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401.

If `sin^(-1)(x/5)+cose c^(-1)(5/4)=pi/2`then the value of x is:(1) 1 (2)3 (3) 4(4) 5

Answer» `sin^-1(x)=cosec^-1(1/x)`
`cosec^-1 5/4= sin^(-1)4/5=0`
`sin^-1(x/5)+theta=pi/2`
`sin^-1(x/5)+theta=pi/2`
`sin^-1(x/5)=pi/2-theta`
`x/5=sin(pi/2,-theta)`
`x/5=costheta`
`x/5=3/5`
`x=3`
Discussion
402.

If `x in[-1/2,1] then sin^(-1)(sqrt(3)/(2)x-1/2sqrt(1-x^(2)))`A. `sin^(-1)1/2 -sin^(-1)x`B. `sin^(-1)x-(pi)/(6)`C. `sin^(-1)x+(pi)/(6)`D. none of these

Answer» Let `x=sin theta` then
`-1/2lexle1rarr-1/2lesin theta le 1 rarr -(pi)/(6)le theta le (pi)/(2)`
Now
`sin^(-1)sqrt(3)/(2)x-1/2sqrt(1-x^(2))`
`=sin^(-1){sin(theta-(pi)/(6))}`
`=theta-(pi)/(6)`
`=sin^(-1)x-(pi)/(6)`
Discussion
403.

If `x_1, x_2, x_3,a n dx_4`are the roots of the equations `x^4-x^3sin2beta+x^2cos2beta-xcosbeta-sinbeta=0,`prove that `tan^(-1)x_1+tan^(-1)x_2+tan^(-1)x_3+tan^(-1)x_4=npi+(pi/2)-beta`, where `n`is an integer.

Answer» Given equation is,
`x^4-x^3sin2beta+x^2cos2beta - xcosbeta - sinbeta = 0`
As `x_1,x_2,x_3 and x_4` are the roots of the above equation.
`:. sum x_1 = sin2beta`
`sumx_1x_2 = cos2beta`
`sumx_1x_2x_3 = cosbeta`
`x_1x_2x_3x_4 = -sinbeta`
Now, `tan(tan^-1x_1+tan^-1x_2+tan^-1x_3+tan^-1x_4) = (sumx_1-sum x_1x_2x_3)/(1-sumx_1x_2+x_1x_2x_3x_4)`
`=(sin2beta-cosbeta)/(1+cos2beta-sinbeta)`
`=(2sinbetacosbeta-cosbeta)/(2sin^2beta - sinbeta)`
`=(cosbeta(2sinbeta-1))/(sinbeta(2sinbeta -1))`
`=cot beta = tan(pi/2-beta)`
`:. tan(tan^-1x_1+tan^-1x_2+tan^-1x_3+tan^-1x_4) = tan(pi/2-beta)`
`:. tan^-1x_1+tan^-1x_2+tan^-1x_3+tan^-1x_4 = npi+pi/2-beta`, where `n in Z`.
Discussion
404.

The value of `cot(cose c^(-1)5/3+tan^(-1)2/3)`is:(1) `6/(17)`(2) `3/(17)`(2) `4/(17)`(4) `5/(17)`

Answer» `cosec^-1 5/3 = theta_1`
`cosec theta_1 = 5/3`
`tan^-1 (2/3) = theta_2`
`tan theta_2 = 2/3`
now, `cot(tan^-1(3/4) + tan^-1 (2/3) )`
`= cot(tan^-1((3/4 + 2/3)/(1 - 3/4 * 2/3)))`
`= cot(tan^-1((17/2)/(1/2)))`
`= cot(tan^-1(17/6))`
`= cot theta_3`
`= 6/17`
option 1 is correct
Discussion
405.

`tan^(-1)sqrt(3)-cot^(-1)(-sqrt(3))`is equal to (A) `pi` (B) `-pi/2` (C) 0 (D) `2sqrt(3)`

Answer» We know, `tan (pi/3) = sqrt(3)`
and `cot(5pi/6) = -sqrt(3)`
So, putting these values in the given expression
`tan^-1(tan(pi/3)) cot^-1(cot((5pi)/6))`
`pi/3-(5pi)/6 = -pi/2`
Discussion
406.

If `f(x)=sin^(-1)x and lim_(xto1//2+)f(3x-4x^3)=a-3lim_(xto1//2+)f(x)`, then [a] is equal to {where [] denotes G.I.F}

Answer» Correct Answer - 3
Discussion
407.

Find the principal value of: `cot^(-1)(sqrt(3))`

Answer» As `cot (pi/6) = sqrt3`, we can write our expression as:
`cot^-1(cot (pi/6))``=pi/6` which is the required principal value.
Discussion
408.

Find the number of positive integral solution of the equation `tan^(-1)x+cos^(-1)(y/(sqrt(1+y^2)))=sin^(-1)(3/(sqrt(10)))`

Answer» `tan^-1x+cos^-1(y/(1+y^2)) = sin^-1(3/sqrt10)`
We will convert this equation in `tan^-1` form.
`=> tan^-1x + tan^-1(1/y) = tan^-1 3`
`=>tan^-1(1/y) = tan^-1 3 - tan^-1 x`
`=>tan^-1(1/y) = tan^-1((3-x)/(1+3x))`
`=>1/y = (3-x)/(1+3x)`
`=> y = (1+3x)/(3-x)`
As, we have to find positive integral values, so `x` can have only two values, `1` and `2`.
When `x = 1, y= (1+3)/2 => y= 2`
When `x = 2, y= (1+6)/1 => y= 7`
So, these are two possible solutions for the given equation.
Discussion
409.

Find the domain for `f(x)=sin^(-1)((1+x^2)/(2x))`

Answer» `f(x) = sin^-1((1+x^2)/(2x))`
We know `sin theta` lies between `-1` and `1`.
`:. -1 le (1+x^2)/(2x) le 1.`
`|1+x^2| le 2x`
`=>x^2-2|x| + 1 le 0`
`=>(|x| - 1)^2 le 0`
`=>|x| = 1`
`:. x = -1 or x = 1.`
`:.` Domain of `f(x)` will be `x in {-1,1}.`
Discussion
410.

If `y=tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))`, find `(dy)/(dx)`.

Answer» `x=cos2theta`
`1+x=1+cos2theta=2cos^2theta`
`sqrt(1+x)=sqrt2costheta`
`1-x=1-cos2theta=2sin^2theta`
`sqrt(1-x)=sqrt2sintheta`
`y=tan^(-1)((sqrt2costheta-sqrt2sintheta)/(sqrt2costheta+sqrt2sintheta))`
`=tan^(-1)((costheta-sintheta)/(costheta+sintheta))`
`=tan^(-1)((1-tantheta)/(1+tantheta))`
`=tan^(-1){tan(pi/4-theta)}`
`y=pi/4-theta`
diff. with respect to x
`dy/dx=d/dx(pi/4-a)=-dy/dx`
`-(-1/2)*(1/sqrt(1-x^2))`
`=1/(2sqrt(1-x^2)`.
Discussion
411.

Find the sum `cos e c^(-1)sqrt(10)+cos e c^(-1)sqrt(50)+cos e c^(-1)sqrt(170)++cos e c^(-1)sqrt((n^2+1)(n^2+2n+2))`

Answer» Let `S = cosec^-1sqrt10+cosec^-1sqrt50+cosec^-1sqrt170+...+cosec^-1sqrt((n^2+1)(n^2+2n+2))`
Here, `T_n = cosec^-1sqrt((n^2+1)(n^2+2n+2))`
Let `cosec^-1sqrt((n^2+1)(n^2+2n+2)) = theta`
`=>cosec theta = sqrt((n^2+1)(n^2+2n+2))`
`=>cosec^2theta = (n^2+1)(n^2+2n+2)`
`=>cosec^2theta = (n^2+1)(n^2+1+2n+1)`
`=>1+cot^2theta = (n^2+1)(n^2+1)+2n(n^2+1)+n^2+1`
`=>1+cot^2theta = (n^2+1+n)^2+1`
`=>cot^2theta = (n^2+1+n)^2`
`=>tan theta = 1/(1+n^2+n) = ((n+1) - n)/(1+n(n+1))`
`=>theta = tan^-1(((n+1) - n)/(1+n(n+1)))`
`=>theta = tan^-1(n+1) - tan^-1(n)`
`:. T_n = tan^-1(n+1) - tan^-1(n)`
Now, `S = tan^-1(2) - tan^-1(1) + tan^-1(3) - tan^-1(2)+tan^-1(4) - tan^-1(3)+...+tan^-1(n+1) - tan^-1(n)`
`=>S = tan^-1(n+1) - tan^-1(1)`
`=>S = tan^-1(n+1) - pi/4.`
Discussion
412.

The sum of series `sec^(-1)sqrt(2)+sec^(-1)(sqrt(10))/3+sec^(-1)(sqrt(50))/7++sec^(-1)sqrt(((n^2+1)(n^2-2n+2))/((n^2-n+1)^2))`is`tan^(-1)1`(b) `n``tan^(-1)(n+1)`(d) `tan^(-1)(n-1)`

Answer» Let `S = sec^-1sqrt2+sec^-1(sqrt10/3)+sec^-1(sqrt50/7)+...+sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2)`
Here, `T_n = sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2)`
Let `sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2) = theta`
`=>sec theta =sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2`
`=>sec^2theta = ((n^2+1)(n^2-2n+2))/(n^2-n+1)^2`
`=>sec^2theta = ((n^2+1)(n^2+1-2n+1))/(n^2-n+1)^2`
`=>1+tan^2theta = ((n^2+1)(n^2+1)-2n(n^2+1)+n^2+1)/(n^2-n+1)^2`
`=>1+tan^2theta = ((n^2+1-n)^2+1)/(n^2-n+1)^2`
`=>1+tan^2theta = 1+1/(n^2+1-n)^2`
`=>tan^2theta = 1/(n^2+1-n)^2`
`=>tan theta = 1/(1+n^2-n) = (n - (n-1))/(1+n(n-1))`
`=>theta = tan^-1((n - (n-1))/(1+n(n-1)))`
`=>theta = tan^-1(n) - tan^-1(n-1)`
`:. T_n =tan^-1(n) - tan^-1(n-1)`
Now, `S = tan^-1(1) - tan^-1(0) + tan^-1(2) - tan^-1(1)+tan^-1(3) - tan^-1(2)+...+tan^-1(n) - tan^-1(n-1)`
`=>S = tan^-1(n) - tan^-1(0)`
`=>S = tan^-1(n) - 0`
`=>S = tan^-1(n)`
Discussion
413.

If `f(x) = sin^(-1) (sin ("log"_(2) x))`, then find the value of `f(300)`

Answer» Correct Answer - `3 pi - "log"_(2) 300`
`f(300) = sin^(-1) (sin (log._(2) 300))`
`log_(2) 256 = 8 and log_(2) 512 = 9`
`:. (log_(2) 300) in (8,9)`
`:. Sin^(-1) (sin (log_(2) 300) = sin^(-1) (sin (3pi - log_(2) 300))`
`= 3pi - log_(2) 300`
Discussion
414.

Find the value of `sin(1/4)cos^(-1)((-1)/9)`

Answer» Correct Answer - `(1)/(sqrt6)`
Let `theta = co^(-1). ((-1)/(9))`
`rArr cos theta = (-1)/(9)`
`rArr 2 cos^(2). (theta)/(2) = 1 - (1)/(9)`
`rArr cos. (theta)/(2) = (2)/(3)`
`rArr 1 - 2 sin^(2). (theta)/(4) = (2)/(3)`
`rArr 2 sin^(2). (theta)/(4) = (1)/(3)`
`rArr sin.(theta)/(4) = (1)/(sqrt6)` (as `(pi)/(2) lt theta lt pi rArr (pi)/(8) lt (theta)/(4) lt (pi)/(4)`)
`rArr sin. ((1)/(4) cos^(-1) ((-1)/(9)))) = (1)/(sqrt6)`
Discussion
415.

Let `tan^(-1)y=tan^(-1)x+tan^(-1)((2x)/(1-x^2))`, where `|x|

Answer» We have
`tan^(-1)y=tan^(-1)x+tan^(-1)(2x)/(1-x^(2))` ltMbrgt `rarr tan^(-1)y=tan^(-1)x+2 tan^(-1)x=3 tan^(-1)x=tan^(-1)(3x-x^(3))/(1-3x^(22))`
`rarr y=(3x-x^(3))/(1-3x^(2))`
Discussion
416.

Statement -1: If `xin[-1//sqrt(3),1//sqrt(3)]` then `cot^(-1)((3x-x^(3))/(1-3x^(2)))=cos^(-1)((1-x^(2))/(1+x^(2)))rarrx=sqrt(25-10sqrt(5))/(5)` statement -2: `sin18^(@)=sqrt(5-1)/(4) and cos18^(@)=sqrt(10+2sqrt(5))/4 `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» Clearly statement 2 is true
now `cot^(-1)(3x-x^(3))/(1-3x^(2))=cos^(-1)(1-x^(2))/(1+x^(2))`
`rarr (pi)/(2) tan^(-1)(3x-x^(3))/(1-3x^(2))=cos^(-1)(1-x^(2))/(1+x^(2))`
`rarr (pi)/(2) -3 tan^(-1)x=2 tan^(-1)x`
`rarr tan^(-1)x=(pi)/(10)`
`rarr x=tana(pi)/(10)=(sin 18^(@))/(cso 18^(@))=(sqrt(5)-1)/(sqrt(10+2sqrt(5))`
`rarr x=(sqrt(5)-1sqrt(10-2)sqrt(5))/(sqrt(100-20) rarr x=(sqrt(10-2)sqrt(5)(6-2sqrt(5))/(4sqrt(5))`
`rarr x =sqrt(80-32sqrt(5))/(4sqrt(5))=sqrt(5-2sqrt(5))/sqrt(5)=sqrt(25-10)sqrt(5)/(5)`
so statement -1 is true
Also statemnet -2 is correct explanation for statement -1
Discussion
417.

The sum of series `sec^(-1)sqrt(2)+sec^(-1)(sqrt(10))/3+sec^(-1)(sqrt(50))/7++sec^(-1)sqrt(((n^2+1)(n^2-2n+2))/((n^2-n+1)^2))`is`tan^(-1)1`(b) `n``tan^(-1)(n+1)`(d) `tan^(-1)(n-1)`A. `tan^(-1) 1`B. `tan^(-1) n`C. `tan^(-1) (n + 1)`D. `tan^(-1) (n -1)`

Answer» Correct Answer - B
`T_(n) = "sec"^(-1) sqrt(((n^(2) +1) (n^(2) -2n + 2))/((n^(2) -n +1)^(2)))`
`rArr sec^(2) T_(n) = ((n^(2) + 1) (n^(2) -2n + 2))/((n^(2) -n + 1)^(2))`
`rArr sec^(2) T_(n) = ((n^(2) +1)^(2) + (n^(2) +1) -2n(n^(2) + 1))/((n^(2) - n+ 1)^(2))`
`rArr sec^(2) T_(n) = (1 + (n^(2) + 1 -n)^(2))/((n^(2) -n + 1)^(2))`
`rArr tan T_(n) = (1)/(n^(2) -n + 1)`
`rArr tan T_(n) = (n -(n -1))/(1 + n(n -1)) = tan^(-1) n - tan^(-1) (n -1)`
`:. S = T_(1) + T_(2) + T_(3) +..+ T_(n)`
`:. S = tan^(-1) n`
Discussion
418.

If `-1 le x le 0` then `cos^(-1)(2x^(2)-1)` equalsA. `2 cos^(-1)x`B. `pi-2 cos^(-1)x`C. `2pi-2 cos^(-1)x`D. `-2 cos^(-1)x`

Answer» Correct Answer - D
Discussion
419.

If `sin(sin^(-1)1/5+cos^(-1)x)` =1 then x is equal toA. 1B. 0C. `4//5`D. `1//5`

Answer» Correct Answer - D
Discussion
420.

If `(1)/(2) sin^(-1) [(3 sin 2 theta)/(5 + 4 cos 2 theta)] = tan^(-1) x, " then " x =`A. `tan 3 theta`B. `3 tan theta`C. `(1//3) tan theta`D. `3 cot theta`

Answer» Correct Answer - C
`(1)/(2) sin^(-1).(3 sin 2 theta)/(5 + 4 cos 2 theta) = tan^(-1) x`
`rArr sin^(-1).((6 tan theta)/(1 + tan^(2) theta))/(5 + 4 ((1-tan^(2) theta)/(1 + tan^(2) theta))) = sin^(-1).(2x)/(1 + x^(2))`
`rArr sin^(-1).(6 tan theta)/(9 + tan^(2) theta) = sin^(-1).(2x)/(1 + x^(2))`
`rArr sin^(-1).(2((tan theta)/(3)))/(1 + ((tan theta)/(3))^(2)) = sin^(-1).(2x)/(1 + x^(2))`
`rArr x = (tan theta)/(3)`
Discussion
421.

If `-1/2 le x le 1/2` hence `sin^(-1)(3x-4x^(3))` equalsA. `3 sin^(-1)x`B. `pi-3 sin^(-1)x`C. `-pi - 3 sin^(-1)x`D. none of these

Answer» Correct Answer - D
Discussion
422.

Value of x satisfying ` tan(sec^(- 1)x)=sin(cos^(- 1)(1/(sqrt(5))))`A. `pm sqrt(5)/(3)`B. `pm (3)/sqrt(5)`C. `pm sqrt(3)/(5)`D. `pm 3/5`

Answer» We have
`tan(sec^(-1)x)=sin(cos^(-1)(1)/sqrt(5))`
`rarr sqrt(sec^(2)(sec^(-1)x)-1)=sin(sin^(-1))(2)/sqrt(5)`
`rarr sqrt(x^(-1))=(2)/sqrt(5)rarrx=pm (3)/sqrt(5)`
Discussion
423.

Statement -1: If `xltsqrt(e )`then `cot^(-1){log(e//x^(2))/log(ex^(2))}+cot^(-1){log(ex^(4))/log(e^(2)//x^(2)))}=pi-tan^(-1)3` statement 2:`tan^(-1)(x+y)/(1-xy)=tan^(-1)x+tan^(-1)y if xylt1`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» statement 2 is true (see history )
Now
`cot^(-1){log(e//x^(1))/log(ex^(2))}+cot^(-1){logex^(4))/log(e^(2)//x^(2)))}`
`=cot^(-1){(1-2logx)/(1+2 logx)}+cot^(-1){(1+4logx)/(2-2logx)}`
`=tan^(-1)(1+2 logx)/(1-23logx)+(pi)/(2)tan^(-1)((1/2+2logx)/(1-1/2xx2logx)`
`=tan^(-1)(1+2logx)/(1-2logx)+(pi)/(2)-tan^(-1)((1/2+2logx)/(1-1/2xx2logx))`
`=tan^(-1)(1)+tan^(-1)(2 logx)+(pi)/(2)-tan^(-1)1/2-tan^(-1)(2 logx)`
if `2 log x l 1 and log x lt 1`
`=(3pi)/(4) -tan^(-1)1/2 if x lt sqrt(e )`
`=pi -(tan^(-1)1+tan^(-1)1/2)=pi tan^(-1)3`
So statement 1 is alos true also statement -2 is a correct explanation fro statement-1
Discussion
424.

If `A=tan^(-1) x ,x in R` then the value of sin 2A isA. `(2x)/(1-x^(2))`B. `(2x)/sqrt(1-x^(2))`C. `(2x)/(1+x^(2))`D. `(1-x^(2))/(1+x^(2))`

Answer» Correct Answer - C
Discussion
425.

If `sin^(-1) ((2a)/(1+a^2))+ sin^(-1) ((2b)/(1+b^2)) = 2 tan^(-1)x` then `x=`A. `(a - b)/(1 + ab)`B. `(b)/(1 + ab)`C. `(b)/(1 - ab)`D. `(a + b)/(1 - ab)`

Answer» Correct Answer - D
Since `sin^(-1) ((2x)/(1 + x^(2))) = 2 tan^(-1) x " for " x in (-1, 1)`
`sin^(-1) ((2a)/(1 + a^(2))) + sin^(-1) ((2b)/(1 + b^(2))) = 2 tan^(-1) x`
`rArr 2 tan^(-1) a + 2 tan^(-1) b = 2 tan^(-1) x`
or `tan^(-1) a + 2 tan^(-1) b = 2 tan^(-1) x`
or `tan^(-1) ((a + b)/(1 -ab)) = tan^(-1) x`
or `x = (a + b)/(1 - ab)`
Discussion
426.

If `|a|lt1| b|lt1and|x|lt1` then the solution of `sin^(-1)((2a)/(1+a^(2)))-cos^(-1)((1-b^(2))/(1+b^(2)))=tan^(-1)((2x)/(1-x^(2)))` isA. `(a-b)/(1-ab)`B. `(1+ab)/(a-b)`C. `(ab-1)/(a+b)`D. `(a-b)/(a+ab)`

Answer» We have
`sin^(-1)(2a)/(1+a^(2))-cos^(-1)(1-b^(2))/(1+b^(2))=tan^(-1)(2x)/(1-x^(2))`
`rarr 2tan^(-1)a-2tan^(-1)b=2tan^(-1)x`
`rarr tan^(-1)a-tan^(-1)b=tan^(-1)x`
`rarr tan^(-1)x=tan^(-1)(a-b)/(1+ab)`
`rarr x=(a-bb)/(1+ab)`
Discussion
427.

The value of `sin^(-1)`(sin 10) isA. 10B. `10-3pi`C. `3pi-10`D. none of these

Answer» Correct Answer - C
Discussion
428.

The value of `"tan"(sin^(-1)("cos"(sin^(-1)x)))"tan"(cos^(-1)(sin(cos^(-1)x))),w h e r ex in (0,1),`is equal to0 (b) 1(c) `-1`(d) none of these

Answer» we have
`cos(sin^(-1)x)=sqrt(1-x^(2)) and sin cos^(-1)x=sqrt(1-x^(2))`
`therefore tan{sin^(-1)(cos sin^(-1)x)}tan{cos^(-12)(sin(cos^(-1)x)}`
`=tan{sin^(-1)sqrt(1-x^(2))}tan{cos^(-1)sqrt(1-x^(2))`
`=tan(sin^(-1)sqrt(1-x^(2))tan{(pi)/(2)-sin^(-1)sqrt(1-x^(2))=1`
`=tan(sin^(-1)sqrt(1-x^(2))cot(sin^(-1))sqrt(1-x^(2))=1`
Discussion
429.

`sec^(-1)(sin x)` exist ifA. `x on (-oo,oo)`B. `x in [-1,1]`C. `x=(2n+1)(pi)/(2),x in Z`D. `x=npi , x in z`

Answer» Correct Answer - C
Discussion
430.

`[cot^(-1)x][cos^(-1) x]` =0 where x is non negative and [.] is he greatest integer function then the value of x is / areA. (cos1,1]B. (cos 1, cos 1)C. (cot 1,1]D. none of these

Answer» Correct Answer - C
Discussion
431.

In any triangle ABC `sum (sin^2A+sinA+1)/sinA` is always greater than or equal

Answer» In triangle ABC,
`sum (sin^2A + sin A+1)/sinA = (sin^2A + sin A+1)/sinA+ (sin^2B + sin B+1)/sinB+ (sin^2C + sin C+1)/sinC`
`=sinA+1/sinA+sinB+1/sinB+sinC+1/sinC+3`
As, `sin theta` in any triangle is always greater than `0`.
`:. sin theta + 1/sin theta ge 2`.
`:. sinA+1/sinA+sinB+1/sinB+sinC+1/sinC ge6 ``=>sinA+1/sinA+sinB+1/sinB+sinC+1/sinC+3 ge 9`
Thus,
`sum (sin^2A + sin A+1)/sinA ge 9`.
Discussion
432.

The value of `tan^(-1)1+tan^(-1)2+tan^(-1)3` is

Answer» Correct Answer - C
Discussion
433.

Find the following values : (a) `tan^(1) tan.(13pi)/(5)` (b) `sec^(-1) sec.(13pi)/(3)` (c) `sin^(-1) sin.(33pi)/(5)` (d) `sin^(-1) (sin 8)` (e) `tan^(-1) (tan 10)` (f) `sec^(-1) (sec 9)` (g) `cot^(-1) (cot 6)` (h) `cosec^(-1) (cosec 7)`

Answer» Correct Answer - (a) `-(2pi)/(5)` (b) `(pi)/(3)` (c) `(2pi)/(5)` (d) `3 pi -8` (e) `10 - 3 pi` (f) `9 - 2pi` (g) `6 - pi` (h) `7 - 2pi`
(a) `tan^(-1) tan. (13 pi)/(5) = tan^(-1) tan {3pi + (-(2pi)/(5))}`
`= tan^(-1) tan.(-(2pi)/(5))`
`= -(2pi)/(5)`
(b) `sec^(-1) sec. (13pi)/(3) = sec^(-1) sec.(3pi + (pi)/(3))`
`= sec^(-1) sec. (pi)/(3)`
`= (pi)/(3)`
(c) `sin^(-1) sin.(33 pi)/(5) = sin^(-1) sin. (7 pi - (2pi)/(5))`
`= sin^(-1) sin. (2pi)/(5) = (2pi)/(5)`
(d) `sin^(-1) (sin 8) = sin^(-1) (sin (3pi -8)) = 3 pi -8`
(e) `tan^(-1) (tan 10) = tan^(-1) (tan(10 - 3 pi)) = 10 - 3 pi`
(f) `sec^(-1) (sec 9) = sec^(-1) (sec(9 - 2pi)) = 9 - 2pi`
(g) `cot^(-1) (cot 6) = cot^(-1) (cot(6 - pi)) = 6 - pi`
(h) `cosec^(-1) (cosec7) = cosec^(-1) (cosec(7 - 2 pi)) = 7 - 2pi`
Discussion
434.

The value of `cot(pi/4-2 cot^(-1)3)` isA. 1B. 7C. -1D. none of these

Answer» Correct Answer - B
Discussion
435.

The value of `tan[cos^(-1)(4/5)+tan^(-1)(2/3)]`is`6/(17)`(b) `7/(16)`(c) `(16)/7`(d) none of these

Answer» LHS
`tan[tan^(-1)3/4+tan^(-1)2/3]`
`tan[tan^(-1)((3/4)+(2/3))/(1-(3/4)(2/3))]`
`tan[tan^(-1)(17/6)]`
`=17/6`.
Discussion
436.

If `tan^(-1)a+tan^(-1)b+tan^(-1)c=pi` then prove tjhat `a+b+c=abc`A. a+b+c=abcB. ab+bc+ca=abcC. `(1)/(a)+(1)/(b)+(1)/(c )-(1)/(abc)=0`D. `ab+bc+ca=a+b+c`

Answer» Correct Answer - C
Discussion
437.

The value of `sin^(-1)[cos((33pi)/5)]` isA. `(3pi)/(5)`B. `(7pi)/(5)`C. `(pi)/(10)`D. `(-pi)/(10)`

Answer» Correct Answer - D
Discussion
438.

Find the real values of x for which the function `f(x) = cos^(-1) sqrt(x^(2) + 3 x + 1) + cos^(-1) sqrt(x^(2) + 3x)` is defined

Answer» Correct Answer - `x = 0, -3`
We have `f(x) = cos^(-1) sqrt(x^(2) + 3x + 1) + cos^(-1) sqrt(x^(2) + 3x)`
We must have `0 le x^(2) + 3x + 1 le 1 and 0 le x^(2) + 3x le 1`
`rArr x^(2) + 3x = 0`
`rArr x = 0, -3`
Discussion
439.

Find the value of x for which `sin^(-1) (cos^(-1) x) lt 1 and cos^(-1) (cos^(-1) x) lt 1`

Answer» Correct Answer - `cos sin 1 lt x lt cos cos 1`
We have
`sin^(-1) (cos^(-1) x) lt 1`
`rArr -(pi)/(2) le sin^(-1) (cos^(-1) x) lt 1`
`rArr - 1 le cos^(-1) x lt sin 1`
`rArr 0 le cos^(-1) x lt sin 1`
`rArr cos (sin 1) lt x le 1`..(i)
Also, `cos^(-1) (cos^(-1) x) lt 1`
`rArr 0 le cos^(-1) (cos^(-1) x) lt 1`
`rArr cos 1 lt cos^(-1) x le 1`
`rArr cos 1 le x lt cos (cos 1)`....(ii)
`rArr cos 1 lt cos (sin 1)`
Hence, from (i) and (ii), we have
`cos (sin 1) lt x lt cos (cos 1)`
Discussion
440.

The value of `cot[cos^(-1)(7/25)]` isA. `25/24`B. `25/7`C. `24/25`D. none of these

Answer» Correct Answer - D
Discussion
441.

The value of `cot(c o s e c^(- 1)5/3+tan^(- 1)(2/3))i s`A. `4/17`B. `5/17`C. `6/17`D. `3/17`

Answer» Correct Answer - C
Discussion
442.

The value of `tan[cos^(-1)(4/5)+tan^(-1)(2/3)]`is`6/(17)`(b) `7/(16)`(c) `(16)/7`(d) none of theseA. `6/17`B. `7/16`C. `17/6`D. none of these

Answer» Correct Answer - C
Discussion
443.

The equation `sin^-1x-cos^-1x=cos^-1(sqrt3/2)` hasA. no solutionB. unique solutionC. infinite number of solutionD. none of these

Answer» Correct Answer - B
Discussion
444.

Find the smallest and the largest values of `tan^(-1) ((1 - x)/(1 + x)), 0 le x le 1`A. `0,pi`B. `0,(pi)/(4)`C. `-(pi)/(4),(pi)/(4)`D. `(pi)/(4),(pi)/(2)`

Answer» Correct Answer - B
Discussion
445.

Find the smallest and the largest values of `tan^(-1) ((1 - x)/(1 + x)), 0 le x le 1`

Answer» Correct Answer - `[0, (pi)/(4)]`
`f(x) = tan^(-1).((1 - x)/(1 + x)), 0 le x le 1` ltbr gt Now `(1 - x)/(1 + x) = (2)/(1 + x) - 1`
Given `0 le x le 1`,
`rArr (2)/(1 + x) -1 in [0,1]`
`rArr tan^(-1).((1 - x)/(1 + x)) in [tan^(-1) 0, tan^(-1) 1] " or " [0, (pi)/(4)]`
Discussion
446.

Find all possible values of x and y for which `cos^(-1)sqrtx+cos^(-1)sqrt(1-x)+cos^(-1)sqrt(1-y)=((3pi)/4)`

Answer» `cos^(-1)sqrtx=cos^(-1)sqrt(1-x)=cos^(-1)sqrtx+sin^(-1)sqrtx=pi/2`
The given equation can be rewritten as
`pi/2+cos^(-1)sqrt(1-y)=(3pi)/4`
`rArr cos^(-1)sqrt(1-y)=pi/4`
`sqrt(1-y)=1/sqrt2`
`rArr 1-y=1/2`
`rArry=1-1/2=1/2`
For `cos^(-1)sqrtx` to be defined `-1 le sqrtx le 1`
But `sqrtx` by definition, is non-negative i.e., `sqrtx ge0`
Combining the two `0 le sqrtx le 1i.e., 0 le xle1`
Thus the possible values of x, y are `0 le x le 1, y=1/2`
Discussion
447.

The number of solution of equation `sin^(-1)x+nsin^(-1)(1-x)=(mpi)/2,w h e r en >0,mgeq0,`is3 (b)1 (c) 2(d) None of theseA. 3B. 1C. 2D. none of these

Answer» Correct Answer - D
`sin^(-1) x` is defined if `-1 le x le 1 and sin^(-1) (1- x)` is defined if
`- 1 le 1 -x le 1 rArr 0 le x le 2`
Therefore, `sin^(-1) x + n sin^(-1) (1 -x)` is defined if `0 le x le 1` when `0 le x le 1, " also " 0 le 1 - x le 1`. So,
`0 le sin^(-1) x le (pi)/(2) and 0 le sin^(-1) (1-x) le (pi)/(2)`
`L.H.S. ge 0 and R.H.S. le 0`
Which equally holds if L.H.S. = R.H.S. = 0
But L.H.S. `= 0 " if " sin^(-1) x and n sin^(-1) (1 -x)` are simultaneously zero, which is not possible
Discussion
448.

The value of `sin^(-1) (sin 12) + cos^(-1) (cos 12)` is equal toA. zeroB. `24 - 2pi`C. `4pi -24`D. none of these

Answer» Correct Answer - A
`sin^(-1) (sin 12) + cos^(-1) (cos 12)`
`= sin^(-1) (sin (12 - 4pi)) + cos^(-1) (cos (4 pi - 12))`
`= 12 - 4pi + 4pi - 12 = 0`
Discussion
449.

`sin^(-1)(3/5)+tan^(-1)(1/7)=`A. `(pi)/(4)`B. `(pi)/(2)`C. `cos^(-1)(4/5)`D. `pi`

Answer» Correct Answer - A
Discussion
450.

Prove the following : `sin^(-1)4/5+2 tan^(-1)1/3=pi/2`A. `(pi)/(3)`B. `(pi)/(4)`C. `(pi)/(4)`D. 0

Answer» Correct Answer - C
Discussion