RD Sharma Class 12 Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.10 (Updated for 2021-22)

RD Sharma Solutions Class 12 Chapter 4 Exercise 4.10 is a part of RD Sharma Class 12 Solutions. Subject experts have developed the solutions in a step-by-step format in accordance with the CBSE syllabus from the exam perspective. When students practise the steps on a regular basis, they build understanding and confidence. The topic of properties of inverse trigonometric functions is explained in this exercise. Students can learn more by downloading RD Sharma Solutions for Class 12 Maths Chapter 4 Exercise 4.9 in PDF format.

Download RD Sharma Solutions Class 12 Chapter 4 Exercise 4.10 Free PDF

RD Sharma Solutions Class 12 Chapter 4 Exercise 4.10

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

Evaluate:

1. Cot (sin^-1 (3/4) + sec^-1 (4/3))

2. Sin (tan^-1 x + tan^-1 1/x) for x < 0

3. Sin (tan^-1 x + tan^-1 1/x) for x > 0

4. Cot (tan^-1 a + cot^-1 a)

5. Cos (sec^-1 x + cosec^-1 x), |x| ≥ 1

Solution:

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

2. Given Sin (tan^-1 x + tan^-1 1/x) for x < 0

3. Given Sin (tan^-1 x + tan^-1 1/x) for x > 0

4. Given Cot (tan^-1 a + cot^-1 a)

5. Given Cos (sec^-1 x + cosec^-1 x), |x| ≥ 1

6. 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

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. 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

9. 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

• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.1
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.2
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.3
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.4
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.5
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.6
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.7
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.8
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.9
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.11
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.12
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.13
• RD Sharma Solutions for Class 12 Chapter 4 Exercise 4.14

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