Mathematics Properties of Inverse Trigonometric Functions

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Properties of Inverse Trigonometric Functions

`sin (sin^-1 x) = x, x ∈ [– 1, 1]` and `sin^-1 (sin x) = x, x ∈ [ - pi/2, pi/2]`

`cos (cos^-1 x) = x, x ∈ [– 1, 1]` and `cos ^-1 (cos x) = x, x ∈ [ 0,pi]`

`tan (tan^-1 x) = x, x ∈ R` and `tan^-1 (tan x) = x, x ∈ ( - pi/2, pi/2)`

`cosec (cosec^-1 x) = x, |x| ge 1` and `cosec^-1 (cosec x) = x, x ∈ [ - pi/2, pi/2] , xne0`

`sec (sec^-1 x) = x, |x| ge 1` and `sec^-1 (sec x) = x, x ∈ [ 0, pi], x ne pi/2`

`cot (cot^-1 x) = x, x ∈ R` and `cot^-1 (cot x) = x, x ∈ ( 0,pi)`

Q 1871567426

Find the value of each of the expresion

`sin^(-1) (sin ((2 pi)/3))`
Class 12 Exercise 2.2 Q.No. 16

Solution:

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

So answer is not correct.

So we can write ` sin^(-1) [sin (pi - pi/3)]`

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

Q 2615191960

Find the value of `sin(sin^(-1) (pi/4)) `

Solution:

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

because `pi/4 in [-1,1]`

Property 1 :

• `sin ^-1 \ \1/x = cosec^-1 x , x >= 1 ` or ` x <= -1`

• `cos^-1 \ \ 1/x = sec^-1 x, x ≥ 1` or `x ≤ – 1`

• `tan ^-1 \ \ 1/x = cot^-1 x , x > 0`

To prove the first result, we put `cosec^-1 x = y`, i.e., x = cosec y

Therefore `1/x = sin y`

Hence `sin^-1 \ \1/x = y`

or ` sin ^-1 \ \1/x = cosec^-1 x`

Similarly, we can prove the other parts.

Property 2 :

(i) `sin^-1 (–x) = – sin^-1 x, x ∈ [– 1, 1]`

(ii) `tan^-1 (–x) = – tan^-1 x, x ∈ R`

(iii) `cosec^-1 (–x) = – cosec^-1 x, | x | ≥ 1`

Let `sin^-1 (–x) = y`, i.e., `–x = sin y` so that `x = – sin y`, i.e., `x = sin (–y)`.

Hence `sin^-1 x = – y = – sin^-1 (–x)`

Therefore `sin^-1 (–x) = – sin^-1 x`

Similarly, we can prove the other parts.

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`

Property 3 :

(i) `cos^-1 (–x) = π – cos^-1 x, x ∈ [– 1, 1]`

(ii) `sec^-1 (–x) = π – sec^-1 x, | x | ≥ 1`

(iii) `cot^-1 (–x) = π – cot^-1 x, x ∈ R`

Let `cos^-1 (–x) = y` i.e., `– x = cos y` so that `x = – cos y = cos (π – y)`

Therefore `cos^-1 x = π – y = π – cos^-1 (–x)`

Hence `cos^-1 (–x) = π – cos^-1 x`

Similarly, we can prove the other parts.

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`

Property 4 :

(i) `sin^-1 x + cos^-1 x = pi/2, x ∈ [– 1, 1]`

(ii) `tan^-1 x + cot^-1 x = π/2, x ∈ R`

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

Let `sin^-1 x = y`. Then `x = sin y = cos ( pi/2 - y)`

Therefore `cos^-1 x = pi/2 - y = pi/2 - sin ^-1 x`

Hence `sin^-1 x + cos^-1 x = pi/2`

Similarly, we can prove the other parts.

Q 2615280160

`sin^(-1) (1 - x)- 2 sin^(-1 ) x = pi/2`, then `x` is equal to
Class 12 Exercise 2.mis Q.No. 16

(A)

`0, 1/2`

(B)

`1,1/2`

(C)

`0`

(D)

`1/2`

Solution:

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

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

or `sin^(-1) (1-x) -2 sin^(-1) x = sin^(-1) (1-x) + cos^(-1)`

`(1-x) => -2 sin^(-1) x = cos^(-1) (1-x)`

Let `sin^(-1) x = alpha :. sin alpha = x`

`:. -2 sin^(-1) x = -2 alpha = cos^(-1) (1-x)`

or `cos 2 alpha =1 -x :. cos (- theta) = cos theta`

`:. 1-2 sin^2 alpha = 1-x`

Putting `sin alpha = x => 1-2x^2 =1-x`

or `2x^2 -x= 0 ` `x (2x -1) =0 :. x=0, 1/2`

But `x= 1/2` does not satisfy the equation

`:. x= 0`

`:.` Option (c) is correct.

Correct Answer is `=>` (C) `0`

Property 5

(i) `tan^-1x + tan^-1 y = tan^-1 \ \( x +y)/( 1 - xy) , xy < 1`

(ii) `tan^-1x – tan^-1 y = tan^-1 \ \( x - y)/( 1 + xy), xy > -1`

(iii) `2tan^-1x = tan^-1\ \ (2x ) /( 1- x^2) , | x | < 1`

Let `tan^-1 x = θ` and `tan^-1 y = phi`. Then `x = tan θ, y = tan phi`

Now `tan ( theta + pi) = ( tan theta + tan phi )/( 1- tan theta tan phi ) = ( x + y) /( 1- xy)`

This gives ` theta + phi = tan ^-1 \ \ ( x+y) /(1 -xy)`

Hence ` tan ^-1 a + tan ^-1 y = tan^-1 \ \ ( x +y) /(1 -xy)`

In the above result, if we replace y by – y, we get the second result and by replacing y by x, we get the third result.

Q 1841567423

If `tan^(-1) ((x-1)/(x-2)) + tan^(-1) ((x+1) /(x+2)) =pi/4`, then find the value of `x`.
Class 12 Exercise 2.2 Q.No. 15

Solution:

L.H.S. `tan^(-1) ((x-1)/(x-2)) +tan^(-1) ((x+1)/(x+2))`

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

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

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

`= tan^(-1) [ (2x^2 -4) / -3]`

R.H.S `= pi/4 = tan^(-1) (1)`

`:.` `(2x^2 -4)/-3 =1` or `2x^2 -4 =-3`

or `2x^2 = 4 - 3 =1 => x^2 =1/2 => x = pm 1/sqrt2`

Property 6 :

(i) `2tan^-1 x = sin^-1 \ \( 2x) /( 1 + x^2) , | x| <=1`

(ii) `2tan^-1 x = cos^-1 \ \( 1- x^2 ) /( 1 +x^2 ) , x >= 0`

(iii) `2 tan^-1 x = tan^-1 \ \ (2 x) /(1 -x^2 ) . -1 < x < 1`

Let `tan^-1 x = y`, then `x = tan y`. Now

`sin^-1 \ \ ( 2 x) /( 1 + x^2) = sin^-1 \ \ ( 2 tan y) /( 1 + tan^2 y)`

`= sin^-1 (sin 2y) = 2y = 2tan^-1 x`

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

(iii) Can be worked out similarly.

We now consider some examples.

Examples

Q 1821145921

Prove the following

`3 sin^(-1) x = sin^(-1) (3x- 4x^3); x in [-1/2,1/2]`
Class 12 Exercise 2.2 Q.No. 1

Solution:

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

`:.` `sin theta =x, sin 3 theta = 3 sin theta -4 sin^(3) theta`

`:` `sin 3 theta = 3x - 4x^3`

`3 theta = sin^(-1) (3x- 4x^3)`

or `3 sin^(-1) x = sin^(-1) (3x-4x^3), x in [-1/2, 1/2]`

Q 1811856720

Write the function in the simplest form

`tan^(-1) (x/ sqrt (a^(2) - x^2))`
Class 12 Exercise 2.2 Q.No. 9

Solution:

Let `x = a sin theta => x/a = sin theta`

`=> theta = sin^(-1) (x/a)`

`:.` `tan^(-1) ( x/ (sqrt (a^2 -x^2))) = tan^(-1) ( (a sin theta)/(a^2 - a^2 sin^2 theta) )`

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

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

`= tan^(-1) ( (a sin theta) /(a cos theta) ) = tan^(-1) (tan theta)`

`= theta = sin^(-1)( x/a)`.

Q 1811567420

If `sin (sin^(-1) (1/5)+cos^(-1) x) =1` then find the value of `x`.
Class 12 Exercise 2.2 Q.No. 14

Solution:

`sin (sin^(-1) (1/5) = cos^(-1) x) =sin pi/2`

`sin^(-1) (1/5)+ cos^(-1) x =pi/2`

or `(sin^(-1) (1/5) + cos^(-1) (1/5)) + (- cos^(-1) (1/5)+ cos^(-1) x) =pi/2`

or `pi/2 -cos^(-1) (1/5)+ cos^(-1) x =pi/2`

or `cos^(-1) (1/5)-cos^(-1) x =0`

or `cos^(-1) x = cos^(-1) 1/5 => x =1/5`.

Q 2615180069

`sin (tan^(-1) x), |x| < 1` is equal to

Class 12 Exercise 2.mis Q.No. 15

(A)

`x/(sqrt (1-x^2) )`

(B)

`1/(sqrt (1-x^2) )`

(C)

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

(D)

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

Solution:

Let `tan^(-1) x = alpha :. tan alpha =x, sin alpha = x/(sqrt (1+x^2) )`

or `alpha = sin^(-1) ( x/( sqrt (1+x^2) ) )`

Now, `sin tan^(-1) x = sin alpha`

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

Option (d) is correct.

Correct Answer is `=>` (D) `x/(sqrt (1+x^2) )`