Revision: Section A >> Inverse Trigonometric Functions Mathematics ISC (Commerce) Class 12 CISCE

The inverse trigonometric functions are the inverse forms of trigonometric functions after suitable domain restriction. They are written as:

• \[\sin^{-1} x\]

• \[\cos^{-1} x\]

• \[\tan^{-1} x\]

• \[\cot^{-1} x\]

• \[\sec^{-1} x\]

• \[\csc^{-1} x\]

Important:

• \[\sin^{-1} x\]

does not mean 1/sin⁡x. It means the angle whose sine is x.

The value returned by an inverse trigonometric function is called its principal value. It is the unique angle chosen from the standard restricted interval for that function.

Direct Identities

• sin⁻¹(sin θ) = θ, if −π/2 ≤ θ ≤ π/2
• cos⁻¹(cos θ) = θ, if 0 ≤ θ ≤ π
• tan⁻¹(tan θ) = θ, if −π/2 < θ < π/2

Inverse Identities

• sin(sin⁻¹x) = x, if −1 ≤ x ≤ 1
• cos(cos⁻¹x) = x, if −1 ≤ x ≤ 1
• tan(tan⁻¹x) = x, for all real x

Other Important Ones

• sec⁻¹(sec θ) = θ, if 0 ≤ θ ≤ π, θ ≠ π/2
• cosec⁻¹(cosec θ) = θ, if −π/2 ≤ θ ≤ π/2, θ ≠ 0
• cot⁻¹(cot θ) = θ, if 0 < θ < π

(A) Direct identities

• sin(sin⁻¹x) = x, |x| ≤ 1

• cos(cos⁻¹x) = x, |x| ≤ 1

• tan(tan⁻¹x) = x, x ∈ ℝ

• cot(cot⁻¹x) = x, x ∈ ℝ

• sec(sec⁻¹x) = x, |x| ≥ 1

• cosec(cosec⁻¹x) = x, |x| ≥ 1

(B) Inverse of trigonometric expressions

Valid ONLY in principal value range:

• sin⁻¹(sin θ) = θ, θ ∈ [−π/2, π/2]

• cos⁻¹(cos θ) = θ, θ ∈ [0, π]

• tan⁻¹(tan θ) = θ, θ ∈ (−π/2, π/2)

• cosec⁻¹ x = sin⁻¹1 (1/x), x ∈ R − (−1, 1)

• sec⁻¹ x = cos⁻¹ (1/x), x ∈ R − (−1, 1)

• cot⁻¹ x = tan⁻¹ (1/x), for x > 0

• cot⁻¹ x = π + tan⁻¹ (1/x), for x < 0
  [only if tan⁻¹ is taken in (−π/2, π/2)]

(A) tan⁻¹ formulas

• tan⁻¹x + tan⁻¹y = tan⁻¹( (x+y)/(1−xy) ), if xy < 1

• tan⁻¹x + tan⁻¹y = π + tan⁻¹( (x+y)/(1−xy) ), if x,y > 0 & xy > 1

• tan⁻¹x − tan⁻¹y = tan⁻¹( (x−y)/(1+xy) ) if x,y> -1

(B) sin⁻¹ formulas  

• sin⁻¹x + sin⁻¹y
  = sin⁻¹( x√(1−y²) + y√(1−x²) )

• sin⁻¹x − sin⁻¹y
  = sin⁻¹( x√(1−y²) − y√(1−x²) )

(C) cos⁻¹ formulas

• cos⁻¹x + cos⁻¹y
  = cos⁻¹( xy − √(1−x²)√(1−y²) )

• cos⁻¹x − cos⁻¹y
  = cos⁻¹( xy + √(1−x²)√(1−y²) )

• sin⁻¹x = tan⁻¹( x / √(1−x²) ), |x| < 1

• cos⁻¹x = tan⁻¹( √(1−x²) / x ), x > 0

• tan⁻¹x = sin⁻¹( x / √(1+x²) ), ∀ x

• tan⁻¹x = cos⁻¹( 1 / √(1+x²) ), x ≥ 0

• sin⁻¹x + cos⁻¹x = π/2, |x| ≤1

• tan⁻¹x + cot⁻¹x = π/2, x ∈ ℝ

• sec⁻¹x + cosec⁻¹x = π/2, |x| ≥ 1

sin⁻¹ x = cos⁻¹ (√(1 − x²)), 0 ≤ x ≤ 1

cos⁻¹ x = sin⁻¹ (√(1 − x²)), 0 ≤ x ≤ 1

cos(sin⁻¹ x) = sin(cos⁻¹ x) = √(1 − x²), |x| ≤ 1

2 sin⁻¹ x = sin⁻¹ (2x√(1 − x²))

3 sin⁻¹ x = sin⁻¹ (3x − 4x³)

2 cos⁻¹ x = cos⁻¹ (2x² − 1)

3 cos⁻¹ x = cos⁻¹ (4x³  − 3x)

3 tan⁻¹ x = tan⁻¹ ((3x − x³ )/(1 − 3x²))

• sin⁻¹(−x) = −sin⁻¹x, |x| ≤1

• tan⁻¹(−x) = −tan⁻¹x, x ∈ ℝ

• cosec⁻¹(−x) = −cosec⁻¹x, |x| ≥ 1

• cos⁻¹(−x) = π − cos⁻¹x, |x| ≤ 1

• sec⁻¹(−x) = π − sec⁻¹x, |x| ≥ 1

• cot⁻¹(−x) = π − cot⁻¹x, x ∈ ℝ

Let x = sin θ.

Then, sin⁻¹ x = θ.

We have,

R.H.S = sin⁻¹ (3x – 4x³) = sin⁻¹ (3 sin θ – 4 sin³θ)

= sin⁻¹ (sin 3θ) = sin⁻¹ (3 sin θ – 4 sin³θ)

= 3θ = sin⁻¹ (3 sin θ – 4 sin³θ)

= 3 sin⁻¹ x = sin⁻¹ (3 sin θ – 4 sin³θ)

R.H.S = L.H.S

Let x = cos θ.

Then, cos^–1x = θ.

We have,

R.H.S. = cos^–1(4x³ – 3x)

= cos^–1(4 cos³ θ – 3 cos θ)

= cos^–1(cos 3θ)

= 3θ

= 3cos^–1x

= L.H.S.

Let `sin^-1  8/17 = x`.

Then, `sin x = 8/17`

 ⇒ `cos x = sqrt(1 - (8/17)^2`

= `sqrt((225)/(289)`

= `15/17`

∴ `tan x= 8/15` ⇒ `x = tan^-1  8/15`

∴ `sin^-1  8/17 = tan^-1  8/15`   ...(1)

Now, let `sin^-1  3/5 = y`.

Then, `sin y = 3/5`

⇒ `cos y = sqrt(1 - (3/5)^2`

= `sqrt(16/25)`

= `4/5`

∴ `tan y = 3/5` ⇒ `y = tan^-1  3/4`

∴ `sin^-1  3/5 = tan^-1  3/4`   ...(2)

Now, we ahve:

L.H.S. = `sin^-1  8/17 + sin^-1  3/5`

= `tan^-1  8/15 + tan^-1  3/4`   ...[Using (1) and (2)]

= `tan^-1  (8/15 + 3/4)/(1 - 8/15 xx 3/4)`

= `tan^-1 ((32 + 45)/(60 - 24))`   ...`[tan^-1x + tan^-1y = tan^-1  (x + y)/(1 - xy)]`

= `tan^-1  77/36` = R.H.S.

Let `cos^(-1)  4/5 = x`.

Then, `cos x = 4/5`

⇒ `sin x = sqrt (1 - (4/5)^2)`

⇒ `sin x = sqrt(1 - 16/25)`

⇒ `sin x = sqrt(9/25)`

⇒ `sin x = 3/5`

∴ `tan x = 3/4` ⇒ `x = tan^(-1)  3/4`

∴ `cos^(-1)  4/5 = tan^(-1)  3/4`   ...(1)

Now, let `cos^(-1)  12/13 = y`.

Then, `cos y = 12/13`

⇒ `sin y = 5/13`

∴ `tan y = 5/12` ⇒ `y = tan^(-1)  5/12`

∴ `cos^(-1)  12/13 = tan^(-1)  5/12`  ...(2)

Let `cos^(-1)  33/65 = z`.

Then, `cos z = 33/65`

⇒ `sin z = 56/65`

∴ `tan z = 56/33` ⇒ `z = tan^(-1)  56/33`

∴ `cos^(-1)  33/65 = tan^(-1)  56/33`  ...(3)

Now, we will prove that:

L.H.S = `cos^(-1)  4/5 + cos^(-1)  12/13`

= `tan^(-1)  3/4 + tan^(-1)  5/12`  ...[Using (1) and (2)]

= `tan^(-1)  (3/4 + 5/12)/(1 - 3/4 * 5/12)`   ...`[tan^(-1) x + tan^(-1) y = tan^(-1)  (x + y)/(1 - xy)]`

= `tan^(-1)  (36+20)/(48-15)`

= `tan^(-1)  56/33`

= `tan^(-1)  56/33`   ...[By (3)]

= R.H.S.

Let `sin^-1  3/5 = x`.

Then, `sin x = 3/5`

⇒ `cos x = sqrt(1 - (3/5)^2`

= `sqrt(16/25)`

= `4/5`

∴ `tan x = 3/4` ⇒ `x = tan^-1  3/4`

∴ `sin^-1  3/5 = tan^-1  3/4`   ...(1)

Now, let `cos^-1  12/13 = y`.

Then, `cos y = 12/13` ⇒ `sin y = 5/13`.

∴ `tan y = 5/12` ⇒ `y = tan^-1  5/12`

∴ `cos^-1  12/13 = tan^-1  5/12`   ...(2)

Let `sin^-1  56/65 = z`.

Then, `sin z = 56/65` ⇒ `cos z = 33/65`.

∴ `tan z = 56/33` ⇒ `z = tan^-1  56/33`

∴ `sin^-1  56/65 = tan^-1  56/33`   ...(3)

Now, we have:

L.H.S. = `cos^-1  12/13 + sin^-1  3/5`

= `tan^-1  5/12 + tan^-1  3/4`   ...[Using (1) and (2)]

= `tan^-1  (5/12 + 3/4)/(1 - 5/12 * 3/4)`   ...`[tan^-1x + tan^-1y = tan^-1  (x + y)/(1 - xy)]`

= `tan^-1  (20 + 36)/(48 - 15)`

= `tan^-1  56/33`

= `sin^-1  56/65` = R.H.S.   ...[Using (3)]

Let `sin^(-1)  5/13 = x`.

Then, `sin x = 5/13` ⇒ `cos x = 12/13`.

∴ `tan x = 5/12` ⇒ `x = tan^-1  5/12`

∴ `sin^-1  5/13 = tan^-1  5/12`   ...(1)

Let `cos^-1  3/5 = y`.

Then, `cos y = 3/5` ⇒ `sin y = 4/5`.

∴ `tan y = 4/3` ⇒ `y = tan^-1  4/3`

∴ `cos^-1  3/5 = tan^-1  4/3`   ...(2)

Using (1) and (2), we have

R.H.S. = `sin^-1  5/13 + cos^-1  3/5`

= `tan^-1  5/12 + tan^-1  4/3`

= `tan^-1 ((5/12 + 4/3)/(1 - 5/12 xx 4/3))`   ...`[tan^-1x + tan^-1y = tan^-1  (x + y)/(1 - xy)]`

= `tan^-1  ((15 + 48)/(36 - 20))`

= `tan^-1  63/16`

= L.H.S.

Let x = tan² θ.

Then, `sqrt(x) = tan θ`

⇒ `θ = tan^(-1) sqrtx`

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

= cos 2θ

Now, we have:

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

= `1/2 cos^(-1) (cos 2θ)`

= `1/2 xx 2θ`

= θ

= `tan^(-1) sqrt(x)`

= L.H.S.

Consider `(sqrt(1 + sinx) + sqrt(1 - sin x))/(sqrt(1 + sinx) - sqrt(1 - sinx))`

= `((sqrt(1 + sinx) + sqrt(1 - sinx))^2)/((sqrt(1 + sin x))^2 - (sqrt(1 - sin x))^2)`   ...(By rationalizing)

= `((1 + sinx) + (1 - sinx) + 2sqrt((1 + sinx)(1 - sinx)))/(1 + sinx - 1 + sinx)`

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

= `(1 + cosx)/(sin x)`

= `(2 cos^2  x/2)/(2sin  x/2 cos  x/2)`

= `cot  x/2`

∴ L.H.S = `cot^(-1) ((sqrt(1 + sin x) + sqrt(1 - sinx))/(sqrt(1 + sin x) - sqrt(1 - sinx)))`

= `cot^(-1) (cot  x/2)`

= `x/2` = R.H.S.

• Inverse trigonometric functions give angles corresponding to known trigonometric values.

• Their domains are restricted because ordinary trigonometric functions are not one-one on full domains.

• Principal value means the standard angle selected from a fixed interval.

i. \[\sin^{-1}\frac{1}{x}=\mathrm{cosec}^{-1}\] if x ≥ 1 or x ≤ −1
\[\cos^{-1}\frac{1}{x}=\sec^{-1}x\] if x ≥ 1 or x ≤ −1
\[\tan^{-1}\frac{1}{x}=\cot^{-1}x\] if x > 0

ii. sin⁻¹(−x) = −sin⁻¹x, for x ∈ [−1, 1]
tan⁻¹(−x) = −tan⁻¹x, for x ∈ R
cosec⁻¹(−x) = −cosec⁻¹x, for x ≥ 1
cos⁻¹(−x) = π − cos⁻¹x, for x ∈ [−1, 1]
sec⁻¹(−x) = π − sec⁻¹x, for x ≥ 1
cot⁻¹(−x) = π − cot⁻¹x, for x ∈ R

\[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\] for x ∈ [−1, 1]

\[\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2},\] for x ∈ R

\[\sec^{-1}x+\cos\sec^{-1}x=\frac{\pi}{2},\] for |x| ≥ 1

\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x > 0, y > 0 and xy < 1

\[\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x, y > 0 and xy > 1

\[\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right),\] for x, y > 0

\[2\tan^{-1}x=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right),\] if −1 ≤ x ≤ 1

\[2\tan^{-1}x=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right),\] if x > 0

\[2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^{2}}\right),\] if −1 < x < 1