Mathematics Advance Concepts of Inverse Trigonometric Fuction of JEE part-II :
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Advance Concepts of Inverse Trigonometric Fuction of JEE part-II :

Formula of `tan^(-1) x + tan^(-1) y`

(1) `tan^(-1) x +tan^(-1) y`

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



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



(3) `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` and `xy+yz+zx < 1` and `xy < 1 , yz < 1 , zx < 1`)


Proof of (`I`): Let `tan^(-1) x = A` and `tan^(-1)y = B` where `A , B ∈ ( -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) tanalpha` where `alpha ∈ (-pi , pi)`

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

` = { tt (( alpha+pi , -pi < alpha < -pi/2) , (alpha , -pi/2 le alpha le pi/2) , (alpha-pi , pi/2 < alpha < pi )) = { tt ((tan^(-1) x+tan^(-1)y+pi , -pi < tan^(-1) x+tan^(-1)y < -pi/2) , (tan^(-1)x+tan^(-1)y , -pi/2 le tan^(-1) x+tan^(-1)y le pi/2) , (tan^(-1) x+tan^(-1)y - pi , pi/2 < tan^(-1)x+tan^(-1)y < 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 - II : `pi/2 < tan^(-1)x +tan^(-1) y < pi => x , y > 0`

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


Case- III : `-pi/2 le tan^(-1)x +tan^(-1)y le pi/2 => xy < 1`




Q 2645491363

The value of `tan^(-1) sqrt((a(a+b+c))/(bc))+ tan^(-1) sqrt((b(a+b+c))/(ca))+ tan^(-1) sqrt((c(a+b+c))/(ab))` is equal to

(This question may have multiple correct answers)

(A) `pi/4`
(B) `pi/2`
(C) `pi`
(D) `0`
Solution:

let `(a+b+c)/(abc)=t^2`

`:. tan^(-1) sqrt((a(a+b+c))/(bc))+ tan^(-1) sqrt((b(a+b+c))/(ca))+ tan^(-1) sqrt((c(a+b+c))/(ab))`

`=tan^(-1) at+tan^(-1) bt +tan^(-1) ct`

`=tan^(-1) {(at+bt+ct-at * bt * ct)/(1-t^2(ab+bc +ca))}`

`=tan^(-1)[ ((a+b+c)t -(a+b+c) t)/(1-t^2(ab+bc+ca))]`

`=tan^(-1) (0)`

`=n pi , n in I`

`:. 0 , pi, 2 pi,............`
Correct Answer is `=>` (C)

Formula of `sin^(-1)A + sin^(-1)B`

(I) `sin^(-1) x+sin^(-1) y `

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

(II) `sin^(-1) x - sin^(-1)y = sin^(-1) { x sqrt(1-y^2) - y sqrt(1-x^2)} , x > 0 ; y > 0`

(III) `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`


Proof of I :

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

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

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

`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) \ \ 12/13` is acute I obtuse will be unduly difficult using the above


However if we convert it into `tan^(-1) \ \ 3/4 +tan^(-1) \ \ 12/5` it becomes simple.
Q 1361701625

Solve the equation `sin^-1x + sin^-1 2x =pi/3`

(A)

`1/2 sqrt(3/7)`

(B)

0

(C)

`1/2 sqrt(7/3)`

(D)

Non of theses

Solution:

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

`sin^-1 2x = sin^-1``sqrt3/2 - sin^-1 x = sin^-1[sqrt3/2 sqrt(1-x^2) -x sqrt(1-3/4)]`

`=> 2x = sqrt3/2 sqrt(1-x^2) -x/2`

`=>((5x)/2)^2 =3/4(1-x^2) => 28x^2 =3`

`=>x = sqrt(3/28) =1/2 sqrt(3/7) quad quad quad quad quadquad ( :. x= -1/2 sqrt(3/7) text ( make L.HS of 1 negative) )`
Correct Answer is `=>` (A) `1/2 sqrt(3/7)`

SIMPLIFICATION & TRANSFORMATION OF INVERSE FUNCTIONS BY ELEMENTRY SUBSTITUTION AND THEIR GRAPHS :

1 . `sin^(-1) ((2x)/(1+x^2)) = [ tt (( 2 tan^(-1) x , -1 le x le 1) ,(pi-2 tan^(-1) x , if x ge 1),(-pi- 2 tan^(-1) x , x le -1))`













2. `cos ^(-1) ((1-x^2)/(1+x^2)) = [ tt (( 2 tan^(-1) x , x ge 0),( -2 tan^(-1) x , x < 0))`












3. `tan^(-1) ((2x)/(1-x^2)) = [ tt (( pi+ 2 tan^(-1) x , x < -1),( 2 tan^(-1) x , -1 < x < 1),(2 tan^(-1) x- pi , x > 0))`

Q 2685391267

If `2 tan^(-1) x + sin^(-1) ((2x)/(1 + x^2))` is independent of `x`, then

(A)

`x in [1, oo)`

(B)

`x in [-1,1]`

(C)

`x in (- oo , -1 ]`

(D)

None of these

Solution:

Let `x = tan theta`

Then, `2 tan^(-1) x + sin^(-1) ((2x)/(1 +x^2))`

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

` = 2 theta + sin^(-1) ( sin 2 theta)`

If ` - pi/2 <= 2 theta <= pi/2`

Then, ` 2 tan^(-1) x + sin^(-1) ((2x)/(1 + x^2)) = 2 theta + 2 theta `

` = 4 theta `

` = 4 tan^(-1) x`

Which is not independent of `x`

and if `- pi/2 <= pi - 2 theta <= pi/2`

Then, ` 2 tan^(-1) x + sin^(-1) ((2x)/(1 + x^2))`

` = 2 theta + sin^(-1) sin (pi - 2 theta)`

`= 2 theta + pi - 2 theta`

`= pi =` independent of `x`

` :. theta in [ pi/4 , (3pi)/4 ]` principal Value of

` theta in ( - pi/2 , pi/2)`

`:. theta in [ pi/4 , pi/2 )`

Hence, `x in [1, oo)`
Correct Answer is `=>` (A) `x in [1, oo)`

Additional Formula

1. `sin^(-1) (3x-4x^3)`

` = [ tt(( -(pi + 3 sin^(-1) x) , if , -sqrt3/2 le x le -1//2) , ( 3 sin^(-1) x , if , -1//2 le x le 1//2) , (pi- 3 sin ^(-1) x , if , 1//2 le x le sqrt3/2))`








2. ` cos ^(-1) (4x^3- 3x)`

`= [ tt (( 3 cos^(-1) x-2pi , if , -1 le x le -1//2) , (2 pi - 3 cos^(-1) x , if , -1//2 le x le 1//2) , ( 3 cos ^(-1) x , if , 1//2 le x le 1))`







3 . `tan^(-1) (3x-x^3)/(1-3x^2)`

` = [ tt (( 3 tan^(-1) x , if , -1/sqrt 3 < x < 1/sqrt 3), (-pi+ 3 tan^(-1) x , if , x > 1/sqrt 3), (pi+ 3 tan^(-1) x , if , x < -1/sqrt 3))`
 
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