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

Content of this article has been merged with Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 as per the revised syllabus of NCERT.

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.  

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 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  , 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.

Question 16. 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 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. 

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 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. 

(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.  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