Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.2

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

Write the following functions in simplest forms: 

Question 3.  

Solution:

Let us assume that x = tanθ, so θ = tan^-1x 

Substitute the value of x in question. 

So equation becomes 

We know that, 1 + tan²θ = sec²θ 

Replacing 1 + tan²θ with sec²θ in the equation 

So equation becomes, 

We know, tanθ = sinθ/cosθ and sec = 1/cosθ 

Replacing value of tanθ and secθ in 

We know, 1 – cosθ = 2sin²θ/2​ and sinθ = 2sinθ/2cosθ/2  

So the equations after replacing above value becomes 

We know 

= θ/2           [tan^-1(tanθ) = θ]

= 1/2 tan^-1x            [θ = tan^-1x]   

Question 4. 

Solution:

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

 Substituting above formula in question

= tan^-1(tanx/2)         

 = x/2         [tan^-1(tanθ) = θ] 

Question 5. 

Solution:

Divide numerator and denominator by 

We know, 

This can also be written as   – (1)

We know        – (2)

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

So we can say that 

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

Question 6. 

Solution:

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

Substitute the value of x in question.

Taking a² common from denominator

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

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

 = θ        

= sin^-1x/a 

Question 7. 

Solution:

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

Substitute the value of x in question

Taking a³common from numerator and denominator

We know 

So, 

= 3θ           [ tan^-1(tanθ) = θ]

= 3tan ^-1x/a

Find the values of each of the following: 

Question 8. 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 9.  

Solution:

We know, 2tan^-1x =  and 2tan^-1y =  

= tan(1/2)​[2(tan⁻¹x + tan⁻¹y)]

= tan[tan⁻¹x + tan⁻¹y]

Also, tan⁻¹x + tan⁻¹y =

Therefore, tan[tan⁻¹x + tan⁻¹y] = 

= (x + y)/(1 – xy)

Question 10. sin ^{− 1}(sin2π/3​)  

Solution:

We know that sin⁻¹(sinθ) = θ when θ ∈ [-π/2, π/2], but 

So, sin ^{− 1}(sin2π/3​) can be written as 

 sin ^{− 1}(sinπ/3​)  here 

Therefore, sin ^{− 1}(sinπ/3​) = π/3

Question 11. 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 12. 

Solution:

Let us assume  = x , so sinx = 3/5 

We know, 

cosx = 4/5

We know, 

So, 

tanx = 3/4

Also, 

Hence, 

tan^-1x + tan^-1y = 

So, 

= 17/6

Question 13.  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 14. 

(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 15.  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

Deleted Questions

Solution:

We know, 

Now put x = 2/11 and y = 7/24

So, 

= R.H.S

Solution:

We have to first write 2tan^-1x in terms of  tan^-1x

We know that 2tan^-1x = 

Put x = 1/2 in the above formula

So,  

Now we can replace with 

So equation in L.H.S become 

We know , 

So,  

= R.H.S

 , |x| > 1

Solution:

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

Substitute the value of x in question with 

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

= 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​]

 cot(tan⁻¹a + cot⁻¹a) 

Solution:

We know, tan⁻¹x + cot⁻¹x = π​/2

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

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

If  , then find the value of x

Solution:

We know, tan⁻¹x + tan⁻¹y = 

2x² – 4 = -3

2x² – 4 + 3 = 0

2x² – 1 = 0

x² = 1/2

x = 1/√2, -1/√2

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