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251.

The principal value of `cos^(-1)(-1/2)` isA. `pi/6`B. `pi/3`C. `(2pi)/3`D. `(5pi)/6`

Answer» Correct Answer - C
Discussion
252.

Solve `sin^(-1) [sin((2x^(2) + 4)/(1 + x^(2)))] lt pi -3`

Answer» `sin^(-1)[sin((2x^(2) + 4)/(1 + x^(2)))] lt pi - 3`
Now, `(2x^(2) + 4)/(1 + x^(2)) = (2x^(2) + 2 + 2)/(1 + x^(2)) = 2 + (2)/(1 + x^(2))`
So, `2+(2)/(1 + x^(2)) in (2, 4]`
Therefore, Eq. (i) reduces to
`pi-(2x^(2) + 4)/(1 + x^(2)) lt pi - 3`
or `(2x^(2) + 4)/(1 + x^(2)) gt 3`
or `2x^(2) + 4 gt 3 + 3x^(2)`
or `x^(2) - 1 lt 0`
or `-1 lt x lt1`
Discussion
253.

Solve `sin^(-1) (sin 6x) = x, x in [0,pi]`

Answer» `sin^(-1) (sin 6x) = x`
`rArr sin 6 x = sin x`
`rArr sin 6 x - sin x = 0`
`rArr 2 sin.(5x)/(2) cos.(7x)/(2) = 0`
`rArr (5x)/(2) n pi " or " (7x)/(2) = (2n + 1) (pi)/(2), n in Z`
`rArr x = (2n pi)/(5) " or " x = (2n + 1) (pi)/(7)`
`rArr x = (pi)/(7) , (3pi)/(7), (5pi)/(7), pi, 0, (2pi)/(5), (4pi)/(5)`
But `x = (5 pi)/(7), pi, (4pi)/(5)` are not possible as any solution `[-(pi)/(2), (pi)/(2)]`
Discussion
254.

STATEMENT-1 : The solution of `sin^-1 6x+sin^-1 6sqrt3x=pi/2` is , `x= +- 1/12.` and STATEMENT - 2 As, `sin^-1 x` is defined for `|x|

Answer» Correct Answer - D
Discussion
255.

If `sin^(- 1)(3/x)+sin^(- 1)(4/x)=pi/2` then `x=`A. -5B. 5C. 25D. -25

Answer» Correct Answer - B
Discussion
256.

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

Answer» Correct Answer - D
Discussion
257.

Find the principal value of `cosec^(-1)(-2/sqrt3)`.

Answer» Correct Answer - `-pi/3`
Discussion
258.

The principal value of `cosec^(-1)(-1)` isA. `-pi/2`B. 0C. `pi/2`D. `(3pi)/2`

Answer» Correct Answer - A
Discussion
259.

If `cot^(-1)x+tan^(-1)(1/2)=pi/4` then x isA. `1/3`B. `2/3`C. 2D. 3

Answer» Correct Answer - D
Discussion
260.

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

Answer» `sin^(-1)-sqrt(3)/(2) is an angle theta in [-(pi)/(2),(pi)/(2)]` such that `sin theta =sqrt(3)/(2)` clearly , such an angle is `-(pi)/(3)`
Discussion
261.

The value of `sin^(-1)[cos{sin^(-1)(-sqrt(3)/(2))}]` isA. `(pi)/(3)`B. `(pi)/(6)`C. `-(pi)/(3)`D. `-(pi)/(6)`

Answer» `sin^(-1)[cos{sin^(-1)(-sqrt(3)/(2))}]`
`=sin^(-1){cos(-(pi)/(3))}` [before `sin^(-1)sqrt(3)/(2)=-(pi)//(3)`
`=sin^(-1)(cos(pi)/(3))=sin^(-1)(1/2)=(pi)/(6)`
Discussion
262.

Solve `tan^(-1) x + sin^(-1) x = tan^(-1) 2x`

Answer» `tan^(-1) x + sin^(-1) x = tan^(-1) 2x`
`rArr sin^(-1) x = tan^(-1) 2x - tan^(-1) x`
`rArr tan^(-1).(x)/(sqrt(1 - x^(2))) = tan^(-1) [(2 x - x)/(1 + 2x^(2))]`
`rArr (x)/(sqrt(1 - x^(2))) = (x)/(1 + 2x^(2))`
`rArr 2x^(3) + x = x sqrt(1 - x^(2))`
`rArr 2x^(3) - x sqrt(1 - x^(2)) + x = 0`
`rArr x(2x^(2) - sqrt(1 - x^(2)) + 1) = 0`
`rArr x = 0 " or " 2x^(2) + 1 = sqrt(1 - x^(2))`
`rArr x = 0 " or " 4x^(4) + 4x^(2) + 1 = 1 - x^(2)`
`rArr x = 0 " or " 4x^(4) + 5x^(2) = 0`
`rArr x = 0` is the only solution
Discussion
263.

If `a_(1), a_(2), a_(3),...., a_(n)` is an A.P. with common difference d, then prove that `tan[tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_( - 1)a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))`

Answer» We have
`tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_(n -1) a_(n)))`
`= tan^(-1) ((a_(2) -a_(1))/(1 + a_(1) a_(2))) + tan^(-1) ((a_(3) -a_(2))/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((a_(n) - a_(n -1))/(1 + a_(n -1) a_(n)))`
`= (tan^(-1) a_(2) - tan^(-1) a_(1)) + (tan^(-1) a_(3) - tan^(-1) a_(2)) + ...+ (tan^(-1) a_(n) - tan^(-1) a_(n))`
`= tan^(-1) a_(n) - tan^(-1) a_(1) = tan^(-1) ((a_(n) -a_(1))/(1 + a_(n) a_(1))) = tan^(-1) (((n -1))/(1 + a_(1) a_(n)))`
`rArr tan[tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ... + tan^(-1) ((d)/(1 + a_(n -1) a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))`
Discussion
264.

Solve `tan^(-1) x + cot^(-1) (-|x|) = 2 tan^(-1) 6x`

Answer» If `x gt0`
`tan^(-1) x + cot^(-1) (-x) = 2 tan^(-1) 6x`
`rArrr tan^(-1) x + pi - cot^(-1) x = 2 tan^(-1) 6x`
`rArr (pi)/(2) + 2 tan^(-1) x = 2 tan^(-1) 6x`
`rArr tan^(-1) 6x - tan^(-1) x = (pi)/(4)`
`rArr tan^(-1).(6x - x)/(1 + 6x^(2)) = (pi)/(4)`
`rArr (5x)/(1 + 6x^(2)) =1`
`rArr 6x^(2) -5x + 1 = 0`
`rArr x = (1)/(2), (1)/(3)`
If `x lt 0`
`tan^(-1) x + cot^(-1) x = 2 tan^(-1) 6x`
`rArr tan^(-1) 6x = (pi)/(4)`, This is not possible as `x lt 0`
Discussion
265.

The number of real solution of equation `tan^(-1)x+cot^(-1)(-|x|)=2tan^(-1)(6x)` isA. 4B. 3C. 2D. 1

Answer» Correct Answer - C
When x lt 0, wwe have `tan^(-1)(6x)=(pi)/(4)`
`rArr x =(1)/(6)`, which is not possible
When `x ge 0`, we have `(pi)/(4)=tan^(-1)(6x)-tan^(-1)(x)`
`rArr (6x-x)/(1+6x^(2))=1`
`rArr x=(1)/(2),(1)/(3)`
Discussion
266.

Solves `cos^(-1) x lt 2`

Answer» Correct Answer - `x in (cos2, 1)`
`cos^(-1) x gt 2`
or `0 le cos^(-1) x lt 2`
or `cos 0 ge x gt cos 2` (`:. Cos^(-1) x`is decreasing function)
`rArr x in (cos 2, 1]`
Discussion
267.

If `A = 2 tan^(-1) (2 sqrt2 -1) and B = 3 sin^(-1) ((1)/(3)) + sin^(-1) ((3)/(5))`, then which is greater ?

Answer» We have `A = 2 tan^(-1) (2 sqrt2 -1) = 2 tan^(-1) (1.828)`
`rArr A gt 2 tan^(-1) sqrt3`
`rArr A gt (2pi)/(3)`
Now, `sin^(-1) ((1)/(3)) lt sin^(-1) ((1)/(2))`
`rArr sin^(-1) ((1)/(3)) lt (pi)/(6)`
`rArr 3 sin^(-1).(1)/(3) lt (pi)/(2)`
Further, `sin^(-1) ((3)/(5)) = sin^(-1) (0.6) lt sin^(-1) ((sqrt3)/(2))`
`rArr sin^(-1) ((3)/(5)) lt (pi)/(3)`
`rArr B = 3 sin^(-1) ((1)/(3)) + sin^(-1) ((3)/(5)) lt (pi)/(2) + (pi)/(3)`
`rArr B lt (5pi)/(6)`
From this, we really cannot relate A and B.
Now, `3 sin^(-1) ((1)/(3)) = sin^(-1) [3.(1)/(3) - 4((1)/(3))^(3)]`
`= sin^(-1) ((23)/(27))`
`= sin^(-1) (0.852)`
`rArr 3 sin^(-1) ((1)/(3)) lt sin^(-1) ((sqrt3)/(2)) = (pi)/(3)`
Hence, `B = 3 sin^(-1) ((1)/(3)) + sin^(-1) ((3)/(5)) lt (pi)/(3) + (pi)/(3) = (2pi)/(3)`
`:. A gt B`
Discussion
268.

`cos(tan^(-1)3/4)+cos(tan^(-1)x)` is equal toA. `4/5+x/sqrt(1+x^2)`B. `3/5+1/sqrt(1+x^2)`C. `4/5+1/sqrt(1+x^2)`D. `3/5+x/sqrt(1+x^2)`

Answer» Correct Answer - C
Discussion
269.

`If"sin"{cot^(-1)(x+1)}="cos"(tan^(-1)x),`then find `xdot`A. 1B. `1/2`C. 0D. `-1/2`

Answer» Correct Answer - D
Discussion
270.

Find the value `underset(n rarr oo)("lim") underset(k =2)overset(n)sum cos^(-1) ((1 + sqrt((k -1) k(k + 1) (k + 2)))/(k(k + 1)))`

Answer» `cos^-1((1+sqrt(k(k-1)(k+1)(k+2)))/(k(k+1)))`
`=cos^-1((1+sqrt(k(k^2-1)(k+2)))/(k(k+1)))`
`=cos^-1(1/(k(k+1))+sqrt(1-1/k^2)sqrt(1-1/(k+1)^2))`
`=cos^-1 (1/(k+1)) - cos^-1(1/k)`
`:. sum_(k=2)^n cos^-1 (1/(k+1) - 1/k) = cos^-1(1/3) - cos^-1(1/2) +cos^-1(1/4) - cos^-1(1/3)+cos^-1(1/5) - cos^-1(1/4)+...+cos^-1 (1/(n+1)) - cos^-1(1/n)`
`=cos^-1 (1/(n+1)) - cos^-1(1/2)`
`:. Lim_(n->oo) sum_(k=2)^n cos^-1 (1/(k+1) - 1/k) = Lim_(n->oo) cos^-1 (1/(n+1)) - cos^-1(1/2) `
`=cos^-1(0) - pi/3`
`=pi/2-pi/3 = pi/6`
`:. Lim_(n->oo) sum_(k=2)^n cos^-1((1+sqrt(k(k-1)(k+1)(k+2)))/(k(k+1))) =pi/6.`
Discussion
271.

`tan^(-1)(1/2)+tan^(-1)(1/3)` is equal toA. `pi/6`B. `pi/4`C. `pi/3`D. `(5pi)/12`

Answer» Correct Answer - B
Discussion
272.

If `alpha=tan^(-1)((4x-4x^3)/(1-6x^2+x^2)),beta=2sin^(-1)((2x)/(1+x^2))`and `tanpi/8=k ,`then`alpha+beta=piforx in [(1,1)/k]``alpha+betaforx in (-k , k)``alpha+beta=piforx in [(1,1)/k]``alpha+beta=0forx in [-k , k]`A. `alpha + beta = pi " for " x in [1, (1)/(k))`B. `alpha = beta " for " x in (-k, k)`C. `alpha + beta = - pi " for " x in [1, (1)/(k))`D. `alpha + beta = 0 " for " x in (-k, k)`

Answer» Correct Answer - A::B
Put `x = tan theta`
`rArr alpha = tan^(-1) (tan 4 theta)`
`= 4 theta - pi " for " x in [1, (1)/(k))`
`= 4 theta " for " x in (-k, k)`
Also, `beta - 2(pi - 2 theta) " for " x in [1, (1)/(k))`
`= 4 theta " for " x in (-k, k)`
Discussion
273.

Find the value of`2cos^(-1)3/(sqrt(13))+cot^(-1)(16)/(63)+1/2cos^(-1)7/(25)`

Answer» `2cos^-1 (3/sqrt13) + cot^-1 (16/63) +1/2cos^-1 (7/25)`
Now, `cos^-1(3/sqrt13) = tan^-1(2/3)`
`:. 2cos^-1(3/sqrt13) = 2 tan^-1(2/3) = tan^-1((2*(2/3))/(1-(2/3)^2)) = tan^-1((4/3)/(5/9)) = tan^-1(12/5)`
`cot^-1(16/63) = tan^-1(63/16)`
Let `1/2cos^-1(7/25) = tan^-1x`
`=> cos^-1(7/25) = 2tan^-1x`
`=> tan^-1(24/7) = tan^-1((2x)/(1-x^2))`
`=>(2x)/(1-x^2) = 24/7`
Solving it , we get , ` x =3/4`
`:. 1/2cos^-1(7/25) = tan^-1(3/4)`
Putting these values in the given equation,
`tan^-1(12/5)+tan^-1(63/16)+tan^-1(3/4)`
`=(tan^-1(12/5)+tan^-1(3/4))+tan^-1(63/16)`
As, `12/5 gt 0, 3/4 gt 0 and 12/5*3/4 gt 1`,
So, it becomes,
`=pi+ tan^-1((12/5+3/4)/(1-(12/5)(3/4)))+tan^-1(63/16)`
`=pi- tan^-1(63/16)+tan^-1(63/16)`
`=pi`
`:. 2cos^-1 (3/sqrt13) + cot^-1 (16/63) +1/2cos^-1 (7/25) = pi`
Discussion
274.

The principal value of `sin^-1 {tan ((-5pi)/4)}` isA. `(pi)/(4)`B. `-(pi)/(4)`C. `(pi)/(2)`D. `(pi)/(2)`

Answer» `sin^(-1){tan(-(5pi)/(4))}=sin^(-1)=(pi)/(2)`
Discussion
275.

If `sin^(-1)(1-x) -2sin^(-1)x=(pi)/(2)` then x equalA. `0,1/2`B. `0,1/2`C. 0D. none of these

Answer» Correct Answer - C
Discussion
276.

If `(tan^(-1) x)^2 + (cot^(-1) x)^2 = (5pi^2)/8` then `x` equalsA. -1B. 1C. 0D. 2

Answer» Correct Answer - A
Discussion
277.

If `cos^(-1)(6x)/(1+9x^2)=-pi/2+tan^(-1)3x ,`then find the value of`xdot`

Answer» `cos^(-1)((6x)/(1+9x^2))=-pi/2+2tan^(-1)3x`
`pi/2-sin^(-1)((6x)/(1+9x^2))=-pi/2+2tan^(-1)x`
`pi-sin^(-1)((6x)/(1+9x^2))=2tan^(-1)3x`
`pi-2tan^(-1)3x=sin^(-1)((2(3x))/(1+(3x)^2))`
`pi-2tan^(-1)3x=sin^(-1)((2(3x))/(1+(3x)^2))`
When `3x>=1,x>1/3`
`x in(1/3,oo)`.
Discussion
278.

If `sin^(-1)(x/a)+sin^(-1)(y/b)=alpha` thenA. `(x^(2))/(a^(2))-(2xy)/(ab) cos alpha+(y^(2))/(b^(2))=sin^(2)alpha`B. `(x^(2))/(a^(2))-(2xy)/(ab) sin alpha+(y^(2))/(b^(2))=cos^(2)alpha`C. `(x^(2))/(a^(2))+(2xy)/(ab) cos alpha+(y^(2))/(b^(2))=sin^(2)alpha`D. `(x^(2))/(a^(2))+(2xy)/(ab) sin alpha+(y^(2))/(b^(2))=cos^(2)alpha`

Answer» We have
`sin^(-1)(x)/(a)+sin^(-1)(y)/(b)=alpha`
`rarr (pi)/(2)-cos^(-1)(x)(alpha)+(pi)/(2)-cos^(-1)(y)/(b)=alpha`
`rarr (xy)/(ab)-sqrt(1-(x^(2))/(z^(2))sqrt(1-y^(2))/(b^(2))=cos(pi-alpha)`
`rarr (x^(2))/(a^(2))+(2xy)/(ab)cos alpha+(y^(2))/(b^(2))=sin^(2)alpha`
Discussion
279.

If `cos^(-1)(x/m)+cos^(-1)(y/n)=theta`, then `x^2/m^2-(2xy)/(mn)costheta+y^2/n^2` is equal toA. `theta`B. `sin^2theta`C. `cos^2theta`D. 1

Answer» Correct Answer - B
Discussion
280.

Solve `tan^(-1)("x"+1)+tan^(-1)("x"-1)=tan^(-1)" "8/(31)`A. `-1/4,8`B. `1/4,-8`C. `-4,1/8`D. `4,-1/8`

Answer» Correct Answer - B
Discussion
281.

If `alpha = sin^(-1) (cos(sin^-1) x)) and beta = cos^(-1) (sin (cos^(-1) x))`, then find `tan alpha. tan beta.`

Answer» `beta = cos^(-1) (sin ((pi)/(2) - sin^(-1) x))`
`= cos^(-1) [cos (sin^(-1) x)]`
Also, `alpha = sin^(-1) [cos (sin^(-1) x)]`
`:. alpha + beta = (pi)/(2)`
`rArr tan alpha = cot beta " or " tan alpha. tan beta = 1`
Discussion
282.

The value of `sin^(-1)[xsqrt(1-x)-sqrt(x)sqrt(1-x^2)]`is equal to`sin^(-1)x+sin^(-1)sqrt(x)``sin^(-1)x-sin^(-1)sqrt(x)``sin^(-1)sqrt(x)-sin^(-1)x`none of theseA. `sin^(-1) x + sin^(-1) sqrtx`B. `sin^(-1) x - sin^(-1) sqrtx`C. `sin^(-1) sqrtx - sin^(-1) x`D. none of these

Answer» Correct Answer - B
Let `x = sin theta and sqrtx = sin phi, " where " x in [0, 1]`
`rArr theta, phi in [0, pi//2]`
`rArr theta - phi in [(-pi)/(2), (pi)/(2)]`
Now, `sin^(-1) (x sqrt(1 -x) - sqrtx sqrt(1 - x^(2)))`
`= sin^(-1) (sin theta sqrt(1 - sin^(2) phi) - sin phi sqrt(1 - sin^(2) theta))`
`= sin^(-1) (sin theta cos phi - sin phi cos theta)`
`= sin^(-1) sin (theta - phi) = theta - phi`
`= sin^(-1) (x) - sin^(-1) (sqrtx)`
Discussion
283.

If `alpha = sin^(-1)(sqrt(3)/2)+sin^(-1)(1/3) , beta =cos ^(-1)(sqrt(3)/2)+cos^(-1)(1/3)` thenA. `alpha gt beta`B. `alpha = beta`C. `alpha lt beta`D. `alpha + beta =2pi`

Answer» Correct Answer - C
Discussion
284.

If `cos^(- 1)x-cos^(- 1)(y/2)=alpha` then `4x^2-4xycosalpha+y^2=`A. 4B. `2 sin^(2) alpha`C. `-4 sin^(2) alpha`D. `4 sin^(2) alpha`

Answer» Correct Answer - D
We have `cos^(-1) x - cos^(-1).(y)/(2) = alpha`
or `x = cos (cos^(-1).(y)/(2) + alpha)`
`= cos (cos^(-1).(y)/(2)) cos alpha - sin (cos^(-1).(y)/(2)) sin alpha`
`= (y)/(2) cos alpha - sqrt(1 - (y^(2))/(4)) sin alpha`
or `2x = y cos alpha - sin alpha sqrt(4 -y^(2))`
or `2x - y cos alpha = - sin alpha sqrt(4 -y^(2))`
Squaring, we get
`4x^(2) + y^(2) cos^(2) alpha - 4xy cos alpha = 4 sin^(2) alpha - y^(2) sin^(2) alpha`
or `4x^(2) - 4xy cos alpha + y^(2) = 4 sin^(2) alpha`
Discussion
285.

Statement-1: `tan{cos^(-1)(1)/sqrt(82)-sin^(-1)(5)/sqrt(26)}=29/3` Statement-2: `[x cos(cot^(-1))^(2)=51/50rarr x -(1)/5sqrt(2)`A. Statement-1 is is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.

Answer» we have tan `{cos^(-1)(1)sqrt(82)-sin^(-1)(5)/sqrt(26)}`
`=tan (tan^(-1)9-tan^(-1)5)=tan{tan^(-1)(9-5)/(1+9xx5)}`
`=tan(tan^(-1)2/23)=2/23`
so statement 1 is not true
`rarr (x^(2))/sqrt(x^(2)+1)+(1)sqrt(x^(2)+1)^(2)=51/50 rarr x^(2)+1=51/50 rarr x=(1)/(5sqrt(2))`
so statement 2 is true
Discussion
286.

The trigonometric equation `sin^(-1)x=2sin^(-1)a`has a solution forall real values (b) `|a|

Answer» Correct Answer - C
`sin^(-1) x = 2 sin^(-1) a`
Now `-(pi)/(2) le sin^(-1) a le(pi)/(2)`
`rArr -(pi)/(4) le sin^(-1) a le (pi)/(4)`
`rArr -(1)/(sqrt2) le a le (1)/(sqrt2)`
`rArr |a| le (1)/(sqrt2)`
Discussion
287.

The trigonometric equation `sin^(-1)x=2sin^(-1)a`has a solution forall real values (b) `|a|

Answer» Let `alpha` a solution of `sin^(-1)x=2 sin^(-1) alpha` Then `sin^(-1) alpha =2 sin^(-1) alpha`
`rarr -(pi)/(2) le 2 sin^(-1) alpha le (pi)/(2)`
` rarr -(pi)/(4) le sin^(-1) a le (pi)/(4)`
`rarr sin((pi)/(4)) le alpha le sin(pi)/(4) rarr -(1)/sqrt(2) le alpha le (1)/sqrt(2) rarr |alpha| le (1)/sqrt(2)`
Discussion
288.

The trigonometric equation `sin^(-1)x=2sin^(-1)a`has a solution forall real values (b) `|a|

Answer» `-pi/22<=sin^(-1)x<=pi/2`
`-pi/2<=2sina<=pi/2`
`-pi/4<=sina<=pi/4`
`-1/sqrt2<=a<=1/sqrt2`
`|a|<=1/sqrt2`.
Discussion
289.

`tan^-1[(cos x)/(1+sin x)]` is equal toA. `(pi)/(4) - (x)/(2)`, for `x in (-(pi)/(2), (3pi)/(2))`B. `(pi)/(4) -(x)/(2), " for " x in (-(pi)/(2), (pi)/(2))`C. `(pi)/(4), (x)/(2), " for " x in ((3pi)/(2), (5pi)/(2))`D. `(pi)/(4) -(x)/(2), " for " x in (-(3pi)/(2), (pi)/(2))`

Answer» Correct Answer - A
`tan^(-1) [(cos x)/(1 + sin x)] = tan^(-1) [(sin [(pi//2) -x])/(1 + cos [(pi//2) - x])]`
`= tan^(-1) [((2sin[(pi//4) - (x//2)]),(cos [(pi//4) - (x//2)]))/(2 cos^(2) [(pi//4) - (x//2)]`
`= tan^(-1) tan ((pi)/(4) - (x)/(2)) = (pi)/(4) - (x)/(2)`
`rArr -(pi)/(2) lt (pi)/(4) - (x)/(2) lt (pi)/(2)`
`rArr -(3pi)/(4) lt -(x)/(2) lt (pi)/(4)`
`rArr-(pi)/(4) lt (x)/(2) lt (3pi)/(4)`
`rArr -(pi)/(2) lt x lt (3pi)/(2)`
Discussion
290.

If `f(x)=x^(11)+x^9-x^7+x^3+1`and `f(sin^(-1)(sin8)=alpha,alpha`is constant, then `f(tan^(-1)(t a n8)`is equal to`alpha`(b) `alpha-2`(c) `alpha+2`(d) `2-alpha`A. `alpha`B. `alpha -2`C. `alpha + 2`D. `2 - alpha`

Answer» Correct Answer - D
`f(x) + f(-x) = 2`
Now `(sin^(-1) (sin 8)) = 3pi - 8 = y`
And `(tan^(-1) (tan 8)) = (8 - 3pi) = -y`
Hence, `f(y) + f(-y) = 2`
Given `f(y) = alpha`, we have `f(-y) = 2 - alpha`
Discussion
291.

If `cos(2sin^(-1)x)=1/9,`then find the values of `xdot`

Answer» `cos(2sin^(-1)x)=1/9`
Let ` sin^(-1)x=alpha`
`x=sinalpha``cos(2alpha)=1/9`
`1-2sin^2alpha=1/9`
`1-1/9=2sin^2alpha`
`2sin^2alpha=8/9`
`sin^2alpha=4/9`
`x=sinalpha=pm2/3`.
Discussion
292.

The algebraic expression for `f(x)=tan(sin^(-1)(cos("tan"^(-1)(x)/(2))))` isA. `(2)/(x)`B. `(x)/(2)`C. `(1)/(x)`D. `(2)/(|x|)`

Answer» Correct Answer - D
Let `"tan"^(-1) (x)/(2)=theta`
`rArr tan. theta=(x)/(2)`
`rArr cos("tan"^(-1)(x)/(2))=cos. theta=(2)/(sqrt(4+x^(2)))`
`rArr f(x)=tan["sin"^(-1)(2)/(sqrt(4+x^(2)))]=(2)/(x)`
If `x gt 0`, then `f(x)=tan("tan"^(-1)(2)/(x))=(2)/(x)`
If `x lt 0`, then `f(x)=tan(tan^(-1)((-2)/(x)))=(-2)/(x)`
`rArr f(x)=(2)/(|x|)`
Discussion
293.

Which of the following is not true ?A. `sin cos^(-1)tan cot^(-1)x=sqrt(1-(1)/(x^(2)))`B. `cos tan^(-1)cot sin^(-1)x = x`C. `tan cot^(-1)sin cos^(-1)x = (1)/(sqrt(1-x^(2)))`D. `cot sin^(-1)cos tan^(-1)x=sqrt(1-x^(2))`

Answer» Correct Answer - D
(a) `sin cos^(-1)tan cot^(-1)x=sin "cos"^(-1)(1)/(x)=sqrt(1-(1)/(x^(2)))`
(b) `cos tan^(-1)cot sin^(-1)x = "cos tan"^(-1)(sqrt(1-x^(2)))/(x)=x`
(c ) `tan cot^(-1)sin cos^(-1)x=tan cot^(-1) sqrt(1-x^(2))=(1)/(sqrt(1-x^(2)))`
(d) `cot sin^(-1)cos tan^(-1)x="cot sin"^(-1)(1)/(sqrt(1+x^(2)))=x`
Discussion
294.

`sin(1/4sin^(- 1)(sqrt 63/8)i s`A. `(1)/(2)`B. `(1)/(3)`C. `(1)/(2sqrt(2))`D. `(1)/(5)`

Answer» Correct Answer - C
We have sin `((1)/(4)"sin"^(-1)(sqrt(63))/(8))`
Let `"sin"^(-1)(sqrt(63))/(8)=theta`
`therefore (sqrt(63))/(8)=sin theta`
`rArr cos. theta = (1)/(8)`
`therefore cos.(theta)/(2)=sqrt((1+cos theta)/(2))=sqrt((1+(1)/(8))/(2))=(3)/(4)`
`rArr = sin.(theta)/(4)=sqrt((1-cos.(theta)/(2))/(2))=sqrt((1-(3)/(4))/(2))=(1)/(2sqrt(2))`
Discussion
295.

The value of `sin^(-1)("cos"(cos^(-1)(cosx)+sin^(-1)(sinx))),`where `x in (pi/2,pi)`, is equal to`pi/2`(b) `-pi`(c) `pi`(d) `-pi/2`A. `(pi)/(2)`B. `-pi`C. `pi`D. `-(pi)/(2)`

Answer» Correct Answer - D
`sin^(-1) (cos (cos^(-1) (cos x) + sin^(-1) (sin x)))`
`= sin^(-1) (cos (x + pi - x)) " " ["as " x in (pi//2)]`
`= sin^(-1) (cos pi) = sin^(-1) (-1) = -(pi)/(2)`
Discussion
296.

The solution of `sin^(-1)|sin x|=sqrt(sin^(-1)|sin x|)` isA. `n pi pm 1, n pi, n in Z`B. `n pi+1, n pi, n in Z`C. `n pi-1, n pi, n in Z`D. `2n pi+1, n pi, n in Z`

Answer» Correct Answer - A
Solution of `y=sqrt(y)` is y = 1 and y = 0
`rArr sin^(-1)|sin x|=0` or 1
`sin^(-1)|sin x|` is periodic with period `pi`
In `(0, pi)`, if `sin^(-1)|sin x|=1, x = 1` or `x = pi -1`
`therefore` General solution is `x = n pi 1, n in Z`
If `sin^(-1)|sin x| =0 rArr x = n pi, n in Z`
Discussion
297.

Maximum value of function `f(x)=(sin^(-1)(sinx)^(2)-sin^(-1)(sinx)` is:A. `(pi)/(4)[pi+2]`B. `(pi)/(4)[pi-2]`C. `(pi)/(2)[pi+2]`D. `(pi)/(2)[pi-2]`

Answer» Correct Answer - A
`y=(sin^(-2)(sin x))^(2)-sin^(-1)(sin x)`
`=(sin^(-1)(sin x)-(1)/(2))^(2)-(1)/(4)`
For maximum value of `y, sin^(-1)(sin x)=-(pi)/(2)`
`rArr y =((pi)/(2)+(1)/(2))^(2)-(1)/(4)=(pi)/(4)(pi+2)`
Discussion
298.

The vale of `sec(sin^-1(sin((-50pi)/9))+cos^-1(cos(31pi)/9))`A. `sec.(10pi)/(9)`B. `sec.(pi)/(9)`C. 1D. `-1`

Answer» Correct Answer - D
`sin^(-1)(-sin.(50pi)/(9))=sin^(-1)(sin(-6pi+(4pi)/(9)))`
`=sin^(-1)(sin.(4pi)/(9))`
`=(4pi)/(9)`
`cos^(-1)cos(-(31pi)/(9))=cos^(-1)cos((31pi)/(9))`
`=cos^(-1)cos(4pi-(5pi)/(9))=cos^(-1)cos.(5pi)/(9)=(5pi)/(9)`
Hence, given equals to sec `((4pi)/(9)+(5pi)/(9))=sec pi =-1`.
Discussion
299.

The value of `lim_(|x| rarr oo) cos (tan^(-1) (sin (tan^(-1) x)))` is equal toA. `-1`B. `sqrt2`C. `-(1)/(sqrt2)`D. `(1)/(sqrt2)`

Answer» Correct Answer - D
`underset(|x| rarr oo)("lim") cos (tan^(-1) (sin (tan^(-1) x)))`
`= cos (tan^(-1) (sin (tan^(-1) oo)))`
`= cos (tan^(-1) (sin ((pi)/(2))))`
`= cos (tan^(-1) (1)) = cos ((pi)/(4))`
`= (1)/(sqrt2)`
Discussion
300.

`2 tan^(-1) (-2)` is equal toA. `-cos^(-1) ((-3)/(5))`B. `-pi + cos^(-1).(3)/(5)`C. `-(pi)/(2) + tan^(-1) (-(3)/(4))`D. `-pi + cot^(-1) (-(3)/(4))`

Answer» Correct Answer - A::B::C
Let `tan^(-1) (-2) = theta " or " tan theta = -2`
`rArr theta in (-pi//2, 0) " or " 2 theta in (-pi, 0)`
`cos (-2 theta) = cos 2 theta = (1 - tan^(2) theta)/(1 + tan^(2) theta) = (-3)/(5)`
or `-2 theta = cos^(-1) ((-3)/(5)) = pi - cos^(-1).(3)/(5)`
or `2 theta = pi + cos^(-1).(3)/(5) = - pi + tan^(-1).(4)/(3)`
`= - pi + cot^(-1).(3)/(4) = - pi + (pi)/(2) - tan^(-1). (3)/(4)`
`= - (pi)/(2) - tan^(-1).(3)/(4) = - (pi)/(2) + tan^(-1) (-(3)/(4))`
Discussion