Q`
`=>P = 1/(2^(999))`
`:. cosacos2acos3a...cos999a = 1/(2^(999))`
Discussion
346.

Show that: `cotx-cot2x=2`

Answer» `L.H.S. = cotx - cot2x = cosx/sinx -(cos2x)/(sin2x)`
`=(cosxsin2x - cos2xsinx)/(sinxsin2x)`
`=(sin(2x-x))/(sinxsin2x) = sinx/(sinxsin2x)`
When, `x = 15^@`
`1/(sin2x) = 1/(sin30^@) = 2 = R.H.S.`
So, given equation is true for `x = 15^@`
Discussion
347.

`nCr + 2. nC(r-1) + nC(r-2) = (n+2)Cr (2

Answer» Here, we will use the property of binomial coefficients,
`n_(C_r) + n_(C_(r-1)) = (n+1)_(C_r)`
`L.H.S. = n_(C_r)+2*n_(C_(r-1))+n_(C_(r-2))`
`= n_(C_r)+n_(C_(r-1))+n_(C_(r-1))+n_(C_(r-2))`
Using the above property,
`=(n+1)_(C_r)+(n+1)_(C_(r-1))`
Again using the above property,
`=(n+2)_(C_r) = R.H.S.`
Discussion
348.

The general solution of equation `sin^(50)x-cos^(50)x=1`

Answer» `0 <= sin^50 x <= 1; 0 <= a <= 1`
`0 <= cos^50 x <=1 ; 0 <= b <= 1`
`a-b =1`
`a=1`
`sin^50 x = 1`
`sin x = +- 1`
`x= n pi +- pi /2`
C option is correct
answer
Discussion
349.

The value of the expression`cos^3theta+cos^3(120+theta)+cos^3(240+theta)` is;

Answer» `cos 3 theta = 4 cos theta - 3 cos theta`
`cos^3 theta = ( cos 3 theta + 3 cos theta)/4`
`(cos(3 theta) + 3 cos theta + cos ( 360 + 3 theta) + 3 cos( theta + 120) + cos(120+ 3 theta) + 3cos( theta + 240))/4`
`= (cos 3 theta + 3 cos theta + cos 3 theta + 3 cos ( theta+ 120) + cos 30 + 3cos( theta-120))/4`
`= (3 cos 3 theta + 3 cos theta + 3 xx 2 cos theta cos 120)/4`
`= ( 3 cos 3 theta + 3 cos theta + 6 xx (-1/2) cos theta)/4`
`= 3/4 cos 3 theta`
option C is correct
Answer
Discussion
350.

If `max{5sin theta+3sin(theta-alpha)}=7` then the set of possible vaues of `alpha` is `theta in R`

Answer» `5 sin theta + 3 sin( theta - alpha)`
`= ( 5 + 3 cos alpha) sin theta - 3sin alpha cos theta`
`= {(5 + 3cos alpha)^2 + 9 sin^2 alpha}`
`= sqrt((5+ 3 cos alpha)^2 + 9 sin^2 alpha) sin(theta - phi) `
`max{5sin theta + 3 cos(theta- alpha)} `
`= sqrt((5+3cos alpha)^2 + 9sin^2alpha)`
=`7`
`25 + 30 cos alp-ha + 9 = 49`
`30 cos alpha = 15`
`cos alpha = 1/2`
`alpha = 2n pi + - pi/3`
option A is correct
answer
Discussion