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Exercise 2.1 Page No: 41
Find the principal values of the following:
1. π¬π’π§βπ (βπ
π)
2.
3. Cosec -1 (2)
4.
5.
6. tan-1(-1)
7.
8.
9.
10.
Solution 1: Consider y = sinβ1 (β1
2)
Solve the above equation, we have sin y = -1/2 We know that sin Ο/6 = Β½ So, sin y = - sin Ο/6
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Since range of principle value of sin-1 is
Principle value of sinβ1 (β1
2) is - Ο/6.
Solution 2:
Let y =
Cos y = cos Ο/6 (as cos Ο/6 = β3 / 2 )
y = Ο/6
Since range of principle value of cos-1 is [0, Ο]
Therefore, Principle value of is Ο/6
Solution 3: Cosec -1 (2)
Let y = Cosec -1 (2)
Cosec y = 2
We know that, cosec Ο /6 = 2
So Cosec y = cosec Ο /6
Since range of principle value of cosec-1 is
Therefore, Principle value of Cosec -1 (2) is Ξ /6.
Solution 4:
Let y =
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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tan y = - tan Ο/3
or tan y = tan (-Ο/3)
Since range of principle value of tan-1 is
Therefore, Principle value of is β Ο/3.
Solution 5:
y =
cos y = -1/2
cos y = cos(Ο β Ο/3) = cos (2Ο/3)
Since principle value of cos-1 is [0, Ο]
Therefore, Principle value of is 2Ο/3.
Solution 6: tan-1(-1)
Let y = tan-1(-1)
tan (y) = -1
tan y = -tan Ο/4
Since principle value of tan-1 is
Therefore, Principle value of tan-1(-1) is - Ο/4.
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Solution 7:
y =
sec y = 2/β3
Since principle value of sec-1 is [0, Ο]
Therefore, Principle value of is Ο/6
Solution 8:
y =
cot y = β3
cot y = Ο/6
Since principle value of cot-1 is [0, Ο]
Therefore, Principle value of is Ο/6.
Solution 9:
Let y =
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Since principle value of cos-1 is [0, Ο]
Therefore, Principle value of is 3 Ο/ 4.
Solution 10.
Ley y =
Since principle value of cosec-1 is
Therefore, Principle value of is β Ο/4
Find the values of the following:
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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13. If sinβ1 x = y, then
(A) 0 β€ y β€ Ο
(B) βπ
πβ€ π² β€
π
π
(C) 0 < y < Ο
(D) βπ
π< π² <
π
π
14. tanβ1 ( βπ) - sec -1 (-2) is equal to
(A) Ο
(B) - Ο/3
(C) Ο/3
(D) 2 Ο/3
Solution 11.
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Solution 12:
Solution 13: Option (B) is correct.
Given sinβ1 x = y,
The range of the principle value of sin-1 is
Therefore, βΟ
2β€ y β€
Ο
2
Solution 14:
Option (B) is correct.
tanβ1 ( β3) - sec -1 (-2) = tan-1 (tan Ο/3) β sec-1 (-sec Ο/3)
= Ο/3 - sec-1 (sec (Ο - Ο/3))
= Ο/3 - 2Ο/3 = - Ο/3
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Exercise 2.2 Page No: 47
Prove the following
1.
Solution:
(Use identity: )
Let then
Now, RHS
= sinβ1(3π₯ β 4π₯3)
= sinβ1(3 sin ΞΈ β 4 sin3 ΞΈ)
= sinβ1(sin 3 ΞΈ)
= 3 ΞΈ
= 3 sin-1 x
= LHS
Hence Proved
2.
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Solution:
Using identity:
Put x = cos ΞΈ
ΞΈ = cos-1 (x)
Therefore, cos 3 ΞΈ = 4x3 β 3x
RHS:
= cos-1 (cos 3 ΞΈ)
= 3 ΞΈ
= 3 cos-1 (x)
= LHS
Hence Proved.
3.
Solution:
Using identity:
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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LHS =
= tan -1 (125/250)
= tan -1 (1/2)
= RHS
Hence Proved
4.
Solution:
Use identity:
LHS
=
=
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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= tan-1(4/3) + tan-1(1/7)
Again using identity:
We have,
= π‘ππβ1(28+3
21β4 )
= tan-1 (31/17)
RHS
Write the following functions in the simplest form:
5.
Solution:
Letβs say x = tan ΞΈ then ΞΈ = tan -1 x
We get,
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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This is simplest form of the function.
6.
Solution:
Let us consider, x = sec ΞΈ, then ΞΈ = sec-1 x
= tan-1(cot ΞΈ)
= tan-1 tan(Ο/2 - ΞΈ)
= (Ο/2 - ΞΈ)
= Ο/2 β sec-1 x
This is simplest form of the given function.
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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7.
Solution:
8.
Solution:
Divide numerator and denominator by cos x, we have
πππβπ(
πππ(π)
πππ(π)β
πππ(π)
πππ(π)
πππ(π)
πππ(π)+
πππ(π)
πππ(π)
)
= πππβπ(π β
πππ(π)
πππ(π)
π +πππ(π)
πππ(π)
)
=
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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= tan -1 tan(Ο/4 β x)
= Ο/4 β x
9.
Solution:
Put x = a sin ΞΈ, which implies sin ΞΈ = x/a and ΞΈ = sin-1(x/a)
Substitute the values into given function, we get
= tan-1(tan ΞΈ)
= ΞΈ
= sin-1(x/a)
10.
Solution: After dividing numerator and denominator by a^3 we have
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Put x/a = tan ΞΈ and ΞΈ = tan-1(x/a)
=
= tan-1 (tan 3 ΞΈ)
= 3 ΞΈ
= 3 tan-1(x/a)
Find the values of each of the following:
11.
Solution:
= tan-1 (2 cos Ο/3)
= tan-1(2 x Β½)
= tan-1 (1)
= tan-1 (tan (Ο/4))
= Ο/4
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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12. cot (tanβ1a + cotβ1a)
Solution:
cot (tanβ1a + cotβ1a) = cot Ο/2 = 0
Using identity: tanβ1a + cotβ1a = Ο/2
13.
Solution:
Put x = tan ΞΈ and y = tan Π€, we have
= tan1/2[sin-1 sin 2 ΞΈ + cos-1 cos 2 Π€]
= tan (1/2) [2 ΞΈ + 2 Π€]
= tan (ΞΈ + Π€)
=
= (x+y) / (1-xy)
14. If , then find the value of x.
Solution: We know that, sin 90 degrees = sin Ο/2 = 1
So, given equation turned as,
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Using identity: = Ο/2
We have,
Which implies, the value of x is 1/5.
15. If , then find the value of x.
Solution:
We have reduced the given equation using below identity:
or
or
ππ 2π₯2 β 4
π₯2 β 4 β π₯2 + 1= tan (
π
4)
or (2x^2 - 4)/-3 = 1
or 2x^2 = 1
or x = Β± 1
β2
The value of x is either 1
β2ππ β
1
β2
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Find the values of each of the expressions in Exercises 16 to 18.
16. π¬π’π§βπ(π¬π’π§ (ππ
π))
Solution:
Given expression is sinβ1(sin (2π
3))
First split 2π
3ππ
(3πβπ)
3 ππ π β
π
3
After substituting in given we get,
sinβ1(sin (2π
3)) = sinβ1(sin (π β
π
3)) =
π
3
Therefore, the value of sinβ1 (sin (2π
3)) is
π
3
17. πππβπ(πππ§ (ππ
π))
Solution:
Given expression is π‘ππβ1(tan (3π
4))
First split 3π
4ππ
(4πβπ)
4 ππ π β
π
4
After substituting in given we get,
π‘ππβ1(tan (3π
4)) = tanβ1(tan (π β
π
4 )) = β
π4
The value of π‘ππβ1(tan (3π
4)) is
βπ
4.
18. πππ§ (πππβπ (π
π) + ππ¨πβπ π
π)
Solution:
Given expression is tan (π ππβ1 (3
5) + cotβ1 3
2)
Putting, π ππβ1 (3
5) = π₯ πππ cotβ1(
3
2) = π¦
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Or sin(x) = 3/5 and cot y = 3/2
Now, sin(x) = 3/5 => cos x = β1 β sin2 π₯ = 4/5 and sec x = 5/4
(using identities: cos x = β1 β sin2 π₯ and sec x = 1/cos x)
Again, tan x = βsec2 π₯ β 1 = β25
16β 1 = ΒΎ and tan y = 1/cot(y) = 2/3
Now, we can write given expression as,
tan (π ππβ1 (3
5) + cotβ1 3
2) = tan(x + y)
=
= 17/6
19. ππ¨π¬βπ(ππ¨π¬ππ
π) is equal to
(A) 7Ο/6 (B) 5 Ο/6 (C) Ο/3 (D) Ο/6
Solution:
Option (B) is correct.
Explanation:
cosβ1(cos7Ο
6) = cosβ1(cos (2Ο β
7Ο
6)
(As cos (2 Ο β A) = cos A)
Now 2Ο β 7Ο
6=
12Οβ7Ο
6=
5Ο
6
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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20. is equal to
(A) Β½ (B) 1/3 (C) ΒΌ (D) 1
Solution:
Option (D) is correct
Explanation:
First solve for: sinβ1 (β1
2)
= - Ο/6
Again,
= sin(Ο/2)
= 1
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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21. tan-1 βπ β cot-1 ( -βπ) is equal to
(A) Ο (B) -Ο/2 (C) 0 (D) 2βπ
Solution:
Option (B) is correct.
Explanation:
tan-1 β3 β cot-1 ( -β3) can be written as
=π
3β (π β
π
6)
= π
3β
5π
6
=β3π
6
= - π/2
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Miscellaneous Exercise Page No: 51
Find the value of the following:
1. ππ¨π¬βπ(ππ¨π¬πππ
π)
Solution:
First solve for, cos13π
6= cos(2 π +
π
6) = cos
π
6
Now: cosβ1(cos13π
6) = cosβ1(cos
π
6) =
π
6 β [0, Ο]
[As cos-1 cos(x) = x if x β [0, Ο] ]
So the value of cosβ1(cos13π
6) is
π
6.
2. πππβπ(πππ§ππ
π)
Solution:
First solve for, tan7π
6= tan(π +
π
6) = tan
π
6
Now: π‘ππβ1(tan7π
6) = tanβ1(tan
π
6) =
π
6 β (-Ο/2, Ο/2)
[As tan-1 tan(x) = x if x β (-Ο/2, Ο/2) ]
So the value of π‘ππβ1(tan7π
6) is
π
6.
3. Prove that π π¬π’π§βπ π
π= πππ§βπ ππ
π
Solution: Step 1: Find the value of cos x and tan x
Let us considersinβ1 3
5= π₯, then sin x = 3/5
So, cos x = β1 β sin2 π₯ = β1 β (3
5)
2
= 4/5
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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tan x = sin x/ cos x = ΒΎ
Therefore, x = tan-1 (3/4), substitute the value of x,
sinβ1 3
5= tanβ1 (
3
4) β¦..(1)
Step 2: Solve LHS
2 sinβ13
5= 2 tanβ1
3
4
Using identity: 2tan-1 x = tanβ1 = tanβ1(2π₯
1βπ₯2), we get
= tanβ1(2(
3
4)
1β(3
4)
2)
= tan-1(24/7)
= RHS
Hence Proved.
4. Prove that π¬π’π§βπ π
ππ+ π¬π’π§βπ π
π= πππ§βπ ππ
ππ
Solution:
Let sinβ1 (8
17) = π₯ then sin x = 8/17
Again, cos x = β1 β sin2 π₯ = β1 β64
289 = 15/17
And tan x = sin x / cos x = 8/ 15
Again,
Let sinβ1 (3
5) = π¦ then sin y = 3/5
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Again, cos y = β1 β sin2 π¦ = β1 β9
25 = 4/5
And tan y = sin y / cos y = ΒΎ
Solve for tan(x + y), using below identity,
= 32+45
60β24
= 77/36
This implies x + y = tan-1(77/36)
Substituting the values back, we have
sinβ1 8
17+ sinβ1 3
5= tanβ1 77
36 (Proved)
5. Prove that ππ¨π¬βπ (π
π) + ππ¨π¬βπ (
ππ
ππ) = ππ¨π¬βπ (
ππ
ππ)
Solution:
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Solve the expression, Using identity: cos (ΞΈ + Ο) = cosΞΈ cos Ο - sinΞΈ sin Ο
= 4/5 x 12/13 β 3/5 x 5/13
= (48-15)/65
= 33/65
This implies cos (ΞΈ + Ο) = 33/65
or ΞΈ + Ο = cos-1 (33/65)
Putting back the value of ΞΈ and Ο, we get
cosβ1 (4
5) + cosβ1 (
12
13) = cosβ1 (
33
65)
Hence Proved.
6. Prove that ππ¨π¬βπ (ππ
ππ) + π¬π’π§βπ (
π
π) = π¬π’π§βπ (
ππ
ππ)
Solution:
Solve the expression, Using identity: sin (ΞΈ + Ο) = sin ΞΈ cos Ο + cos ΞΈ sin Ο
= 12/13 x 3/5 + 12/13 x 3/5
= (20+36)/65
= 56/65
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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or sin (ΞΈ + Ο) = 56/65
or ΞΈ + Ο) = sin -1 56/65
Putting back the value of ΞΈ and Ο, we get
cosβ1 (12
13) + sinβ1 (
3
5) = sinβ1 (
56
65)
Hence Proved.
7. Prove that πππ§βπ (ππ
ππ) = π¬π’π§βπ (
π
ππ) + ππ¨π¬βπ (
π
π)
Solution:
Solve the expression, Using identity:
=
5
12+
4
3
1β5
12Γ
4
3
= 63/16
(ΞΈ + Ο) = tan-1 (63/16)
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Putting back the value of ΞΈ and Ο, we get
tanβ1 (63
16) = sinβ1 (
5
13) + cosβ1 (
3
5)
Hence Proved.
8. Prove that πππ§βπ (π
π) + πππ§βπ (
π
π) + πππ§βπ (
π
π) + πππ§βπ (
π
π) =
π
π
Solution:
LHS = (tanβ1 (1
5) + tanβ1 (
1
7)) + ( tanβ1 (
1
3) + tanβ1 (
1
8) )
Solve above expressions, using below identity:
= tanβ1(1
5+
1
7
1β1
5Γ
1
7
) + tanβ1(1
3+
1
8
1β1
3Γ
1
8
)
After simplifying, we have
= tan-1 (6/17) + tan-1 (11/23)
Again, applying the formula, we get
After simplifying,
= tan-1(325/325)
= tan-1 (1)
= Ο/4
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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9. Prove that πππ§βπ βπ = π
πππ¨π¬βπ πβπ
π+π , x β (0, 1)
Solution:
Let tanβ1 βπ₯ = ΞΈ , then βπ₯ = tan ΞΈ
Squaring both the sides
tan2 ΞΈ = x
Now, substitute the value of x in 1
2cosβ1 1βπ₯
1+π₯ , we get
= Β½ cos-1 (cos 2 ΞΈ)
= Β½ (2 ΞΈ)
= ΞΈ
= tanβ1 βπ₯
10. Prove that cot-1 (βπ+π¬π’π§ π + βπβπ¬π’π§ πβπ+π¬π’π§ π β βπβπ¬π’π§ π
) = ππ, x β (0, Ο/4)
Solution: We can write 1+ sin x as,
And
LHS:
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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= cot-1(2 cos (
π₯
2)
2 sin(π₯
2))
= cot-1 (cot (x/2)
= x/2
11. Prove that πππ§βπ(βπ+π ββπβπ
βπ+π + βπβπ) =
π
πβ
π
πππ¨π¬βπ π , β
π
βπ β€ π β€ π
[Hint: Put x = cos 2 ΞΈ]
Solution:
Divide each term by β2 cos ΞΈ
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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= RHS
Hence proved
12. Prove that ππ
πβ
π
ππ¬π’π§βπ π
π=
π
ππ¬π’π§βπ πβπ
π
Solution:
LHS = 9π
8β
9
4sinβ1 1
3
β¦β¦.(1)
(Using identity: )
Let ΞΈ = cos-1 (1/3), so cos ΞΈ = 1/3
As
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Using equation (1), 9
4sinβ1 2β2
3
Which is right hand side of the expression.
Solve the following equations:
13. 2tan-1 (cos x) = tan-1 (2 cosec x)
Solution:
Cot x = 1
x = Ο/4
14. Solve
Solution:
Put x = tan ΞΈ
This implies
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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Ο/4 β ΞΈ = ΞΈ /2
or 3ΞΈ /2 = Ο/4
ΞΈ = Ο/6
Therefore, x = tan ΞΈ = tan Ο/6 = 1/β3
15. is equal to
Solution:
Option (D) is correct.
Explanation:
Again, Letβs say
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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This implies,
Which shows,
16. then x is equal to
(A) 0, Β½ (B) 1, Β½ (C) 0 (D) Β½
Solution:
Option (C) is correct.
Explanation:
Now,
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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(As x = sin ΞΈ)
After simplifying, we get
x(2x β 1) = 0
x = 0 or 2x β 1 = 0
x = 0 or x = Β½
Equation is not true for x = Β½. So the answer is x = 0.
17. is equal to
(A) Ο/2 (B) Ο/3 (C) Ο/4 (D) -3 Ο/4
Solution:
Option (C) is correct.
Explanation:
Given expression can be written as,
Class - XII _ Maths Ch. 2 - Inverse-Trigonometric-Functions
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