﻿ Class 12 NCERT Math Solution
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TOPICS
Miscellaneous

Example-1 :-  Find the value of sin-1 (sin 3π/5).

Solution :-
```   sin-1 (sin 3π/5)
Here, 3π/5 is not in range of sin-1 i.e., [-π/2, π/2]
= sin-1 [sin (π - 3π/5)]
= sin-1 [sin 2π/5]
= 2π/5
```

Example-2 :-  Show that sin-1 (3/5) - sin-1 (8/17) = cos-1 (84/85).

Solution :-
```   sin-1 (3/5) - sin-1 (8/17) = cos-1 (84/85)
Firstly,
Let sin-1 (3/5) = θ₁
sin θ₁ = 3/5
cos θ₁ = √1 - sin²θ₁
cos θ₁ = √1 - (3/5)²
cos θ₁ = √1 - 9/25
cos θ₁ = √16/25
cos θ₁ = 4/5

Secondly,
Let sin-1 (8/17) = θ₂
sin θ₂ = 8/17
cos θ₂ = √1 - sin²θ₂
cos θ₂ = √1 - (8/17)²
cos θ₂ = √1 - 64/289
cos θ₂ = √225/289
cos θ₂ = 15/17

Now, cos(θ₁ - θ₂) = cos θ₁ cos θ₂ + sin θ₁ sin θ₂
= 4/5 x 15/17 + 3/5 x 8/17
= 12/17 + 24/85
= (60 + 24)/85
= 84/85

cos(θ₁ - θ₂) = 84/85
θ₁ - θ₂ = cos-1 (84/85)
```

Example-3 :-  Show that sin-1 (12/13) + cos-1 (4/5) + tan-1 (63/16) = π.

Solution :-
```   sin-1 (12/13) + cos-1 (4/5) + tan-1 (63/16) = π
By using of L.H.S
Firstly,
Let sin-1 (12/13) = θ₁
sin θ₁ = 12/13
cos θ₁ = √1 - sin²θ₁
cos θ₁ = √1 - (12/13)²
cos θ₁ = √1 - 144/169
cos θ₁ = √25/169
cos θ₁ = 5/13
Now, tan θ₁ = sin θ₁/ cos θ₁ = 12/13 x 13/5 = 12/5
tan θ₁ = 12/5

Secondly,
Let cos-1 (4/5) = θ₂
cos θ₂ = 4/5
sin θ₂ = √1 - cos²θ₂
sin θ₂ = √1 - (4/5)²
sin θ₂ = √1 - 16/25
sin θ₂ = √9/25
sin θ₂ = 3/5
Now, tan θ₂ = sin θ₂/ cos θ₂ = 3/5 x 5/4 = 3/4
tan θ₂ = 3/4

We have, tan (θ₁ + θ₂) = [(tan θ₁ + tan θ₂) ÷ (1 - tan θ₁ tan θ₂)]
= [(12/5 + 3/4) ÷ (1 - 12/5 x 3/4)]
= [(48 + 15)/20 ÷ (1 - 36/20)]
= 63/20 ÷ (-16/20)
= 63/20 x (-20/16)
= -63/16
tan (θ₁ + θ₂) = -63/16

Thirdly,
tan-1 (63/16) = θ₃
tan θ₃ = 63/16

tan (θ₁ + θ₂) = -tan θ₃
i.e., tan (θ₁ + θ₂) = tan (θ₃) or tan (θ₁ + θ₂) = tan (π – θ₃)
Therefore, θ₁ + θ₂ = -θ₃ or θ₁ + θ₂ = π – θ₃
Since θ₁, θ₂ and θ₃ are positive, θ₁ + θ₂ ≠ -θ₃
Hence, θ₁ + θ₂ + θ₃ = π = R.H.S
```

Example-4 :-  Simplify if a/b tan x > -1.

Solution :-
``` ```

Example-5 :-  Solve tan-1 2x + tan-1 3x = π/4.

Solution :-
```   tan-1 2x + tan-1 3x = π/4
tan-1 [(2x + 3x) ÷ (1 - 2x.3x)] = π/4
5x ÷ (1 - 6x²) = tan π/4
5x ÷ (1 - 6x²) = 1
5x = 1 - 6x²
6x² + 5x - 1 = 0
6x² + 6x - x - 1 = 0
6x(x + 1) - 1(x + 1) = 0
(x + 1)(6x - 1) = 0
x = -1, x = 1/6
```
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