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Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 | Set 2
• Difficulty Level : Expert
• Last Updated : 05 Apr, 2021

### 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 = 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

### 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

### (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

### (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

### (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

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