﻿ Class 12 NCERT Math Solution
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TOPICS
Exercise - 2.1

Question-1 :-  Find the principal value of sin-1(-1/2).

Solution :-
```   Let sin-1(1/2) = θ.
sin θ = 1/2
sin θ = sin(π/6)  {Range = [-π/2, π/2]}
θ = π/6
The principal value of sin-1(1/√2) = π/6
```

Question-2 :-  Find the principal value of cos-1(√3/2).

Solution :-
```   Let cos-1(√3/2) = θ.
cos θ = √3/2
cos θ = cot(π/6)  {Range = [0, π]}
θ = π/6
The principal value of cos-1(√3/2) = π/6
```

Question-3 :-  Find the principal value of cosec-1(2).

Solution :-
```   Let cosec-1(2) = θ.
cosec θ = 2
cosec θ = coesec(π/6)  {Range = [0, π]}
θ = π/6
The principal value of cosec-1(2) = π/6
```

Question-4 :-  Find the principal value of tan-1(-√3).

Solution :-
```   Let tan-1(-√3) = θ.
tan θ = -√3
tan θ = -tan(π/3)  {Range = [-π/2, π/2]}
tan θ = tan(-π/3)
θ = -π/3
The principal value of tan-1(-√3) = -π/3
```

Question-5 :-  Find the principal value of cos-1(-1/2).

Solution :-
```   Let cos-1(-1/2) = θ.
cos θ = -1/2
cos θ = -cos(π/3)  {Range = [0, π]}
cos θ = cos(π - π/3)
cos θ = cos(2π/3)
θ = 2π/3
The principal value of cos-1(-1/2) = 2π/3
```

Question-6 :-  Find the principal value of tan-1(-1).

Solution :-
```   Let tan-1(-1) = θ.
tan θ = -1
tan θ = -tan(π/4)  {Range = [-π/2, π/2]}
tan θ = tan(-π/4)
θ = -π/4
The principal value of tan-1(-√3) = -π/4
```

Question-7 :-  Find the principal value of sec-1(2/√3).

Solution :-
```   Let sec-1(2/√3) = θ.
sec θ = 2/√3
sec θ = sec(π/6)  {Range = [0, π]}
θ = π/6
The principal value of sec-1(2/√3) = π/6
```

Question-8 :-  Find the principal value of cot-1(√3).

Solution :-
```   Let cot-1(√3) = θ.
cot θ = √3
cot θ = cot(π/6)  {Range = [0, π]}
θ = π/6
The principal value of cot-1(√3) = π/6
```

Question-9 :-  Find the principal value of cos-1(-1/√2).

Solution :-
```   Let cos-1(-1/√2) = θ.
cos θ = -1/√2
cos θ = -cos(π/4)  {Range = [0, π]}
cos θ = cos(π - π/4)
cos θ = cos(3π/4)
θ = 3π/4
The principal value of cos-1(-1/√2) = 3π/4
```

Question-10 :-  Find the principal value of cosec-1(-√2).

Solution :-
```   Let cosec-1(-√2) = θ.
cosec θ = -√2
cosec θ = -cosec(π/4)  {Range = [-π/2, π/2]}
cosec θ = cosec(-π/4)
θ = -π/4
The principal value of cosec-1(-√2) = -π/4
```

Question-11 :-  Find the value of tan-1(1) + cos-1(-1/2) + sin-1(-1/2).

Solution :-
```   tan-1(1) + cos-1(-1/2) + sin-1(-1/2)

Let tan-1(1) = θ₁.
tan θ₁ = 1
tan θ₁ = tan(π/4)  {Range = [-π/2, π/2]}
θ₁ = π/4

Let cos-1(-1/2) = θ₂.
cos θ₂ = -1/2
cos θ₂ = -cos(π/3)  {Range = [0, π]}
cos θ₂ = cos(π - π/3)
cos θ₂ = cos(2π/3)
θ₂ = 2π/3

Let sin-1(-1/2) = θ₃.
sin θ₃ = -1/2
sin θ₃ = -sin(π/6)  {Range = [-π/2, π/2]}
sin θ₃ = sin(-π/6)
θ₃ = π/6

Now, θ₁ + θ₂ + θ₃
= π/4 + 2π/3 + π/6
= (3π + 8π + 2π)/12
= 13π/12
```

Question-12 :-  Find the value of cos-1(1/2) + 2sin-1(1/2).

Solution :-
```    cos-1(1/2) + 2sin-1(1/2)

Let cos-1(1/2) = θ₁.
cos θ₁ = 1/2
cos θ₁ = cos(π/3)  {Range = [0, π]}
θ₁ = π/3

Let sin-1(1/2) = θ₂
sin θ₂ = 1/2
sin θ₂ = sin(π/6)  {Range = [-π/2, π/2]}
θ₂ = π/6

Now, θ₁ + 2θ₂
= π/3 + 2 x π/6
= π/3 + π/3
= 2π/3
```

Question-13 :-  If sin-1 x = y, then
(A) 0 ≤ y ≤ π   (B) -π/2 ≤ y ≤ π/2   (C) 0 < y < π   (D) -π/2 < y < π/2

Solution :-
```   (B) -π/2 ≤ y ≤ π/2, because sin-1 x range is [-π/2, π/2].
```

Question-14 :-  tan-1(√3) - sec-1(-2) is equal to
(A) π   (B) -π/3   (C) π/3   (D) 2π/3

Solution :-
```   tan-1(√3) - sec-1(-2)

Let tan-1(√3) = θ₁.
tan θ₁ = √3
tan θ₁ = tan(π/3)  {Range = [-π/2, π/2]}
θ₁ = π/3

Let sec-1(-2) = θ₂.
sec θ₂ = -1/2
sec θ₂ = -sec(π/3)  {Range = [0, π]}
sec θ₂ = sec(π - π/3)
sec θ₂ = sec(2π/3)
θ₂ = 2π/3

Now, θ₁ - θ₂
= π/3 - 2π/3
= (π - 2π)/3
= -π/3

Option (B).
```
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