Min menu

Pages

NCERT Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)Exercise 2.2

NCERT Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)Exercise 2.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-2 (Inverse Trigonometric Functions)Exercise 2.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)Exercise 2.1

Exercise 2.2  

Set 1

Prove the following

Question 1. 3sin-1x = sin-1(3x – 4x3), x[-1/2, 1/2] 

Solution:

Let us take x = sinθ, so θ = sin-1x

Substitute the value of x in the equation present on R.H.S. 

The equation becomes sin-1(3sinθ – 3sin3θ) 

We know, sin3θ = 3sinθ – 4sin3θ

So , sin-1(3sinθ – 3sin3θ) = sin-1(sin3θ) 

By the property of inverse trigonometry we know, sin(sin-1(θ)) = θ   

So, sin-1(sin3θ) = 3θ 

And we know θ = sin-1x   

So, 3θ = 3sin-1x = L.H.S

Question 2. 3cos-1x = cos-1(4x3 – 3x), x[-1/2, 1] 

Solution:

Let us take x = cosθ, so θ = cos-1x

Substitute value of x in the equation present on R.H.S. 

The equation becomes cos-1(4cos3θ – 3cosθ) 

We know, cos3θ = 4cos3θ – 3cosθ  

So, cos-1(4cos3θ – 3cosθ) = cos-1(cos3θ) 

By the property of inverse trigonometry we know, cos(cos-1(θ)) = θ    

So, cos-1(cos3θ) = 3θ 

And we know θ = cos-1x     

So, 3θ = 3cos-1x = L.H.S

Question 3.Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

We know, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Now put x = 2/11 and y = 7/24

So, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= R.H.S

Question 4. Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

We have to first write 2tan-1x in terms of  tan-1x
We know that 2tan-1x = Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) 
Put x = 1/2 in the above formula
So,  Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Now we can replace Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)with Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
So equation in L.H.S become Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
We know , Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) 
So,  Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) 
= R.H.S

Write the following functions in simplest forms: 

Question 5.  Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

Let us assume that x = tanθ, so θ = tan-1
Substitute the value of x in question. 
So equation becomes Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
We know that, 1 + tan2θ = sec2θ 
Replacing 1 + tan2θ with sec2θ in the equation Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
So equation becomes, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
We know, tanθ = sinθ/cosθ and sec = 1/cosθ 
Replacing value of tanθ and secθ in Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
We know, 1 – cosθ = 2sin2θ/2​ and sinθ = 2sinθ/2cosθ/2  
So the equations after replacing above value becomes Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
We know Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) 
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
= θ/2           [tan-1(tanθ) = θ]
= 1/2 tan-1x            [θ = tan-1x]

Question 6. Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) , |x| > 1

Solution:

Let us assume that x = cosecθ, so θ = cosec -1

Substitute the value of x in question with cosecθ

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

We know that, 1 + cot2θ = cosec2θ, so cosec2θ = 1 – cot2θ  

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= tan-1(tanθ)          [1/cotθ = tanθ]  

= θ           [tan-1(tanθ) = θ] 

= cosec−1x          [θ = cosec−1x]

= π/2 ​- sec−1x         [cosec−1x + sec−1x = π/2​]

Question 7. Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

We know, 1 – cosx = 2sin2x/2 and 1 + cosx = 2cos2x/2   

 Substituting above formula in question

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= tan-1(tanx/2)         Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

 = x/2         [tan-1(tanθ) = θ] 

Question 8Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

Divide numerator and denominator by cosx

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

We know, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

This can also be written as Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)  – (1)

We know  Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)      – (2)

On comparing equation (1) and (2) we can say that x = 1 and y = tan-1x

So we can say that Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= π/4​ – tan−1x          [tan−11 = π/4​]

Question 9Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

Let us assume that x = asinθ, so θ = sin -1x/a

Substitute the value of x in question.

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Taking acommon from denominator

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

We know that, sin2θ + cos2θ = 1, so 1 – sin2θ = cos2θ 

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= tan-1(tanθ)          [sinθ/cosθ = tanθ]

 = θ        

= sin-1x/a 

Question 10Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

Let us assume that x = atanθ, so θ = tan -1x/a

Substitute the value of x in question

 Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Taking a3common from numerator and denominator

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

We know Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

So, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= 3θ           [ tan-1(tanθ) = θ]

= 3tan -1x/a

Exercise 2.2  

Set 2

Find the values of each of the following: 

Question 11. tan−1[2cos(2sin−11/2​)]

Solution:

Let us assume that sin−11/2 = x

So, sinx = 1/2

Therefore, x = π​/6 = sin−11/2

Therefore, tan−1[2cos(2sin−11/2​)] =  tan−1[2cos(2 * π​/6)]

tan−1[2cos(π​/3)]

Also, cos(π/3​) = 1/2​

Therefore, tan−1[2cos(π​/3)] = tan−1[(2 * 1/2)]

tan−1[1] = π​/4 

Question 12. cot(tan−1a + cot−1a) 

Solution:

We know, tan−1x + cot−1x = π​/2

Therefore, cot(tan−1a + cot−1a) = cot(π​/2) =0

Question 13.  Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

We know, 2tan-1x = Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) and 2tan-1y =  Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

tan(1/2)​[2(tan−1tan−1y)]

tan[tan−1tan−1y]

Also, tan−1tan−1y = Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Therefore, tan[tan−1tan−1y] = Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= (x + y)/(1 – xy)

Question 14. If sin(sin−11/5​+ cos−1x) = 1 then find the value of x

Solution:

sin−11/5​ + cos−1x = sin−11

We know, sin−11 = π/2

Therefore, sin−11/5​ + cos−1x = π/2

sin−11/5​ = π/2 – cos−1x

Since, sin−1x​ + cos−1x = π/2

Therefore, π/2 – cos−1x = sin−1x

sin−11/5​ = sin−1x

So, x = 1/5

Question 15. If Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) , then find the value of x

Solution:


We know, tan−1x + tan−1y = Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

2x– 4 = -3

2x– 4 + 3 = 0

2x– 1 = 0

x= 1/2

x = 1/√2, -1/√2

Find the values of each of the expressions in Exercises 16 to 18.

Question 16. sin − 1(sin2π/3​)  

Solution:


We know that sin−1(sinθ) = θ when θ ∈ [-π/2, π/2], but Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

So, sin − 1(sin2π/3​) can be written as Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

 sin − 1(sinπ/3​)  here Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Therefore, sin − 1(sinπ/3​) = π/3

Question 17. tan−1(tan3π/4​)

Solution:


We know that tan−1(tanθ) = θ when  but 

So, tan−1(tan3π/4​) can be written as tan−1(-tan(-3π/4)​)

tan−1[-tan(π – π/4​)]

tan−1[-tan(π/4​)]

= –tan−1[tan(π/4​)]

= – π/4 where 

Question 18.Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solution:

Let us assume Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) = x , so sinx = 3/5 

We know, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

cosx = 4/5


We know, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

So, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

tanx = 3/4

Also, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Hence, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

tan-1x + tan-1y = Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

So, Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)

= 17/6

Question 19.  cos−1(cos7π/6​) is equal to

(i) 7π/6    (ii) 5π/6    (iii)π/3    (iv)π/6

Solution:

 We know that cos−1(cosθ) = θ, θ  [0, π]

cos−1(cosθ) = θ, θ  [0, π]

Here, 7π/6 > π 

So, cos−1(cos7π/6​) can be written as cos−1(cos(-7π/6)​)

cos−1[cos(2π – 7π/6​)]      [cos(2π + θ) = θ]

cos−1[cos(5π/6​)]       where 5π/6   [0, π]

  Therefore, cos−1[cos(5π/6​)] = 5π/6 

Question 20. Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)
(i) 1/2    (ii) 1/3   (iii) 1/4    (iv) 1

Solution:

Let us assume sin-1(-1/2)= x, so sinx = -1/2 

Therefore, x = -π/6​

Therefore, sin[π/3​ – (-π/6​)]

= sin[π/3​ + (π/6​)]

= sin[3π/6]

= sin[π/2]

= 1

Question 21. Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions) is equal to
(i) π    (ii) -π/2    (iii)0    (iv)2√3

Solution:

We know, cot(−x) = −cotx

Therefore, tan-13 – cot-1(-3) = tan-13 – [-cot-1(3)]

= tan-13 + cot-13

Since, tan-1x + cot-1x = π/2

Tan-13 + cot-13 = -π/2

Chapter-2 (Inverse Trigonometric Functions)

Comments