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^{-1}x, x ≥ 1, x ≤ -1

(b) cos^{-1} \(\frac{1}{x}\) = sec^{-1}x, x ≥ l, x ≤ -1

(c) tan^{-1} \(\frac{1}{x}\) = cot^{-1}x, x > 0

(d) cosec^{-1} \(\frac{1}{x}\) = sin^{-1}x, x ∈ [-1,1]

(e) sec^{-1} \(\frac{1}{x}\) = cos^{-1}x, x ∈ [-1,1]

(f) cot^{-1} \(\frac{1}{x}\) = tan^{-1}x, x > 0.

2. (a) sin^{-1}(-x) = – sin^{-1}x, x ∈ [-1,1]

(b) tan^{-1}(-x) = – tan^{-1}x, x ∈ R

(c) cosec^{-1}(-x) = – cosec^{-1}x, | x | >1

(d) cos^{-1}(-x) = π – cos^{-1}x, x ∈ [-1,1]

(e) sec^{-1}(-x) = π – sec^{-1}x, x ∈ R

(f) cot^{-1}(-x) = π – cot^{-1}x, x ∈ R

3. (a) sin^{-1}x + cos^{-1}x = \(\frac{x}{2}\), x ∈ [-1,1]

(b tan^{-1}x + cot^{-1}x = \(\frac{x}{2}\), x ∈ R

(c) cosec^{-1}x + sec^{-1}x = \(\frac{x}{2}\), | x | ≥ 1

4. (a) tan^{-1}x + tan^{-1}y = tan-1\(\frac{x+y}{1-x y}\), xy < 1

(b) tan^{-1}x – tan^{-1}y = tan-1\(\frac{x-y}{1+x y}\), xy > -1

(c) 2 tan^{-1}x = 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^{-1}x = sin^{-1}[2x\(\sqrt{1-x^{2}}\)]

(e) 2cos^{-1}x = cos^{-1}(2x^{2} – 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^{-1}x |
– ∞ < 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^{-1}x = 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^{-1}x = tan^{-1}\(\frac { 1 }{ x }\), x > 0.

(iii) (a) sin^{-1} (- x) = – sin^{-1}x, x ∈ [-1,11

(b) cos^{-1} (-x) = π – cos^{-1}x, 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^{-1}x = \(\frac{\pi}{2}\) x ∈ R

(c) sec^{-1}x + 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^{-1}x – tan^{-1}y = 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 ^{-1}x ± sin^{-1} y = sin ^{-1} \(\left(x \sqrt{1-y^{2}} \pm y \sqrt{1-x^{2}}\right)\)

(b) cos ^{-1}x ± cos^{-1}y = cos ^{-1} \(\left(x y \mp \sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)\)