Mathematics Must Do Problems of Inverse Trigonometric Functions For NDA

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Must Do Problems of Inverse Trigonometric Functions For NDA

Must Do Problems of Inverse Trigonometric Functions

Must Do Problems

Q 2605891768

If `x+1/x = 2` , the principal value of `sin^(-1) x` is

(A)

`pi/4`

(B)

`pi/2`

(C)

`pi`

(D)

`(3pi)/2`

Solution:

`x+1/x = 2`

`=> ( sqrtx - 1/sqrtx)^2 = 0`

`=> (x-1)/sqrtx = 0`

`therefore x = 1`

So, the principal value of `sin^(-1) x` is `pi/2`

Correct Answer is `=>` (B) `pi/2`

Q 2601580428

Write the principal value of `cos^(-1) ( 1/2) - 2 sin^(-1) ( -1/2)`.
CBSE-12th 2012

Solution:

Principal value of `cos^(-1) (1/2) = pi/3`

Principal value of `sin^(-1) ((-1)/2) = (-pi)/3`

Hence principal value of ` cos^(-1) (1/2) - sin^(-1) ((-1)/2)`

` = pi/3 - 3( (-pi)/6)`

` = (2 pi)/3`

Q 2130645512

Find the principal values of `sec^(-1) (2/sqrt3)`
Class 12 Exercise 2.1 Q.No. 7

Solution:

Let `sec^(-1) (2/sqrt3) =y`

`=> sec y =2/sqrt3 = sec pi/6`

The range of principal value of `sec^(-1)` is

`[0, pi] - (pi/2)`

`:.` The principal value of `sec^(-1) (2/sqrt3)` is `pi/6`

Q 2100545418

Find the principal values of `cosec^(-1) (2)`
Class 12 Exercise 2.1 Q.No. 3

Solution:

Let `cosec^(-1) (2) = theta`

`=> cosec theta =2`

i.e., `cosec theta = cosec (pi/6)`

So, principal value of `cosec^(-1)(2)` is `pi/6` as

principal value of `y = cosec^(-1) x` is

`[-pi/2, pi/2] - {0}`

Q 2569691515

The value of `cos { tan^(-1) ( tan (15 pi)/4 ) }` is
BCECE Mains 2015

(A)

` 1/sqrt2`

(B)

` - 1/sqrt2`

(C)

`1`

(D)

None of these

Solution:

We have, `cos { tan^(-1) ( tan (15 pi)/4 ) }`

` = cos { tan^(-1) ( tan (4 pi - pi/4 )) }`

` = cos { tan^(-1) tan (- pi//4)}`

`= cos (-pi//4)`

`= cos pi//4 = 1//sqrt2`

Correct Answer is `=>` (A) ` 1/sqrt2`

Q 2446201173

The value of `2 tan^(- 1) (cosec tan^(- 1) x - tan cot^(- 1) x)`
is
UPSEE 2014

(A)

`tan^(-1) x`

(B)

`tan x`

(C)

`cot x`

(D)

`cosec^(- 1)x`

Solution:

`2 tan^(- 1) (cosec tan^(- 1) x- tan cot^(- 1) x)`

`= 2 tan^(- 1) [ (cosec { cosec^(-1) sqrt(1-x^2)/x})-(tan {tan^(- 1) (1/x)})]`

` = 2 tan^(-1) [sqrt(1-x^2)/x - 1/x] = 2 tan^(-1) [(sqrt(1-x^2) - 1)/x]`

`= 2 tan^(- 1) [ (sec theta - 1)/(tan theta) ]` (put `x = tan theta` )

`=2 tan^(-1) [(1 - cos theta )/(sin theta)]`

`= 2 tan^(-1) [( 2 sin^2 \ theta/2)/(2 sin \ theta/2 . cos \ theta/2)]`

`= 2 tan^(-1) (tan \ theta/2) = 2 · theta/2 = theta = tan^(-1) x`

Correct Answer is `=>` (A) `tan^(-1) x`

Q 2416801770

If `a le sin^(-1) x + cos^(- 1) x + tan^( -1) x le b`, then
UPSEE 2010

(A)

`a=0, b = pi`

(B)

`a=0, b= pi/2`

(C)

`a=pi/2, b=pi`

(D)

None of these

Solution:

`pi/2 =sin^(-1) +cos^(-1) x=pi/2`...........(i)

and `-pi/2 < tan^(-1) x < pi/2`...............(ii)

Adding eqs. (i) and (ii), we get

`pi/2-pi/2 < sin^(-1) + cos^(-1) x +tan^(-1) x < pi/2 + pi/2`

`=> 0 < sin^(-1) x +cos^(-1) x +tan^(-1) x < pi`

Comparing it with given condition, we get

`a=0, b =pi`

Correct Answer is `=>` (A) `a=0, b = pi`

Q 2417591489

If `cot (cos^(-1) x) = sec ( tan^(-1) \ a/sqrt(b^2 - a^2))` then `x`
is equal to
UPSEE 2009

(A)

` b/sqrt(2b^2 -a^2)`

(B)

` a/sqrt(2b^2 -a^2)`

(C)

`sqrt(2b^2 -a^2)/a`

(D)

`sqrt(2b^2 -a^2)/b`

Solution:

Given, `cot (cos^(-1) x) = sec ( tan^(-1) a/sqrt(b^2 - a^2))`

`:. cot( cot^(-1) ((x)/sqrt(1-x^2))) = sec ( sec^(-1) ((b/sqrt(b^2 - a^2)))`

` => x/sqrt(1 - x^2) = b/sqrt(b^2 - a^2)`

` => x^2 (b^2 - a^2) = b^2 - b^2x^2`

` => x^2 (2b^2 - a^2) = b^2`

` => x = b/sqrt(2b^2 - a^2)`

Correct Answer is `=>` (A) ` b/sqrt(2b^2 -a^2)`

Q 2486734677

If `sin^(-1) x + sin^(-1) y = (2pi)/ 3` and `cos^(-1) x- cos^(-1) y =pi/3` Then, `(x, y)` is equal to
UPSEE 2010

(A)

`(0,1)`

(B)

`(1//2 , 1)`

(C)

`(1, 1//2)`

(D)

`(sqrt 3//2 ,1)`

Solution:

We have,

`sin^(-1) x+ sin^(-1) y= (2 pi)/3`

`(pi/2 -sin^(-1) x)-(pi/2 -sin^(-1) y) =pi/3`

`=> -sin^(-1) x + sin^(-1) y =pi/3`

On adding Eqs. (i) and (ii),

`sin^( -1) y = pi/2 => y = 1`

On subtracting Eq. (i) from Eq. (ii),

`sin^(-1) x= pi/6 => x= 1`

`(x, y) = ( 1/2, 1)`

Correct Answer is `=>` (B) `(1//2 , 1)`

Q 2610812719

If `tan^(-1)\ \ a/x + tan^(-1)\ \ b/x = pi//2`, then `x` is equal to
BCECE Mains 2015

(A)

` sqrt(ab)`

(B)

` sqrt(2ab)`

(C)

`2ab`

(D)

`ab`

Solution:

We have,

` tan^(-1) \ \ a/x + tan^(-1) \ \ b/x = pi/2`

` => tan^(-1) ( (a/x + b/x)/(1 - (ab)/x^2) ) = pi/2`

` => 1 - (ab)/x^2 = 0`

` => x = sqrt(ab)`

Correct Answer is `=>` (A) ` sqrt(ab)`

Q 1510623510

If `sin^-1 (3/x) + sin^-1(4/x) = pi/2` then `x` is equal to
BITSAT 2008

(A)

(B)

(C)

(D)

Solution:

Given that

`sin^-1 (3/x) + sin^-1(4/x) = pi/2`

`sin^-1 (3/x) = pi/2 - sin^-1(4/x)`

`sin^-1(3/x) = cos^-1(4/x)`

`sin^-1(3/x) =sin^-1(sqrt(x^2-16)/x)`

`3/x = sqrt(x^2-16)/x`

`9 = x^2-16`

`x^2=25`

`x= +-5`

`x=5`

`( –5` is not satisfied the given equation`)`

Correct Answer is `=>` (B) 5

Q 2437791682

Number of solutions of the equation
`tan^(-1) (1/(2x + 1)) + tan^(-1) (1/(4x + 1)) = tan^(-1) (2/x^2)` is
UPSEE 2009

(A)

(B)

(C)

(D)

Solution:

`tan^(-1) a+ tan^(-1) b = tan^(-1) (( a+ b)/(1 - ab))`

When, `a, b > 0` and `ab < 1`

Given ,

`tan^(-1) (1/(2x + 1)) + tan^(-1) (1/(4x + 1)) = tan^(-1) (2/x^2)`

` tan^(-1) ( ( 1/(2x + 1) + 1/(4x +1) )/(1 - 1/(2x + 1) xx 1/(4x +1))) = tan^(-1) (2/x^2)`

` => tan^(-1) ((6x + 2)/(8x^2 + 6x))= tan^(-1) (2/x^2)`

` => ( 3x + 1)/(4x + 3) = 2/x`

` => 3x^2 -7x - 6 = 0`

` => (3x + 2) (x - 3) = 0`

`=> x = 3 , - 2/3`

Sometime we have seen that the solving values

of `x` does not satisfy the given condition, so we reject

those values.

Correct Answer is `=>` (B) 2

Q 2446201173

The value of `2 tan^(- 1) (cosec tan^(- 1) x - tan cot^(- 1) x)`
is
UPSEE 2014

(A)

`tan^(-1) x`

(B)

`tan x`

(C)

`cot x`

(D)

`cosec^(- 1)x`

Correct Answer is `=>` (A) `tan^(-1) x`

Set - 2

Q 2655378264

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

Solution:

Let `cot^(-1) ((-1)/sqrt 3) =y` Then,

`cot y= (-1)/ sqrt 3=- cot (pi/3)= cot (pi- pi/3) = cot ((2 pi)/3)`

We know that the range of principal value branch of `cot^(–1)` is `(0, π)` and `cot ((2pi)/3) =(-1)/sqrt 3` Hence, principal value of `cot^(–1) ((-1)/sqrt 3) ` is `(2 pi)/3`

Q 2641156023

Write the principal value of ` tan^(-1) (1) + cos ^(-1) ( - 1/2)`.
CBSE-12th 2013

Solution:

Let ` tan^(-1 ) (1) = y`

` => tan y = 1 = tan ( pi/4) => y = pi/4`

` => tan^(-1 ) (1) = pi/4`

` cos^(-1) (-1/2) = z`

` => cos z = -1/2 = - cos \ pi/3 = cos \ pi/3 = cos ( pi - pi/3) = cos ( (2pi)/3 )`

` => z = (2 pi)/3 = cos^(-1) ( - 1/2) = (2 pi)/3`

` :. tan^(-1) + cos^(-1) ( -1/2) = pi/4 + (2pi)/3 = (11 pi)/(12)`

Q 2686634577

The principal value of `sin^(-1) ( sin ((2 pi)/3) )` is

1986

(A)

` - (2 pi)/3`

(B)

` (2 pi)/3`

(C)

` pi/3`

(D)

` (5 pi)/3`

Solution:

`sin^(-1) ( sin ((2 pi)/3) ) = sin^(-1) [ sin ( pi - pi/3) ]`

` = sin^(-1) ( sin (pi/3) ) = pi/3`

Correct Answer is `=>` (C) ` pi/3`

Q 2685491367

The principal value of ` cos^(-1) ( cos (2pi)/3) + sin^(-1) ( sin (2pi)/3)` is

(A)

`pi`

(B)

`pi/2`

(C)

`pi/3`

(D)

`(4pi)/3`

Solution:

`∵ cos^(-1) ( cos ( (2pi)/3)) + sin^(-1) ( sin ( (2pi)/3))`

` = (2pi)/3 + ( pi - (2pi)/3 ) = pi`

Correct Answer is `=>` (A) `pi`

Q 2417591489

If `cot (cos^(-1) x) = sec ( tan^(-1) \ a/sqrt(b^2 - a^2))` then `x`
is equal to
UPSEE 2009

(A)

` b/sqrt(2b^2 -a^2)`

(B)

` a/sqrt(2b^2 -a^2)`

(C)

`sqrt(2b^2 -a^2)/a`

(D)

`sqrt(2b^2 -a^2)/b`

Correct Answer is `=>` (A) ` b/sqrt(2b^2 -a^2)`

Q 2466701675

The value of `sin [pi/2 -sin^(-1) (- sqrt 3/2)]` is
UPSEE 2010

(A)

`1/2`

(B)

`-1/2`

(C)

`1`

(D)

`-1`

Solution:

We have `sin [ pi/2 -sin^(-1) (-sqrt 3/2)]`

`=sin [pi/2 +sin^(-1) (sqrt 3/2)]`

`=[pi/2 + pi/3]=sin[(5 pi)/6]`

`=cos(pi/3) =1/2 .`

Correct Answer is `=>` (A) `1/2`

Q 1831378222

`sin [ pi/3- sin^(-1) (-1/2)]` is equal to
Class 12 Exercise 2.2 Q.No. 20

(A)

`1/2`

(B)

`1/3`

(C)

`1/4`

(D)

`1`

Solution:

`sin^(-1) (-1/2) = sin^(-1) sin (-pi/6) = -pi/6`

`:.` `sin [pi/3 - sin^(-1) (-1/2)] = sin [pi/3- (-pi/6)]`

`= sin ( pi/3+ pi/6) = sin pi/2 =1`

Correct Answer is `=>` (D) `1`

Q 2655480364

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

Solution:

Let `cot^(-1) ((-1)/(sqrt 3)) =y` , Then

`cot y =(-1)/(sqrt 3) =-cot (pi/3)= cot (pi- pi/3) = cot ((2 pi)/3)`

We know that the range of principal value branch of `cot^(–1)` is `(0, π)` and `cot ((2 pi)/3)=(-1)/(sqrt 3)` Hence, principal value of `cot^(–1) ((-1)/(sqrt 3))` is `(2 pi)/3`

Alternatively :

Using property

`cot^-1 (–x) = π – cot^-1x`

`cot^(-1) ((-1)/(sqrt 3)) = π – cot^-1((1)/(sqrt 3))`

`π –π/3 = (2π)/3`

Q 2486734677

If `sin^(-1) x + sin^(-1) y = (2pi)/ 3` and `cos^(-1) x- cos^(-1) y =pi/3` Then, `(x, y)` is equal to
UPSEE 2010

(A)

`(0,1)`

(B)

`(1//2 , 1)`

(C)

`(1, 1//2)`

(D)

`(sqrt 3//2 ,1)`

Correct Answer is `=>` (B) `(1//2 , 1)`

Q 2500834718

If `x in ( (3pi)/2 , 2pi),` then the value of the expression `sin^(-1) [cos {cos^(-1) ( cos x)} + sin^(-1) (sin x)]`, is
BCECE Stage 1 2013

(A)

` - pi/2`

(B)

` pi/2`

(C)

`0`

(D)

`pi`

Solution:

`∵ x in ( (3pi)/2 , 2 pi)`

Now, `cos^(-1) (cos x) = 2pi - x`

and `sin^(-1) (sin x) = x - 2pi`

`:. cos^(-1) (cos x) + sin^(-1) (sin x) = 0`

Therefore,

`sin^(-1) [cos { cos^(-1) (cos x) + sin^(-1) (sin x)}]`

`= sin^(-1) {cos (0)} = sin^(-1) (1) = pi/2`

Correct Answer is `=>` (B) ` pi/2`

Q 2437791682

Number of solutions of the equation
`tan^(-1) (1/(2x + 1)) + tan^(-1) (1/(4x + 1)) = tan^(-1) (2/x^2)` is
UPSEE 2009

(A)

(B)

(C)

(D)

Correct Answer is `=>` (B) 2

Q 2681723627

If `tan^(-1) x + tan^(-1) y = pi/4` ,` xy < 1` , then write the value of `x + y + xy`.

CBSE-12th 2014

Solution:

Given that `tan^(-1) x + tan^(-1) y = pi/4` and `xy < 1`.

We need to find the value of `x + y + xy`.

`tan^(-1) + tan^(-1) y = pi/4`

`=> tan^(-1) ( ( x + y) /(1 - xy) ) = pi/4`

`=> tan^(-1) [ tan^(-1) ( ( x + y) /(1 - xy) ) ] = tan ( pi/4)`

` => ( x + y) /(1 - xy) = 1`

` => x + y = 1 - xy`

` => x + y + xy = 1`

Q 2433278142

The solution of equation `sin^(-1) (1-x/2)-2 sin^(-1) \ x/2 =pi/2` is
UPSEE 2011

(A)

`x=0`

(B)

`x=1/2`

(C)

`x=0` and `1/2`

(D)

None of these

Solution:

Given, `sin^(-1) (1-x/2)-2 sin^(-1) / / x/2 =pi/2`

Put `x=0, LHS sin^(-1)(1)-2sin^(-1)(0)`

`=pi/2 = RHS`

Put `x =1/2, LHS = sin^( -1) (3/4) -2 sin^( -1)(1/4)`

`=sin^(-1) (3/4) -sin^(-1) (2 xx 1/4 sqrt (1-1/16))`

`=sin^(-1) (3/4)-sin^(-1) (sqrt 15 /8)`

`=sin^(-1) [3/4 xx 7/8 -sqrt 15/8 xx sqrt 7/4]`

`=sin^(-1) [21/32-sqrt 105/32]`

Hence, the required solution is `x = 0`.

Correct Answer is `=>` (A) `x=0`

Q 2433278142

The solution of equation `sin^(-1) (1-x/2)-2 sin^(-1) \ x/2 =pi/2` is
UPSEE 2011

(A)

`x=0`

(B)

`x=1/2`

(C)

`x=0` and `1/2`

(D)

None of these

Correct Answer is `=>` (A) `x=0`