## 1. Basic concepts , Domain and Range ,Principal Branch

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## 1. Basic concepts , Domain and Range ,Principal Branch

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## 1. Basic concepts , Domain and Range ,Principal Branch

- Books Name
- Mathmatics Book Based on NCERT

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 12

- Subject
- Mathmatics

**Chapter-2**

**Inverse Trigonometric functions**

**Basic concepts , Domain and Range ,Principal Branch:**

**Let y = sinx => x = sin ^{-1}y**

- should not be confused with . In fact
- i.e., ¹
- And similarly for other trigonometric functions.
- The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric functions.
- The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table:
- The functions

sin^{−1}x & tan^{−1}x

are increasing functions in their domain.

- The functions

cos^{−1}x & cot^{−1}x

are decreasing functions in over domain.

**For suitable values of domain, we have**

•

•

## 2. Graphs of inverse trigonometric functions

- Books Name
- Mathmatics Book Based on NCERT

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 12

- Subject
- Mathmatics

**Graphs of inverse trigonometric functions**

## 3. Properties of inverse trigonometric functions and important results

- Books Name
- Mathmatics Book Based on NCERT

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 12

- Subject
- Mathmatics

**Properties of inverse trigonometric functions and important results**

**Properties:**

**[1]**

- cos(cos
^{-1}x)=x - tan(tan
^{-1}x)=x - cot(cot
^{-1}x)=x - cosec(cosec
^{-1}x)=x - sec(sec
^{-1}x)=x - Cos
^{-1}(cosx)=x - tan
^{-1}(tanx)=x - cot
^{-1}(cotx)=x - cosec
^{-1}(cosecx)=x - sec
^{-1}(secx)=x

**[2]**

**[3]**

- sin
^{–1}(–x) = – sin^{–1}x - tan
^{–1}(–x) = – tan^{–1}x

**[4]**

**[5]**

- tan
^{–1}x + cot^{–1}x = π/2 - sin
^{–1}x + cos^{–1}x = π/2

**[6]**

**Examples:**

**1. Find the domain of definition of f(x) = cos ^{-1 }(x^{2} – 4)**

**Solution:**

Given f(x) = cos ^{-1 }(x^{2} – 4)

We know that domain of cos^{-1} (x^{2} – 4) lies in the interval [-1, 1]

Therefore, we can write as

-1 ≤ x^{2} – 4 ≤ 1

4 – 1 ≤ x^{2} ≤ 1 + 4

3 ≤ x^{2} ≤ 5

±√ 3 ≤ x ≤ ±√5

– √5 ≤ x ≤ – √3 and √3 ≤ x ≤ √5

*Therefore domain of cos ^{-1} (x^{2} – 4) is [- √5, – √3] *

*∪ [√3, √5]*

**2. Find the domain of f(x) = cos ^{-1} 2x + sin^{-1} x.**

**Solution:**

Given that f(x) = cos^{-1} 2x + sin^{-1} x.

Now we have to find the domain of f(x),

We know that domain of cos^{-1} x lies in the interval [-1, 1]

Also know that domain of sin^{-1} x lies in the interval [-1, 1]

Therefore, the domain of cos^{-1} (2x) lies in the interval [-1, 1]

Hence we can write as,

-1 ≤ 2x ≤ 1

– ½ ≤ x ≤ ½

*Hence, domain of cos ^{-1}(2x) + sin^{-1 }x lies in the interval [- ½, ½]*

**3. Find the principal value of each of the following:**

**(i) sec ^{-1} (-√2)**

**(ii) sec ^{-1} (2)**

**(iii) sec ^{-1} (2 sin (3π/4))**

**(iv) sec ^{-1} (2 tan (3π/4))**

**Solution:**

(i) Given sec^{-1} (-√2)

Now let y = sec^{-1} (-√2)

Sec y = -√2

We know that sec π/4 = √2

Therefore, -sec (π/4) = -√2

= sec (π – π/4)

= sec (3π/4)

Thus the range of principal value of sec^{-1} is [0, π] – {π/2}

And sec (3π/4) = – √2

*Hence the principal value of sec ^{-1} (-√2) is 3π/4*

(ii) Given sec^{-1} (2)

Let y = sec^{-1} (2)

Sec y = 2

= Sec π/3

Therefore the range of principal value of sec^{-1} is [0, π] – {π/2} and sec π/3 = 2

*Thus the principal value of sec ^{-1} (2) is π/3*

(iii) Given sec^{-1} (2 sin (3π/4))

But we know that sin (3π/4) = 1/√2

Therefore 2 sin (3π/4) = 2 × 1/√2

2 sin (3π/4) = √2

Therefore by substituting above values in sec^{-1} (2 sin (3π/4)), we get

Sec^{-1 }(√2)

Let Sec^{-1 }(√2) = y

Sec y = √2

Sec (π/4) = √2

Therefore range of principal value of sec^{-1} is [0, π] – {π/2} and sec (π/4) = √2

*Thus the principal value of sec ^{-1} (2 sin (3π/4)) is π/4.*

(iv) Given sec^{-1} (2 tan (3π/4))

But we know that tan (3π/4) = -1

Therefore, 2 tan (3π/4) = 2 × -1

2 tan (3π/4) = -2

By substituting these values in sec^{-1} (2 tan (3π/4)), we get

Sec^{-1} (-2)

Now let y = Sec^{-1} (-2)

Sec y = – 2

– sec (π/3) = -2

= sec (π – π/3)

= sec (2π/3)

Therefore the range of principal value of sec^{-1} is [0, π] – {π/2} and sec (2π/3) = -2

*Thus, the principal value of sec ^{-1} (2 tan (3π/4)) is (2π/3).*

**4. Evaluate:**

**(i) Cot (sin ^{-1} (3/4) + sec^{-1} (4/3))**

**(ii) Sin (tan ^{-1} x + tan^{-1} 1/x) for x < 0**

**Solution:**

(i) Given Cot (sin^{-1} (3/4) + sec^{-1} (4/3))

(ii) Given Sin (tan^{-1} x + tan^{-1} 1/x) for x < 0

**5. If cos ^{-1} (x/2) + cos^{-1} (y/3) = α, then prove that 9x^{2} – 12xy cos α + 4y^{2} = 36 sin^{2} α**

**Solution:**

Given cos^{-1} (x/2) + cos^{-1} (y/3) = α

**6. If sin ^{-1} (2a/1 + a^{2}) – cos^{-1}(1 – b^{2}/1 + b^{2}) = tan^{-1}(2x/1 – x^{2}), then prove that x = (a – b)/ (1 + a b)**

**Solution:**

Given sin^{-1} (2a/1 + a^{2}) – cos^{-1}(1 – b^{2}/1 + b^{2}) = tan^{-1}(2x/1 – x^{2})

**7. If sin ^{-1} x + sin^{-1} y = π/3 and cos^{-1} x – cos^{-1} y = π/6, find the values of x and y.**

**Solution:**

Given sin^{-1} x + sin^{-1} y = π/3 ……. Equation (i)

And cos^{-1} x – cos^{-1} y = π/6 ……… Equation (ii)

**8. Which of the following is the principal value branch of cos-1 x?**

**(a) [-π/2, π/2] (b) (0, π) (c) [0. π] (d) [0, π] – {π/2}**

**Solution:**

(c) [0. π]

As we know that the principal value branch cos^{-1} x is [0, π].

**9. Which of the following is the principal value branch of cosec ^{-1} x?**

**(a) (-π/2, π/2) (b) [0, π] – {π/2} (c) [-π/2, π/2] (d) [-π/2, π/2] – {0}**

**Solution:**

(d) [-π/2, π/2] – {0}

As the principal branch of cosec^{-1} x is [-π/2, π/2] – {0}.

**10. If 3 tan ^{-1} x + cot^{-1} x = π, then x equals**

**(a) 0 (b) 1 (c) -1 (d) ½**

**Solution:**

(b) 1

Given, 3 tan^{-1} x + cot^{-1} x = π

2 tan^{-1} x + tan^{-1} x + cot^{-1} x = π

2 tan^{-1} x + π/2 = π (As tan^{-1} + cot^{-1} = π/2)

2 tan^{-1} x = π/2

tan^{-1} x = π/4

x = 1

**11. The value of sin ^{-1} cos 33π/5 is**

**(a) 3π/5 (b) -7π/5 (c) π/10 (d) -π/10**

**Solution:**

(d) -π/10