Note that :
(a) `I^(st)` quadrant is common to all the inverse functions.
(b) `3^(rd)` quadrant is not used in inverse functions.
(c) 4th quadrant is used in the CLOCKWISE DIRECTION i.e. `-pi/2 le y le 0`

(1) `y = sin^-1 x, |x| le 1 , y in [ -pi/2, pi/2 ]`

`y= sin^-1 x`



Highlights :


(i) `sin^-1 x` is bounded in `[-pi/2 , pi/2]`

(ii) `sin^-1x` is an odd function. (symmetric about origin)
(iii) `sin^-1x` is an increasing function in its domain.
(iv) Maximum value of `sin^-1x=pi/2` , occurs at `x = 1` and minimum value of `sin^-1 x =-pi/2`, occurs at `x

=-1`

(v) `sin^-1x` is an aperiodic function.

(2) `y = cos^-1 x , |x| le 1, y in [0, pi]`

Highlights : -

(i) `cos^-1x` is bounded in `[0, pi]`.
(ii) `cos^-1x` is a neither odd nor even function.
(iii) `cos^-1x` is a decreasing function in its domain.
(iv) Maximum value of `cos^- 1x= pi`, occurs at `x = - 1` and minimum value of `cos^-1 x = 0`, occurs at `x = 1`.
(v) `cos^-1x` is an aperiodic function.

(3) `Y = tan^-1 x, x in R, y in (-pi/2 , pi/2)`

Highlights : -

(i) `tan^-1 x` is bounded in `(-pi/2, pi/2)`

(ii) `tan^-1 x` is an odd function. (symmetric about origin)

(iii) `tan^-1 x` is an increasing function in its domain.

(iv) `tan^-1 x` is an aperiodic function.

`y = cot^ -1 x, x in R, y in (0 , pi)`

Highlights : -

(i) `cot^-1x` is bounded in `(0, pi)`.
(ii) `cot^-1x` is a neither odd nor even function.
(iii) `cot^-1x` is a decreasing function in its domain.
(iv) `cot^-1x` is an aperiodic function.

(5) `y= cosec^-1 x ,|x| ge 1` . y in [ -pi/2,0) cup (0, pi/2]`


Highlights : -

(i) `cosec^-1x` is bounded in[-pi/2, pi/2]`
(il) `cosec^-1 x` is an odd function. (symmetric about origin)
(iii) Maximum value of `cosec^-1x =pi/2`, occurs at `x = 1` and minimum value of `cosec^-1 x=-pi/2` , occurs at `x =-1`.

(iv) `cosec^-1x` is a decreasing function
(v) `cosec^-1x` is an aperiodic function.

(6) `y = sec^-1x, |x| ge 1`, `y in [ 0, pi/2)cup(pi/2, pi]`

Highlights : -
(i) `sec^-1x` is bounded in `[0, pi]`.
(ii) `sec^-1x` is a neither odd nor even function.
(iii) Maximum value of `sec^-1x = pi`, occurs at `x =-1` and minimum value of `sec^-1 x = 0`, occurs at `x = 1`.

(iv) `sec^-1x` is an increasing function.
(v) `sec^-1x` is an aperiodic function.


Note:

(a) `tan^-1(x)` and `cot^-1(x)` are continuous and monotonic on `R` does not implies that their range is `R`
(b) If `f(x)` is continuous and has a range `R` does not implies it is monotonic. e.g. `y = x^3 - 3x`.

Property -1 :

(i) `sin (sin^- 1 x) = x , - 1 le x le 1` (ii) `cos(cos^-1 x)=x , - 1 le x le1`
(iii) `tan (tan^- 1 x) = x , x in R` (iv) `cot (cot^-1 x) = x , x in R`
(v) `cosec(cosec^- 1 x) = x , |x | ge 1` (vi) `sec (sec^- 1 x) = x , | x | ge 1`

Property-2 :

( 1 ) `cosec^-1x = sin^-1 1/x ; |x| ge 1`

(2) `sin^-1x = cosec^-1 1/x , |x| le 1, x ne 0`

(3) `sec^-1 x = cos^-1 1/x ; |x| ge 1`

(4) `cos^-1 x = sec^-1 1/x , |x| le 1, x ne 0`

(5) `cot^-1x = tan^-1 1/x ; x >0`

`= pi+ tan ^-1 1/x ; x < 0`

Note:

(i) `cosec^-1x` and `sin^-1 1/x` are identical function.
(ii) `sin^-1 x` and `cosec^-1 1/x` are not identical because domain of `sin^-1 x` and `cosec^-1 1/x` is not equal.
(iii) `sec^-1x` and `cos^-1 1/x` are identical function.
(iv) `cos^-1 x` and `sec^-1 1/x` are not identical because domain of `cos^-1 x ` and `sec^-1 1/x` is not equal.

Property-3 :
(i) `sin^-1 (- x) = - sin^-1x , - 1 le x le 1`
(ii) `tan^- 1 (- x) = - tan^- 1 x , x in R`
(iii) `cos^- 1 (- x) = pi - cos^- 1 x , - 1 le x le 1`
(iv) `cot^-1 (- x) = pi - cot^-1 x , x in R`
(v) `cosec^-1 (- x) = - cosec^- 1 x , | x | ge 1`
(vi) `sec^- 1 (- x) = pi - sec^- 1 x, | x | ge 1`

Property-4 :
(i) `sin^- 1 x + cos^- 1 x= pi/2, - 1 le x le 1` (ii) `tan^-1 x + cot^-1 x = pi/2 , x in R`

(iii) `cosec^- 1 x + sec^- 1 x = pi/2 , |x| ge1`

Property-5 :

(1) `tan^-1 x + tan^-1 y` `= {tt( ( tan^-1((x+y)/(1-xy) ) , text(if) xy < 1) ,(pi+tan^-1((x+y)/(1-xy) ) , text(if) x > 0, y>0 text(and) xy > 1) ,(-pi+ tan^-1 ((x+y)/(1-xy))) )`

Proof:
Let `tan^-1x= A` and `tan^-1 y= B` , where `A,B in ( -pi/2, pi/2)`.


Now, `tan(A+B) = (tanA+tanB)/(1-tanAtanB) =(x+y)/(1-xy)`

`=> tan^-1 ((x+y)/(1-xy)) = tan^-1 tan (A+B)`

`= tan^-1 tan alpha ` where `alpha in (-pi, pi)`
`tan^-1 ((x+y)/(1-xy))= tan^-1 (tan alpha)`
`= {tt((alpha+pi, -pi < alpha < -pi/2),(alpha , -pi/2 le alpha le pi/2),(alpha-pi, pi/2 < alpha < pi) ) ={tt((tan^-1x+tan^-1y + pi, -pi < tan^-1x+tan^-1y< -pi/2),(tan^-1x+tan^-1y, -pi/2 le tan^-1x +tan^-1y le pi/2),(tan^-1x+tan^-1y -pi, pi/2 < tan^-1 x +tan^-1y < pi)) `

Case-I :

`- pi < tan^-1 x+tan^-1 y pi/2 ` `=> x < 0 , y < 0`

Also , `tan^-1 x < -pi/2 - tan^-1 y`

`=> tan^-1 x < - (pi/2 -tan^-1 (-y))` `=> x < -(-1/y) => x < 1/y => xy>1`

Case-ll:
`pi/2 < tan^-1 x +tan^-1 y < pi => x,y >0`

Also, `tan^-1x > pi/2 -tan^-1 y => tan^-1x > tan^(-1) 1/y=>x > 1/y => xy >1`


Case-Ill :

`-pi/2 le tan^-1 x +tan^-1 y le pi/2 => xy <1`

(2) `x > 0` and `y > 0, tan^-1 x - tan^- 1y = tan^-1 (x-y)/(1+xy)` (with no other restriction)

`x > 0` and `y > 0, tan^-1 x - tan^- 1y = tan^-1 (x-y)/(1+xy)` (with no other restriction)
(Remember)
(i) `tan^(-1) 1 + tan^(-1) 2 + tan^(-1)3 = pi`
(ii) `tan^(-1)1 +tan^(-1) 1/2+tan^(-1) 1/3 =pi/2`

(iii) ` (tan^(-1)1 +tan^(-1)2+tan^(-1)3) /(cot^(-1 ) 1 +cot^(-1) 2+cot^(-1) 3)=2`

Sol .(i) `tan^(-1 ) 1 +tan^(-1) 2+tan^(-1)3 = tan^(-1) + (pi+tan^-1 ((2+3)/(1-2*3)))`
`=tan^(-1) 1 + (pi +tan^(-1) (-1))`
`pi/4 +pi - pi/4 =pi`


(ii) `tan^(-1) 1 +tan^(-1) (1/2) +tan^(-1) 1 +tan^-1 ((1/2+1/3)/(1-1/2*1/3)) `

`= tan^(-1) 1 + tan^-1 (5/5)= tan^(-1 ) 1 +tan^(-1) 1= pi/4+pi/4=pi/2`

(iii) `(tan^(-1) 1 +tan^(-1) 2+ tan^(-1) 3)/(cot^(-1) 1 +cot^(1-) 2 + cot^(-1) 3) = (tan^(-1)1 +tan^(-1)2 +tan^(-1)3)/(tan^(-1) 1 + tan^(-1) 1/2+tan^(-1) 1/3)= (pi)/(pi/2) =2`

Property-6 :

(I) `sin^(-1) x +sin^(-1) y` `= [tt( (sin^(-1) (root x (1-y^2)+root y (1-x^2)) text(if) x ge 0; y ge 0 text(and) x^2 +y^2 le 1) , (pi-sin^(-1) (root x (1-y^2)+ root y (1-x^2) ) text(if) x ge 0; y ge 0 text(and) x^2+y^2 >1))`

note that `x^2 + y^2 le 1` `=> 0 ge sin^(- 1)x + sin^(-1)y le pi/2`

Let `sin^(-1) x= alpha` and `sin^(-1) y = beta `; `alpha beta in [0,pi/2]`

now `x^2+y^2 le 1`

`sin^2alpha+sin^2beta le 1` `=> sin^2alpha .e cos^2beta`

`sin^2 alpha le sin^2 (pi/2 -beta) => alpha le pi/2 - beta => alpha+beta le pi/2`

`0 le sin^(-1) x + sin^(-1) y le pi/2 ]`

and `x^2+y^2 >1 => pi/2 < sin^(-1) x+sin^(-1) y < pi`

This formula should normally be used in establishing the identities.

e.g. find whether `sin^(-1) (3/5) +sin^(-1) (12/5) `it becomes simple.

(II)
we have `sin^(-1) x -sin^(-1)y=sin^(-1) (root x (1-y^2) -root y (1-x^2)) ,x > 0; y > 0`

and `cos^(-1) x pm cos^(-1) y =cos^(-1) (xy pm sqrt (1-x^2) sqrt (1-y^2)),x> 0 ,y >0, x < y`

Property -7 :
`tan^(-1)x + tan^(-1)y + tan^(-1)z = tan^(-1) [ (x+y+z -xyz)/(1- (xy+yz+zx))]`

where `x > 0, y > 0, z > 0` Solution and `xy + yz + zx < 1` and `xy < 1 , yz < 1 , zx < 1`

Solution

`tan^(-1)x + tan^(- 1)y + tan^(-1)z = tan^(-1) ((x+y)/(1-xy))+ tan^(-1) z`

`= tan^(-1) ((((x+y)/(1-xy))+z)/(1-((x+y)/(1-xy))z)) = tan^(-1) (x+y+z-xyz)/(1-xy-(x+y)z)= tan^(-1) (x+y+z-xyz)/(1- (xy +yz +zx))`

(I) `2tan^(-1) (tan (pi/4 -alpha)tan (beta/2) ) =cos^(-1) ( (sin2alpha +cos beta)/(1+sin2 alpha cos beta))`

Proof: Let `x= tanc(pi/4 -alpha)tan (beta/2)`

`x= (1-tan alpha)/(1+tan alpha) tan(beta/2) => x = ( (cos alpha-sin alpha)/(cos alpha+sin alpha)) ((sin beta/2)/(cos beta/2))`

`x^2 = ((1-sin2alpha)/(1+sin2 alpha)) ((sin^2 beta/2)/(cos^2 beta/2)) `

`x^2 = ((1-si 2 alpha) (1-cos beta))/((1+sin2 alpha )(1=cos beta)) =(1-sin2 alpha-cosbeta+sin2alpha*cosbeta)/(1=sin2 alpha +cos beta +sin2alpha*cos beta)`

`(x^2-1)/(x^2+1) =-(sin 1 alpha +cos beta)/(1+ sin2 alpha * cos beta)`
(By applying componendo and dividendo)

` (1-x^2)/(1+x^2) =(sin2 alpha+cos beta)/(1+sin2 alpha* cos beta)`
We know that

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

`=> 2tan^(-1) (tan (pi/4 -alpha) tan beta/2) =cos^(-1) ((sin2alpha+ cos beta)/(1+ sin 2 alpha cos beta))`


(II) `tan^(-1)x = 2 tan^(-1) [ cosec (tan^(-1)x)-tan (cot^(-1) x) ]` `( x ne 0)`
Sol. R.H.S. `2tan^(-1) [cosec(tan^(-1) x) - tan(cot^(-1) x) ]`

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

`= 2 tan^(-1) [ cosec (tan^(-1) x) -cot tan^(-1) x]`

Let `tan^(-1) x = theta`

`=> 2 tan^(-1) [cosec theta -cot 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)`