RD Sharma Class 12 Ex 4.1 Solutions Chapter 4 Inverse Trigonometric Functions

Here we provide RD Sharma Class 12 Ex 4.1 Solutions Chapter 4 Inverse Trigonometric Functions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 4.1 Solutions Chapter 4 Inverse Trigonometric Functions book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter4
Exercise4.1
CategoryRD Sharma Solutions

Question 1. Find the principal value of each of the following : 

 (i) Sin-1(- √3/2)

(ii) Sin-1(cos 2π/3)

(iii)Sin-1(-√3 – 1/2√2)

 (iv) Sin-1(√3 + 1/2√2)

 (v) Sin-1(cos 3π/4)

 (vi) Sin-1(tan 5π/4)

 Solution:

   (i) Sin-1 (-√3/ 2)

  = Sin-1 [sin (- π / 3)] 

  = – π / 3     Ans. 

  (ii) Sin-1 (cos 2π / 3)

   = Sin-1 (- 1 / 2)

   = Sin-1 (- π / 6)

   = – π / 6   Ans.

 (iii) Sin-1 (√3 – 1/2√2)

   = Sin-1 ( sin π / 12 ) 

   = π / 12  Ans. 

 (iv) Sin-1 (√3 + 1/2 √2) 

   = Sin-1 (5π / 12)

   = 5π / 12   Ans.

(v) Sin-1 (cos 3π / 4)

   = Sin-1 (- √2 / 2)

   = [Sin-1 (- π / 4)]

   = – π / 4   Ans.

(vi) Sin-1 (tan 5π / 4)

  = Sin-1 (1)

  = Sin-1 [sin (π / 2)]

  = π / 2    Ans.

Question 2. 

 (i) Sin-1 1/2  -2 Sin-1 1/√2

 (ii) Sin-1 {cos (Sin-1 √3 / 2)}

 Solution:

(i) Sin-1 1/2  -2  Sin-1 1/ √2

    = Sin-1 1/2 – Sin-1 [ 2 x 1/ √2 √1- (1 /√2)2  ] 

    =   Sin-1 1/2 – Sin-1 (1)

    =  π/6 –  π /2

    =  π / 3   Ans. 

(ii) Sin-1 { cos ( Sin-1 sin  π / 3 )}

   =   Sin-1 { cos ( π / 3 ) } 

   =    Sin-1 { 1/2 }

   =   Sin-1 { sin  π / 6 }

   =  π / 6   Ans.

 Question 3.  Find the domain of each of the following functions : 

(i) f(x) = Sin-1 x2

(ii) f(x) = Sin-1x + sinx

(iii) f(x) = Sin-1√x2 – 1 

(iv) f(x) = Sin-1x + Sin-1 2x

Solution:

(i) Domain of  Sin-1 lies between the interval [ -1 , 1 ]

and x2 ∈ [ 0 , 1 ] as x2 can not be negative .

 So, x ∈ [ -1 , 1 ] 

Hence, the domain of the function f(x) = [ -1 , 1 ]   Ans. 

(ii) Let f(x) = g(x) + h(x) , where g(x) =  Sin-1x  and h(x) = sinx respectively.

Therefore , the domain of f(x) is given by the intersection of the domain g(x) & h(x) .

The domain of g(x) = [ -1 , 1 ]

The domain of h(x) = [ – ∞ , ∞ ] 

Thus, the interaction of g(x) and h(x) is [ -1 , 1 ] 

Hence , the domain of f(x) is [ -1 , 1 ]    Ans.

(iii) As we know , the domain of Sin-1 x is [ -1 , 1 ]

Therefore , domain of Sin-1  √x2 – 1 will also lies in the interval [ -1 , 1 ]

  :. x2 – 1 ∈ [ 0 , 1 ] as square root cannot be negative .

 => x2 ∈ [ 1 , 2 ] 

 => x ∈ [ – √2 , -1 ] U [ 1 , √2 ] 

Hence, the domain of function f(x) = [ – √2 , -1 ] U [ 1 , √2 ]   Ans. 

(iv) Let f(x) = g(x) + h(x), where g(x) = Sin-1 x  x and h(x) = Sin-1 2x

Therefore, the domain of f(x) will be given by the intersection of g(x) and h(x) .

the domain of g(x) = [ -1 , 1 ] 

lly , the domain of h(x) = [ -1/2 , 1/2 ] 

g(x) ∩ h(x) = [ -1 , 1 ]  ∩ [ -1/2 , 1/2 ]

Hence, the domain of the function f(x) =  [- 1/ 2 , 1/2 ]    Ans. 

Question 4. If sin-1x + sin-1y + sin-1z + sin-1t = 2π, then find the value of x2 + y2 + z2 + t2 .

Solution:

As we already know, Range of sin-1 is [ – π / 2 , π / 2 ] 

 Given: (sin-1x) + (sin-1y) +(sin-1y)+(sin-1t) = 2 π

So, each takes the value of π / 2 

:. x = 1, y = 1, z = 1 & t = 1 

Hence, x2 + y2+ z2 + t2 = 1+ 1 + 1 + 1 = 4    Ans. 

Question 5. If (sin-1x)2 + ( sin-1y )2 + ( sin-1y )2  = 3π2/4, find the value of  x2 + y2 + z2             . 

Solution: 

As we already know , Range of   sin-1 is [ – π / 2 , π / 2 ]

Given:     ( sin-1x )2 + ( sin-1y )2 + ( sin-1y )2 = 3π2/4 

:. each takes the value of  π / 2

x = 1, y = 1 & z = 1 .

Hence ,  x+ y2 + z= 1 + 1 + 1 = 3    Ans.

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