Mathematics `color{red} ✎` Problem Solving Techniques for inverse trigonometric function

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Class 12 Chapter 2 - INVERSE TRIGONOMETRIC FUNCTIONS

`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.

`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))`

`=>` 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` .

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