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
We have provided complete details of RD Sharma Solutions for Class 12 Maths Chapter 4 Exercise 4.10. If you have any queries related to CBSE Class 12 Exam, feel free to ask us in the comment section below.
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