By going through these CBSE Class 12 Maths Notes Chapter 2 Inverse Trigonometric Functions, students can recall all the concepts quickly.

## Inverse Trigonometric Functions Notes Class 12 Maths Chapter 2

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As we have learned in class XI, the domain and range of trigonometric functions are given below:
Functions Domain Range

→ Inverse Function: We know that if f: X →Y such that y = f(x) is one-one and onto, then we define another function g : Y → X such that x = g(y), where x ∈ X and y ∈ Y, which is also one-one and onto.

In such a case, Domain of g = Range of f and Range of g = Domain of f.
g is called the inverse of f or g = f-1.
∴ Inverse of g = g-1 = (f-1)-1 =f.

The graph of the sin-1 function is shown here.

Principal Value Branch of Function sin-1:
It may be noted that for the domain [-1,1}, the range could be any one of the intervals. …..,
[$$\frac{-3 \pi}{2}$$, $$\frac{-\pi}{2}$$], [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$] or [$$\frac{\pi}{2}$$, $$\frac{3\pi}{2}$$] …….

Corresponding to each interval, we get a branch of the function sin-1.
The branch with range [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$] is called the principal value branch.
Thus, sin-1: [-1,1] → [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$]

Principal Value Branch of Function cos-1:
The graph of the function cos-1 is as shown here.

The domain of the function cos-1 is [-1,1]. Its range is one of the intervals. ………., (-π, 0), (0, π), (π, 2π), ………… It is one-one and onto with the range [-1, 1]. The branch with range (0, π) is called the principal value branch of the function cos-1. Thus, cos-1: [-1,1] → [0, π].

Principal Value Branch of Function tan-1:
The function tan-1 is defined whose domain is set of real numbers and range is one of the intervais. [ $$\frac{-3 \pi}{2}$$, $$\frac{-\pi}{2}$$], [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$] or [$$\frac{\pi}{2}$$, $$\frac{3 \pi}{2}$$]…………

The graph of the function is as shown in the adjoining figure.

The branch with range [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$] is called the principal value branch of function tan1. Thus, tan-1: R → [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$].

Principal Value Branch of Function cosec-1:
The graph of function cosec-1 is as shown here. The function cosec-1 defined whose domain is R — (-1, 1) and the range is any one of the intervals …….., [ $$\frac{-3 \pi}{2}$$, $$\frac{-\pi}{2}$$] – {π}, [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$] – {0}, [$$\frac{\pi}{2}$$, $$\frac{3 \pi}{2}$$] – {π}……..

The function corresponding to the range [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$] – {0} is called the principal value branch of cosec-1.
Thus, cosec-1: R – (-1, 1) → [$$\frac{-\pi}{2}$$, $$\frac{\pi}{2}$$] – {0}.

Principal Value Branch of Function sec-1:
The graph of function sec-1 is shown in adjoining figure. The sec-1 is defined as a function whose domain is R – (-1, 1) and the range is one of the intervals……. [- π, 0] – {$$\frac{-\pi}{2}$$}, [0, π] – {$$\frac{\pi}{2}$$}, [π, 2π] – {$$\frac{3 \pi}{2}$$}, ……………

The function corresponding to [0, π] – {$$\frac{\pi}{2}$$} is known as the range [0, π] – {$$\frac{\pi}{2}$$} is known as the principal value branch of sec-1.

Thus, sec-1: R – (-1,1) → [0, π] – {$$\frac{\pi}{2}$$}

Principal Value Branch of Function cot-1:
The graph of function cor’ is as shown here.

The cot-1 function is defined as a function whose domain is R and the range is any of the intervals, …………
(-π, 0), (0, π), (π, 2π), …………

The function corresponding to (0, π) is called the principal value branch of the function cot-1.
Thus, cot-1: R → (0, π).

The principal value branch of trigonometric inverse functions are as follows:

→ Some Important Results:
1. (a) sin-1$$\frac{1}{x}$$ = cosec-1x, x ≥ 1, x ≤ -1
(b) cos-1 $$\frac{1}{x}$$ = sec-1x, x ≥ l, x ≤ -1
(c) tan-1 $$\frac{1}{x}$$ = cot-1x, x > 0
(d) cosec-1 $$\frac{1}{x}$$ = sin-1x, x ∈ [-1,1]
(e) sec-1 $$\frac{1}{x}$$ = cos-1x, x ∈ [-1,1]
(f) cot-1 $$\frac{1}{x}$$ = tan-1x, x > 0.

2. (a) sin-1(-x) = – sin-1x, x ∈ [-1,1]
(b) tan-1(-x) = – tan-1x, x ∈ R
(c) cosec-1(-x) = – cosec-1x, | x | >1
(d) cos-1(-x) = π – cos-1x, x ∈ [-1,1]
(e) sec-1(-x) = π – sec-1x, x ∈ R
(f) cot-1(-x) = π – cot-1x, x ∈ R

3. (a) sin-1x + cos-1x = $$\frac{x}{2}$$, x ∈ [-1,1]
(b tan-1x + cot-1x = $$\frac{x}{2}$$, x ∈ R
(c) cosec-1x + sec-1x = $$\frac{x}{2}$$, | x | ≥ 1

4. (a) tan-1x + tan-1y = tan-1$$\frac{x+y}{1-x y}$$, xy < 1
(b) tan-1x – tan-1y = tan-1$$\frac{x-y}{1+x y}$$, xy > -1
(c) 2 tan-1x = tan-1$$\frac{2 x}{1-x^{2}}$$, | x | <1
= sin-1$$\frac{2 x}{1+x^{2}}$$, | x | <0
= cos-1$$\frac{1-x^{2}}{1+x^{2}}$$ , x ≥ 0
(d) 2sin-1x = sin-1[2x$$\sqrt{1-x^{2}}$$]
(e) 2cos-1x = cos-1(2x2 – 1)

1. Table

 Function Principal Value Branch Domain Range (i) y = sin-1 x – 1 ≤ x ≤ 1 $$-\frac{\pi}{2}$$ ≤ y ≤ $$\frac{\pi}{2}$$ (ii) y = cos-1 x – 1 ≤ x ≤ 1 0 ≤ y ≤ π (iii) y = tan-1x – ∞ < x < ∞ $$-\frac{\pi}{2}$$ < y < $$\frac{\pi}{2}$$ (iv) y = cot-1 x – ∞ < x ≤ -1 0 < y < $$\frac{\pi}{2}$$ (v) y = sec-1 x 1 ≤  x < ∞ -∞ < x ≤ -1 0 ≤ y < $$\frac{\pi}{2}$$ $$\frac{\pi}{2}$$< y < π (vi) y = cosec-1 x 1 ≤ x < ∞ -∞ < x ≤ ∞ 0 < y ≤ $$\frac{\pi}{2}$$ $$-\frac{\pi}{2}$$ ≤ y < 0

2. PROPERTIES :

(i) x = sin-1(sin x) = cos-1(cos x) = tan-1 (tan x); etc.
(ii) (a) cosec-1x = sin-1 $$\frac { 1 }{ x }$$ , x ≥ 1 or x ≤ -1
(b) sec-1 x = cos-1$$\frac { 1 }{ x }$$ ,x ≥ 1 or x ≤ -1
(c) cot-1x = tan-1$$\frac { 1 }{ x }$$, x > 0.

(iii) (a) sin-1 (- x) = – sin-1x, x ∈ [-1,11
(b) cos-1 (-x) = π – cos-1x, x ∈ [- 1, 1]
(c) tan-1(- x) = – tan-1 x, x ∈ R
(d) cot-1(-x) = π – cot-1 x, x ∈ R
(e) sec-1 (-x) = π – sec-1 x, |x| ≥ 1
(f) cosec-1 (- x) = – cosec-1 x, |x| ≥ 1

(iv) (a) sin-1 x + cos-1 x = $$\frac{\pi}{2}$$, x ∈ (-1,1)
(b) tan -1 x + cot-1x = $$\frac{\pi}{2}$$ x ∈ R
(c) sec-1x + cosec-1= $$\frac{\pi}{2}$$, |x| ≥ 1
(d) tan-1 x + tan-1 = tan-1 $$\frac{x+y}{1-x y}$$, xy > -1
(e) tan-1x – tan-1y = tan-1 $$\frac{x-y}{1+x y}$$, xy > -1
(f) 2tan-1 x = sin-1 $$\frac{2 x}{1+x^{2}}$$ = cos-1 $$\frac{2 x}{1-x^{2}}$$ = tan-1

(v) (a) sin -1x ± sin-1 y = sin -1 $$\left(x \sqrt{1-y^{2}} \pm y \sqrt{1-x^{2}}\right)$$
(b) cos -1x ± cos-1y = cos -1 $$\left(x y \mp \sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)$$