RD Sharma Solutions For Class 12 Maths Exercise 4.10 Chapter 4 Inverse Trigonometric Functions

Get the detailed RD Sharma Solutions for Class 12 Maths Exercise 4.10 Chapter 4 Inverse Trigonometric Functions, which is available here. From the exam point of view, subject experts have prepared the solutions in a step by step format in accordance with the CBSE syllabus. Students gain knowledge and increase their confidence by following the steps, when practised regularly. This exercise explains the topic of properties of inverse trigonometric functions. To know more about the construction of graphs using trigonometric functions students can download RD Sharma Solutions for Class 12 Maths.

Download the PDF of RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.10

 

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Access answers to Maths RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.10

Exercise 4.10 Page No: 4.66

1. Evaluate:

(i) Cot (sin-1 (3/4) + sec-1 (4/3))

(ii) Sin (tan-1 x + tan-1 1/x) for x < 0

(iii) Sin (tan-1 x + tan-1 1/x) for x > 0

(iv) Cot (tan-1 a + cot-1 a)

(v) Cos (sec-1 x + cosec-1 x), |x| ≥ 1

Solution:

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 43

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

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 44

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 45

(iii) Given Sin (tan-1 x + tan-1 1/x) for x > 0

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 46

(iv) Given Cot (tan-1 a + cot-1 a)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 47

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 48

(v) Given Cos (sec-1 x + cosec-1 x), |x| ≥ 1

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 49

= 0

2. If cos-1 x + cos-1 y = π/4, find the value of sin-1 x + sin-1 y.

Solution:

Given cos-1 x + cos-1 y = π/4

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 50

3. 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)

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 51

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 52

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 53

RD Sharma Solutions for Class 12 Maths Chapter 4 Inverse Trigonometric Functions Image 54

4. If cot (cos-1 3/5 + sin-1 x) = 0, find the value of x.

Solution:

Given cot (cos-1 3/5 + sin-1 x) = 0

On rearranging we get,

(cos-1 3/5 + sin-1 x) = cot-1 (0)

(Cos-1 3/5 + sin-1 x) = π/2

We know that cos-1 x + sin-1 x = π/2

Then sin-1 x = π/2 – cos-1 x

Substituting the above in (cos-1 3/5 + sin-1 x) = π/2 we get,

(Cos-1 3/5 + π/2 – cos-1 x) = π/2

Now on rearranging we get,

(Cos-1 3/5 – cos-1 x) = π/2 – π/2

(Cos-1 3/5 – cos-1 x) = 0

Therefore Cos-1 3/5 = cos-1 x

On comparing the above equation we get,

x = 3/5

5. If (sin-1 x)2 + (cos-1 x)2 = 17 π2/36, find x.

Solution:

Given (sin-1 x)2 + (cos-1 x)2 = 17 π2/36

We know that cos-1 x + sin-1 x = π/2

Then cos-1 x = π/2 – sin-1 x

Substituting this in (sin-1 x)2 + (cos-1 x)2 = 17 π2/36 we get

(sin-1 x)2 + (π/2 – sin-1 x)2 = 17 π2/36

Let y = sin-1 x

y2 + ((π/2) – y)2 = 17 π2/36

y2 + π2/4 – y2 – 2y ((π/2) – y) = 17 π2/36

π2/4 – πy + 2 y2 = 17 π2/36

On rearranging and simplifying, we get

2y2 – πy + 2/9 π2 = 0

18y2 – 9 πy + 2 π2 = 0

18y2 – 12 πy + 3 πy + 2 π2 = 0

6y (3y – 2π) + π (3y – 2π) = 0

Now, (3y – 2π) = 0 and (6y + π) = 0

Therefore y = 2π/3 and y = – π/6

Now substituting y = – π/6 in y = sin-1 x we get

sin-1 x = – π/6

x = sin (- π/6)

x = -1/2

Now substituting y = -2π/3 in y = sin-1 x we get

x = sin (2π/3)

x = √3/2

Now substituting x = √3/2 in (sin-1 x)2 + (cos-1 x)2 = 17 π2/36 we get,

= π/3 + π/6

= π/2 which is not equal to 17 π2/36

So we have to neglect this root.

Now substituting x = -1/2 in (sin-1 x)2 + (cos-1 x)2 = 17 π2/36 we get,

= π2/36 + 4 π2/9

= 17 π2/36

Hence x = -1/2.


Access other exercises of RD Sharma Solutions For Class 12 Chapter 4 – Inverse Trigonometric Functions

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.4 Solutions

Exercise 4.5 Solutions

Exercise 4.6 Solutions

Exercise 4.7 Solutions

Exercise 4.8 Solutions

Exercise 4.9 Solutions

Exercise 4.11 Solutions

Exercise 4.12 Solutions

Exercise 4.13 Solutions

Exercise 4.14 Solutions

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