Get the detailed RD Sharma Solutions for Class 12 Maths Exercise 4.10 Chapter 4 Inverse Trigonometric Functions, which are available here. For efficient exam preparation, 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 given in the solutions 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 the RD Sharma Solutions for Class 12 Maths.
RD Sharma Solutions for Class 12 Chapter 4 – Inverse Trigonometric Functions Exercise 4.10
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))
(ii) Given Sin (tan-1 x + tan-1 1/x) for x < 0
(iii) Given Sin (tan-1 x + tan-1 1/x) for x > 0
(iv) Given Cot (tan-1 a + cot-1 a)
(v) Given Cos (sec-1 x + cosec-1 x), |x| ≥ 1
= 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
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)
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
By 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
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