Mathematics color{red} ✎ Problem Solving Techniques for inverse trigonometric function
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### Technique 1 : Straight forward & obvious

sin sin^(-1) x = x

cos cos^(-1) x=x

tan tan^(-1) x=x

sec sec^(-1) x=x

cosec cosec^(-1) x=x

cot cot^(-1) x=x

can write always with no thoughts or thinking

color{red}✎color{red}(" Handle with care")

color{purple}("only when x is in principal branch of the inverse function").

If x is not in principle branch then write sin x , cosx , tanx, cosec x, cot x, sec x in such away that x is in principal branch.

 sin^(-1) sinx= x

 cos^(-1) cosx=x

 tan^(-1) tanx=x

 sec^(-1) secx=x

cosec^(-1) cosec x=x

 cot^(-1) cotx=x

If x is -ve then take -ve sign out

e.g., sin^(-1) sin (-x)

sin^(-1) sin \ (-pi/4) = sin^(-1) \[-(sin \ (pi/4)]

= sin^(-1) \ sin \ pi/4

If x ge 2 pi or x ge 360^o , subtract multiple of 2 pi

=> sin^(-1) sin \ (4 pi)/4 = sin^(-1) sin \ (2 pi + pi/4)

= sin^(-1) sin \ pi/4 = pi/4

cos^(-1) cos \ (13 pi)/4 = cos^(-1) cos \ (4 pi + pi/4)

cos^(-1) cos \ pi/4 = pi/4

=> write sin x etc in equation from such that x  is in principal domain of given inverse function.

sin^(-1) sin \ pi/3= pi/3 as pi/3 in [- \ pi/2 , pi/2]

sin^(-1) sin \ (2 pi)/3 = sin^(-1) sin (pi-pi/3)

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

=> Writing in simplest form :

First add large number of practice Problems.

Q 3116180079

Find the value of  sin^(-1) sin \ pi/4+sin^(-1) \ sin \ (3 pi)/4+sin^(-1) sin \ (5 pi)/4+sin^(-1) \ sin \ (7 pi)/4

Solution:

sin^(-1) sin \ pi/4 = pi/4

sin^(-1) sin \ (5 pi)/4 = sin^(-1) sin \(pi + pi/4)= sin^(-1) \ (-sin \ pi/4)

=-sin^(-1) \ sin \ pi/4 = - pi/4

sin^(-1) \ sin \ (3 pi)/4 = sin^(-1) \ sin \ (pi-pi/4) = sin^(-1) \ sin \ (pi)/4 = pi/4

sin^(-1) \ sin \ (7 pi)/4 = sin ^(-1) \ sin (2 pi - pi/4) = sin ^(-1) (- sin \ pi/4)= - sin^(-1) \ sin \ pi/4 = -pi/4

:. sin^(-1) sin \ pi/4+sin^(-1) \ sin \ (3 pi)/4+sin^(-1) sin \ (5 pi)/4+sin^(-1) \ sin \ (7 pi)/4 = (pi/4-pi/4+pi/4-pi/4) =0
Q 3106280178

Find the value of cos^(-1) cos \ pi/4 +cos^(-1) cos \ (3 pi)/4+cos^(-1) \ cos \ (5 pi)/4+

cos^(-1) \ cos \ (7 pi)/4

Solution:

cos^(-1) cos \ pi/4 = pi/4

cos^(-1) cos \ (3 pi)/4 = cos^(-1) \ cos \ (pi-pi/4) = cos^(-1) \ (- cos \ pi/4)

cos^(-1) \ cos \ (3 pi)/4 = (3 pi)/4  { since (3 pi)/4 in (0,pi)}

 cos^(-1) \ cos \ (5 pi)/4 = cos^(-1) \ cos \ (pi + pi/4)

= cos^(-1) \ (- cos \ pi/4)

= pi - cos^(-1) \ cos \ pi/4 = pi - pi/4 = (3 pi)/4

cos^(-1) \ cos \ (7 pi)/4 = cos^(-1) \ cos \ (2 pi - pi/4)

= cos^(-1) \ cos \ pi/4 = pi/4

:. cos^(-1) cos \ pi/4 +cos^(-1) cos \ (3 pi)/4+cos^(-1) \ cos \ (5 pi)/4+cos^(-1) \ cos \ (7 pi)/4 = (pi/4+ (3pi)/4+ pi/4+ (3 pi)/4) = 2pi
Q 3146480373

Find the value of tan^(-1) tan \ pi/4 +tan^(-1) \ tan \ (3 pi)/4+tan^(-1) \ tan \ (5 pi)/4+tan^(-1) \ tan \ (7 pi)/4 = 0

Solution:

tan^(-1) tan \ pi/4= pi/4 quad { "since" pi/4 in (-pi/2 , pi/2)}

tan^(-1) \ tan \ (3 pi)/4 = tan^(-1) \ tan\ (pi- pi/4) = tan^(-1) \ tan (- tan \ pi/4)

= - tan^(-1) \ tan \ pi/4 = -pi/4

= tan^(-1) \ tan \ (5 pi)/4 = tan^(-1) \ tan \ (pi+ pi/4) = tan^(-1) \ tan \ pi/4 = pi/4

tan^(-1) \ tan \ (7 pi)/4 = tan^(-1) \ tan \ (2 pi - pi/4) = - tan ^(-1) \ tan \ pi/4= -pi/4

:. tan^(-1) tan \ pi/4 +tan^(-1) \ tan \ (3 pi)/4+tan^(-1) \ tan \ (5 pi)/4+tan^(-1) \ tan \ (7 pi)/4= (pi/4-pi/4+pi/4-pi/4)=0
Q 3106480378

Find the value of  cosec^(-1) cosec \ pi/4+cosec^(-1) \ cosec \ (3 pi)/4+cosec^(-1) \ cosec \ (5 pi)/4+cosec^(-1) \ cosec \ (7 pi)/4

Solution:

cosec^(-1) cosec \ pi/4 = pi/4

cosec^(-1) \ cosec \ (3 pi)/4= cosec^(-1) \ cosec \ (pi- pi/4)

= cosec^(-1) \ cosec \ pi/4 = pi/4

cosec^(-1) \ cosec \ (5 pi)/4 = cosec \ cosec \ (pi -pi/4) = cosec^(-1) \ (-cosec \ pi/4)

=-pi/4

cosec^(-1) \ cosec \ (7 pi)/4 = cosec^(-1) \ cosec \ (2 pi -pi/4) = cosec \ (- cosec\ (pi/4))

=-pi/4

:. cosec^(-1) cosec \ pi/4+cosec^(-1) \ cosec \ (3 pi)/4+cosec^(-1) \ cosec \ (5 pi)/4+cosec^(-1) \ cosec \ (7 pi)/4 = (pi/4+ pi/4 - pi/4 -pi/4)= 0
Q 3156580474

Find the value of sec^(-1) \ sec \ (pi)/4 +sec^(-1) \ sec \ (3 pi)/4 +sec^(-1) \ sec \ (5 pi)/4+sec^(-1) \ sec \ (7 pi)/4

Solution:

sec^(-1) \ sec \ (pi/4) = pi/4

sec^(-1) \ sec \ (3 pi)/4 = sec^(-1) \ sec \ (pi-pi/4) = sec^(-1) \ (- sec^(-1) \ pi/4)

=pi - sec^(-1) \ sec \ pi/4

= pi- pi/4 = (3 pi)/4

since (3 pi)/4 in [0,pi]

sec^(-1) \ sec \ ( 3pi)/4 = (3 pi)/4

sec^(-1) \ sec \ (5 pi)/4= sec^(-1) \ sec \ (pi+ pi/4)

= sec^(-1) \ (- sec \ pi/4)

=pi- pi/4= (3 pi)/4

sec^(-1) \ sec \ (7 pi)/4 = sec^(-1) \ sec \ ( 2pi - pi/4)

= sec^(-1) \ sec \ (pi/4)

= pi/4

:. sec^(-1) \ sec \ (pi)/4 +sec^(-1) \ sec \ (3 pi)/4 +sec^(-1) \ sec \ (5 pi)/4+sec^(-1) \ sec \ (7 pi)/4 = (pi/4+ (3pi)/4 + (3 pi)/4 + (3 pi)/4 + pi/4) = (11pi)/4
Q 3176580476

Find the value of cot^(-1) cot \ pi/4 + cot^(-1) \ cot \ (3 pi)/4+cot^(-1) \ cot \ (5 pi)/4 + cot^(-1) \ cot \ (7 pi)/4

Solution:

cot^(-1) cot \ pi/4 =pi/4

cot^(-1) \ cot \ (3 pi)/4 = cot^(-1) \ cot \ (pi- pi/4) = cot^(-1) \ (- cot ^(-1) \ pi/4)

pi- pi/4 = (3 pi)/4

since, (3 pi)/4 in (0,pi)

cot^(-1) \ cot \ (3 pi)/4 = (3 pi)/4

cot^(-1) \ cot \ (5 pi)/4 = cot^(-1) \ cot \ (pi+ pi/4) = cot^(-1) \ cot \ pi/4 = pi/4

cot^(-1) \ cot \ (7 pi)/4 = cot^(-1) \ cot \ ( 2pi- pi/4)= cot^(-1) \ (-cot \ pi/4) = pi- pi/4

=(3 pi)/4

:. cot^(-1) cot \ pi/4 + cot^(-1) \ cot \ (3 pi)/4+cot^(-1) \ cot \ (5 pi)/4 + cot^(-1) \ cot \ (7 pi)/4 =(pi/4 + (3 pi)/4 + pi/4 + (3 pi)/4)= (8 pi)/4= 2pi
Q 3137401382

sin^(-1) \ sin \ (37 pi)/4 + cos^(-1) \ cos \ (-13 pi)/4

Solution:

= (sin^(-1) \ sin (8 pi + pi + \ pi/4)) + (pi - cos^(-1) \ cos \ (13 pi)/4) quad { " having multiple of " 2pi}

= sin^(-1) \ sin \ (pi + pi/4) + pi - cos^(-1) \ cos \ (2 pi + (5 pi)/4)

= sin^(-1) \ (-sin \ pi/4) + pi - cos^(-1) \ cos \ (pi + pi/4)

=- sin^(-1) \ sin \ pi/4 + pi - cos^(-1) \ (- cos \ pi/4)

=- pi/4 + pi - [ pi - cos^(-1) \ cos \ pi/4]

= -pi/4 + pi - pi + pi/4

=0

### Technique 2 : By substitution

=> Key in what to substitute:

sin^(-1) (x) =theta

x= sin theta

cos theta= sqrt (1-x^2)

theta= cos^(-1) sqrt(1-x^2)

theta= cos^(-1) sqrt(1-x^2)

tan theta= x/(sqrt(1-x^2))

=> theta=tan^(-1) \ x/(sqrt(1- x^2))

sin^(-1) (tan^(-1) \ 1/7)

tan^(-1) \ 1/7 = theta

tan \ theta = 1/7

=> sin theta= 1/(sqrt 50)

=> theta= sin^(-1) \ 1/(sqrt (50))

=> sin \ (tan ^(-1) \ 1/7) = sin \ sin^(-1) \ 1/(sqrt (50))

= 1/(sqrt (50))
Q 3106191078

Show that sin^(-1) \ (2x sqrt(1-x^2))= 2 sin^(-1) x- 1/(sqrt2) le x le 1/ (sqrt 2)

Solution:

Let x= sin theta => theta = sin^(-1) x

sin^(-1) \ (2x sqrt( 1-x^2)) = sin^(-1) \ (2 sin theta sqrt(1-sin^2 theta))

= sin^(-1) \ (2 sin theta cos theta)= sin^(-1) \ (sin 2 theta)

= 2 theta = 2 sin^(-1)\ x
Q 3156291174

Prove the following: cos[tan^(-1)\ {sin(cot^(-1) \ x)}]= sqrt((1+x^2)/(2+x^2))

Solution:

Let cot^(-1) x = theta => x = cot theta => sin theta = 1/ (sqrt(1+x^2))

:. cos [ tan^(-1) { sin (cot^(-1) \ x )}]= cos[ tan^(-1) \ { sin (sin ^(-1) \ 1/ (sqrt(1+x^2)))}]

= cos [ tan^(-1) \ 1/ (sqrt(1+x^2))]= 1/(sec [ tan ^(-1) \ 1/(sqrt(1+x^2))])

= 1/(sqrt(1+ tan^2 (tan^(-1) \ 1/(sqrt(1+x^2)))

= 1/(sqrt(1+ 1/(1+x^2))

= (sqrt(1+x^2))/(sqrt(2+x^2))
Q 2645180063

Prove that tan^(-1) sqrt (x) = 1/2 cos^(-1) ( (1-x)/(1+x) ) x in [0,1]
Class 12 Exercise 2.mis Q.No. 9
Solution:

Put x = tan^2 theta , theta = tan^(-1) sqrt (x)

R.H.S. = 1/2 cos^(-1) ( (1-x)/(1+x))

=1/2 cos^(-1) ((1-tan^2 theta)/( 1+tan^2 theta)) =1/2 cos^(-1) (cos 2 theta)

=1/2 xx 2 theta = theta

= tan^(-1) sqrt (x)

 = L.H.S.
Q 2665180065

tan^(-1) ( ( sqrt (1+x) - sqrt (1-x) )/( sqrt (1+x) + sqrt (1-x) ) ) = pi/4 -1/2 cos^(-1) x,

-1/2 cos^(-1) x, - 1/sqrt(2) le x le 1

(Hint: Put x = cos 2 theta ]
Class 12 Exercise 2.mis Q.No. 11
Solution:

tan^(-1) ( (sqrt (1+x) +sqrt (1-x) )/( sqrt (1+x) - sqrt (1-x) ) )

{Put x = cos 2 theta => theta = 1/2 cos^(-1) x  }

= tan^(-1) ( ( sqrt (1+cos 2 theta ) +sqrt (1- cos 2 theta ) )/( sqrt (1+ cos2 theta ) - sqrt ( 1- cos 2 theta) ) )

= tan^(-1) ( ( sqrt (2 cos^2 theta ) + sqrt ( 2 sin^2 theta ) )/( sqrt (2 cos^2 theta ) - sqrt (2 sin^2 theta) ) )

 [ :. 1+ cos 2 theta = 2 cos^2 theta  and 1-cos 2 theta= 2 sin^2 theta ]

= tan^(-1) ( (cos theta + sin theta )/( cos theta - sin theta ) )

= tan^(-1) tan ( pi/4 + theta) = pi/4 + theta = pi/4 +1/2 cos^(-1) x
Q 3116178979

Show that sin [ cot^(-1) { cos ( tan^(-1) x ) } ] = sqrt( (x^2 + 1)/( x^2 + 2) ) .
Mock
Solution:

We have

L.H.S. = sin [cot^(-1) (cos (tan^(- 1) x ) } ]

= sin [cot^(-1) {cos A} ] . where tan^(- 1) x = A or x = tan A

 = sin [ cot^(-1) { 1/(sec A)} ]

 = sin [ cot^(-1) { 1/sqrt( 1 + tan A )} ]

 = sin [ cot^(-1) 1/sqrt(1 + x^2) ]

 = sin B  where  cot^(-1) \ 1/sqrt(1 + x^2) = B  or  cot B = 1/sqrt(1 + x^2)

 = 1/(cosec B)

 = 1/sqrt( 1 + cot^2 B)

 = 1/sqrt(1 + 1/(1 + x^2))

 = 1/sqrt( (x^2 + 2)/(1 + x^2))

 = sqrt( (x^2 + 2)/(1 + x^2)) = RHS
Q 3176180076

Solve for  x : 3 sin^(-1) ((2x)/(1 + x^2 )) - 4 cos^(-1) ((1 -x^2)/(1 + x^2) ) + 2 tan^(-1) ((2x)/(1 + x^2 )) = pi/3 .

Mock
Solution:

We have

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

Put x = tan theta  in (1), then

 3 sin^(-1) ( ( 2 tan theta)/( 1 + tan^2 theta)) - 4 cos^(-1) ( (1 - tan^2 theta)/( 1 + tan^2 theta)) + 2 tan^(-1) ( ( 2 tan theta)/( 1 - tan^2 theta)) = pi/3

=> 3 sin^(-1) (sin 2 theta ) - 4 cos^(- 1) (cos 2 theta ) + 2 tan^(- 1) (tan 2 theta ) = pi/3

=> 3(2 theta) - 4 (2 theta ) + 2 ( 2 theta ) = pi/3

 => 10 theta - 8 theta = pi/3

=> 2 theta = pi/3

 => theta = pi/3

=> tan^(-1) x = pi/6

 => x = tan pi/6

 => x = 1/sqrt3

### Technique 3 : By simplification

Q 3103291148

Find the values of x which satisfy the equation sin^(-1) x + sin^(-1) (1 - x) = cos^(-1) x.
CBSE-12-All-India 2016
Solution:

sin^(-1) (.x) + sin^(-1) (1-x) = cos^(-1 ) x

=> sin^(-1) [ x sqrt ( 1- (1-x)^2 ) + (1-x) sqrt (1-x^2) ]

= cos^(-1) x

=> x sqrt (2x -x^2) + (1-x) sqrt (1-x^2 )

= sin (cos^(-1) x) = sqrt (1-x^2)

=> x sqrt ( 2x - x^2) + sqrt (1-x^2) (1-x-1) = 0

=> x [ sqrt (2x -x^2) - sqrt (1-x^2) ] =0

=> x= 0  or 2 x - x^2 =1- x^2

=> x= 1/2 => x = 0 , 1/2 .
Q 3153467344

Write the value of tan^( -1) [ 2 sin ( 2 cos^(-1) sqrt3/2 ) ]

Solution:

Consider tan^(-1) [2 sin (2 cos^(-1) sqrt3/2 ) ] = tan^(-1) [2 sin (2 * pi/6) ] = tan^(-1) [2 sin pi/3]

= tan^(-1) (2 * sqrt3/2) = tan^(-1) (sqrt 3) = pi/3 .
Q 3136391272

If tan^(-1) \ (1/(1+ 1*2)) + tan^(-1) \ (1/(1+ 2*3))+ .... + tan^(-1) (1/(1+n (n+1)))= tan^(-1) theta then find the value of theta.

Solution:

Given , tan^(-1) \ ( 1/ (1+ 1*2)) + tan^(-1) \ (1/ (1+ 2*3))+ ........ + tan^(-1) \ (1/ (1+ n (n+1)) = tan^(-1)\ theta

=> tan^(-1) \ ((2-1)/(1+ 2*1)) + tan^(-1) \ ((3-2)/(1+ 3*2))+ ....+ tan^(-1) (((n+1) -n)/(1+ n(n+1))) = tan^(-1) theta

=> tan^(-1) (2) - tan^(-1) (1) + tan^(-1) (3) - tan^(-1) (2)+........+ tan^(-1) (n+1) - tan^(-1) (n) = tan^(-1) theta

[ tan^(-1) ((x-y)/(1+ x.y)) = tan^(-1) \ x - tan^(-1) \ y]

=> tan^(-1)\ (n+1)- tan^(-1) \ (1) = tan^(-1) \ theta

=> tan^(-1) ((n+1-1)/(1+(n+1) * 1)) = tan^(-1) theta

=> tan^(-1) (n/(1+ n+1))= tan^(-1) theta

theta= n/(2+n)
Q 3106391278

Solve for x, tan^(-1) 3x + tan^(-1) 2x = pi/4

Solution:

Given equation is tan^(-1) 3x + tan^(-1) 2x= pi/4

=> tan^(-1) \ ((3x + 2x)/(1-3x xx 2x)) =pi/4

=> tan^(-1) ((5x )/(1-6x^2)) = pi/4 => (5x)/(1- 6x^2) = tan \ pi/4

[tan^(-1) x = theta => x = tan theta]

=> (5x)/(1-6x^2) =1 => 5x =1 -6x^2

=> 6x^2 + 6x - x - 1 = 0

=>6x (x + 1)-1 (x + 1) = 0

=>(6x-1)(x+1)=0

=> 6x -1 = 0 or x + 1 = 0

x = 1//6 or x = -1

But x = -1 does not satisfy the equation, as LHS becomes negative. So, x =1/6 is the only solution of the given equation.
Q 3173491346

Prove the following: 2 sin^(-1) (3/5) - tan^(-1) (17/31) = pi/4.

Solution:

Consider  2 sin^(-1) (3/5) - tan^(-1) (17/31)

= 2 tan^(-1 ) (3/4) - tan^(-1) (17/31)

= tan^(-1) ( (6/4)/(1- 9/16) ) - tan^(-1) (17/31)

= tan^(-1 ) (24/7) - tan^(-1 ) (17/31)

= tan^(-1) ( 24/7 - 17/31)/( 1+ 24/7 xx 17/31) = tan^(-1) ( (744 -119)/( 217 + 408) )

= tan^(-1) (625/625) = tan^(-1) (1) = pi/4 .