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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. tanâˆ’1[2cos(2sinâˆ’11/2â€‹)]

Solution:

Let us assume that sinâˆ’11/2 = x

So, sinx = 1/2

Therefore, x = Ï€â€‹/6 = sinâˆ’11/2

Therefore, tanâˆ’1[2cos(2sinâˆ’11/2â€‹)] =  tanâˆ’1[2cos(2 * Ï€â€‹/6)]

= tanâˆ’1[2cos(Ï€â€‹/3)]

Also, cos(Ï€/3â€‹) = 1/2â€‹

Therefore, tanâˆ’1[2cos(Ï€â€‹/3)] = tanâˆ’1[(2 * 1/2)]

= tanâˆ’1[1] = Ï€â€‹/4

### Question 12. cot(tanâˆ’1a + cotâˆ’1a)

Solution:

We know, tanâˆ’1x + cotâˆ’1x = Ï€â€‹/2

Therefore, cot(tanâˆ’1a + cotâˆ’1a) = cot(Ï€â€‹/2) =0

### Question 13.

Solution:

We know, 2tan-1x =  and 2tan-1y =

= tan(1/2)â€‹[2(tanâˆ’1x + tanâˆ’1y)]

= tan[tanâˆ’1x + tanâˆ’1y]

Also, tanâˆ’1x + tanâˆ’1y =

Therefore, tan[tanâˆ’1x + tanâˆ’1y] =

= (x + y)/(1 – xy)

### Question 14. If sin(sinâˆ’11/5â€‹ + cosâˆ’1x) = 1 then find the value of x

Solution:

sinâˆ’11/5â€‹ + cosâˆ’1x = sinâˆ’11

We know, sinâˆ’11 = Ï€/2

Therefore, sinâˆ’11/5â€‹ + cosâˆ’1x = Ï€/2

sinâˆ’11/5â€‹ = Ï€/2 – cosâˆ’1x

Since, sinâˆ’1xâ€‹ + cosâˆ’1x = Ï€/2

Therefore, Ï€/2 – cosâˆ’1x = sinâˆ’1x

sinâˆ’11/5â€‹ = sinâˆ’1x

So, x = 1/5

### Question 15. If  , then find the value of x

Solution:

We know, tanâˆ’1x + tanâˆ’1y =

2x2 – 4 = -3

2x2 – 4 + 3 = 0

2x2 – 1 = 0

x2 = 1/2

x = 1/âˆš2, -1/âˆš2

### Question 16. sin âˆ’ 1(sin2Ï€/3â€‹)

Solution:

We know that sinâˆ’1(sinÎ¸) = Î¸ when Î¸ âˆˆ [-Ï€/2, Ï€/2], but

So, sin âˆ’ 1(sin2Ï€/3â€‹) can be written as

sin âˆ’ 1(sinÏ€/3â€‹)  here

Therefore, sin âˆ’ 1(sinÏ€/3â€‹) = Ï€/3

### Question 17. tanâˆ’1(tan3Ï€/4â€‹)

Solution:

We know that tanâˆ’1(tanÎ¸) = Î¸ when  but

So, tanâˆ’1(tan3Ï€/4â€‹) can be written as tanâˆ’1(-tan(-3Ï€/4)â€‹)

= tanâˆ’1[-tan(Ï€ – Ï€/4â€‹)]

= tanâˆ’1[-tan(Ï€/4â€‹)]

= –tanâˆ’1[tan(Ï€/4â€‹)]

= – Ï€/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) 7Ï€/6    (ii) 5Ï€/6    (iii)Ï€/3    (iv)Ï€/6

Solution:

We know that cosâˆ’1(cosÎ¸) = Î¸, Î¸ âˆˆ [0, Ï€]

cosâˆ’1(cosÎ¸) = Î¸, Î¸ âˆˆ [0, Ï€]

Here, 7Ï€/6 > Ï€

So, cosâˆ’1(cos7Ï€/6â€‹) can be written as cosâˆ’1(cos(-7Ï€/6)â€‹)

= cosâˆ’1[cos(2Ï€ – 7Ï€/6â€‹)]      [cos(2Ï€ + Î¸) = Î¸]

= cosâˆ’1[cos(5Ï€/6â€‹)]       where 5Ï€/6 âˆˆ  [0, Ï€]

Therefore, cosâˆ’1[cos(5Ï€/6â€‹)] = 5Ï€/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 = -Ï€/6â€‹

Therefore, sin[Ï€/3â€‹ – (-Ï€/6â€‹)]

= sin[Ï€/3â€‹ + (Ï€/6â€‹)]

= sin[3Ï€/6]

= sin[Ï€/2]

= 1

### (i) Ï€    (ii) -Ï€/2    (iii)0    (iv)2âˆš3

Solution:

We know, cot(âˆ’x) = âˆ’cotx

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

= tan-13 + cot-13

Since, tan-1x + cot-1x = Ï€/2

Tan-13 + cot-13 = -Ï€/2

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