NCERT Exemplar: Inverse Trigonometric Functions

## NCERT Exemplar: Inverse Trigonometric Functions

Q.1. Find the value of
Ans.
We know that

[∵ tan- 1(- x) = - tan - 1 x]

Hence,

Q.2. Evaluate
Ans.

Hence,

Q.3. Prove that
Ans.
L.H.S.

= cot [cot– 1 (7)] = 7  R.H.S.
Hence proved.

Q.4. Find the value of
Ans.

Hence,

Q.5. Find the value of
Ans.
We know that

Hence,

Q.6. Show that

Ans.
L.H.S. 2 tan– 1 (– 3) = – 2 tan– 1 (3)

Hence proved.

Q.7. Find the real solutions of the equation

Ans.

⇒ x2 + x = 0 ⇒ x(x + 1) = 0
⇒ x = 0 or x + 1 = 0 ⇒ x = 0 or x = – 1
Hence the real solutions are x = 0 and x = – 1.
Alternate Method:

⇒ x2 + x + 1 = 1 ⇒ x2 + x = 0
⇒ x(x + 1) = 0 ⇒ x = 0 or x + 1 = 0
∴ x = 0, x = – 1

Q.8. Find the value of the expression

Ans.

Hence,

Q.9. If 2 tan–1 (cos θ) = tan–1 (2 cosec θ), then show that θ =
where n is any integer.
Ans.
2 tan– 1(cos θ) = tan– 1(2 cosec θ)

⇒ cos θ sin θ = sin2θ
⇒ cos θ sin θ – sin2θ = 0 ⇒ sin θ(cos θ - sin θ) = 0
⇒ sin θ = 0  or cos θ - sin θ = 0 ⇒ sin θ = 0  or 1 - tan θ = 0
⇒ θ = 0  or tan θ = 1 ⇒ θ = 0° or
Hence proved.

Q.10. Show that
Ans.
L.H.S.

R.H.S.

L.H.S. = R.H.S.  Hence proved.

Q.11. Solve the following equation
Ans.
Given that

Squaring both sides we get,

Hence,

Q.12. Prove that
Ans.
L.H.S.
Put x2 = cos θ ∴ θ = cos– 1 x2

[Dividing the Nr. and Den. by cos θ/2]

Hence proved.

Q.13. Find the simplified form of

Ans.
Given that
Put

= cos– 1 [cos (y – x)] = y – x

Q.14. Prove that
Ans.
L.H.S
Using sin– 1 x + sin– 1 y =

R.H.S. Hence proved.

Q.15. Show that

Ans.

Now

Hence proved.

Q.16. Prove that
Ans.

Hence,

Q.17. Find the value of
Ans.

Q.18. Show thatand justify why the other value
is ignored?
Ans.
To prove that
[∴ tan (tan– 1 θ) = θ]

Q.19. If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression.
Ans.
If a1, a2, a3, ..., an are the terms of an arithmetic progression
∴ d = a2 – a1 = a3 – a2 = a4 – a3 ....

⇒ tan [tan-1 a2 - tan-1 a1 + tan-1 a3 tan-1 a2 + tan-1 a4 tan-1 a3 + ... + tan-1 an tan-1 an-1]
⇒ tan [tan-1 an tan-1 a1]

## Objective Type Questions

Q.20. Which of the following is the principal value branch of cos–1x?
(a)
(b) (0, π )
(c) [0, π]
(d)
Ans. (c)
Solution.
Principal value branch of cos– 1 x is [0, π]. Hence the correct answer is (c).

Q.21. Which of the following is the principal value branch of cosec–1x?
(a)
(b)
(c)
(d)
Ans. (d)
Solution.
Principal value branch of cosec– 1 x is
as cosec– 1(0) = ∞ (not defined).
Hence, the correct answer is (d).

Q.22. If 3 tan–1 x + cot–1 x = π, then x equals
(a) 0
(b) 1
(c) – 1
(d) 1/2
Ans. (b)
Solution.
Given that 3 tan– 1 x + cot– 1 x = θ
⇒ 2 tan– 1 x + tan– 1 x + cot– 1 x = θ

∴ x = 1
Hence, the correct answer is (b).

Q.23. The value of

(a)
(b)
(c)
(d)
Ans. (d)
Solution.

Hence, the correct answer is (d).

Q.24. The domain of the function cos–1 (2x – 1) is
(a) [0, 1]
(b) [–1, 1]

(c) ( –1, 1)
(d) [0, π]

Ans. (a)
Solution.
The given function is cos– 1(2x – 1)
Let f(x) = cos– 1(2x – 1)
– 1 ≤ 2x – 1 ≤ 1 ⇒ - 1 + 1 ≤ 2x ≤ 1 + 1
0 ≤ 2x ≤ 2 ⇒ 0 ≤ x ≤ 1
∴ domain of the given function is [0, 1].
Hence, the correct answer is (a)

Q.25. The domain of the function defined by f (x) = sin–1
(a) [1, 2]
(b) [–1, 1]

(c) [0, 1]
(d) none of these

Ans. (a)
Solution.
Let

⇒ 0 ≤ x - 1 ≤ 1 ⇒ 1 ≤ x ≤ 2 ⇒ x ∈ [1, 2]
Hence, the correct answer is (a).

Q.26. Ifthen x is equal to
(a) 1/5
(b) 2/5
(c) 0
(d) 1

Ans. (b)
Solution.
Given that

Hence, the correct answer is (b).

Q.27. The value of sin (2 tan–1 (.75)) is equal to
(a) 0.75
(b) 1.5
(c) 0.96
(d) sin 1.5
Ans. (c)
Solution.
Given that sin [2 tan– 1 (0.75)]

= sin [sin– 1 (0.96)]
= 0.96
Hence, the correct answer is (c).

Q.28. The value of is equal to
(a)2/π
(b)3π/2
(c)5π/2
(d)7π/2
Ans. (a)
Solution.

Hence, the correct answer is (a).

Q.29. The value of the expression
(a) π/6
(b) 5π/6
(c) 7π/6
(d) 1
Ans. (b)
Solution.

Hence, the correct answer is (b).

Q.30. If tan–1 x + tan–1y =then cot–1 x + cot–1 y equals
(a)π/ 5
(b)2π/ 5
(c)3π/5
(d) π
Ans. (a)
Solution.
Given that tan– 1 x + tan– 1 y =

Hence, the correct answer is (a).

Q.31. Ifwhere a, x ∈ ]0, 1,
then the value of x is
(a) 0
(b) a/2

(c) a
(d)
Ans. (d)
Solution.

⇒ 4 tan– 1 a = 2 tan– 1 x ⇒ 2 tan– 1 a = tan– 1 x

Hence, the correct answer is (d).

Q.32. The value of
(a) 25/24
(b) 25/7
(c) 24/25
(d) 7/24
Ans. (d)
Solution.
We have,
Let

Hence, the correct answer is (d).

Q.33. The value of the expression
(a)
(b)
(c)
(d)

Ans. (b)
Solution.
We have,
Let

Hence, the correct answer is (b).

Q.34. If | x | ≤  1, then 2 tan–1 x + sin–1is equal to
(a) 4 tan–1
(b) 0
(c) 2/π
(d) π
Ans. (a)
Solution.
Here, we have 2 tan-1 sin -1

Hence, the correct answer is (a).

Q.35. If  cos–1 α + cos–1 β + cos–1 γ = 3π, then α (β + γ) + β (γ + α) + γ (α + β) equals
(a) 0
(b) 1
(c) 6
(d) 12
Ans. (c)
Solution.

We have cos–1 α + cos–1 β + cos–1 γ = 3π
⇒ cos–1 α + cos–1 β + cos–1 γ = π + π + π
⇒ cos–1 α = π, cos–1 β = π and cos–1 γ = π
⇒ α = cos π, β = cos π and γ = cos π
α  = – 1, β = – 1 and γ = – 1
Which gives a = β = γ = –1
So α (β + γ) + β( γ+ α) + γ(α  + β)
⇒ (– 1)(– 1 – 1) + (– 1)(– 1 – 1) + (– 1)(– 1 – 1)
⇒ (– 1)(– 2) + (– 1)(– 2) + (– 1)(– 2) ⇒ 2 + 2 + 2 ⇒ 6
Hence, the correct answer is (c).

Q.36. The number of real solutions of the equation

(a) 0
(b) 1
(c) 2
(d) infinite
Ans. (d)
Solution.

Which does not satisfy for any value of x.
Hence, the correct answer is (d).

Q.37. If cos–1x > sin–1x, then
(a)
(b)
(c)
(d) x > 0
Ans. (c)
Solution.

Here, given that cos– 1 x > sin– 1 x
⇒ sin [cos– 1 x] > x

We know that – 1 ≤ x ≤ 1

Hence, the correct answer is (c).

## Fill in the blanks

Q.38. The principal value ofis______.
Ans.

Hence, Principal value of

Q.39. The value ofis_____.
Ans.

Hence, the value of

Q.40. If cos (tan–1 x + cot–1 √3 ) = 0, then value of x is_____.
Ans.
Given that

Hence, the value of x is √3 .

Q.41. The set of values of  is_____.
Ans.
Let⇒sec x =
Since, the domain of sec– 1 x is R – {– 1, 1} and
Hence, sec-1has no set of values.

Q.42. The principal value of tan–1 √3 is_____.

Ans.

Hence the principal value of tan - 1

Q.43. The value of cos–1is_____.
Ans.

Hence, the value of cos-1

Q.44. The value of cos (sin–1 x + cos–1 x), |x| ≤ 1 is______ .

Ans.

Hence, the value of cos (sin– 1 x + cos– 1 x) = 0.

Q.45. The value of expression tan
is______ .
Ans.

Hence, the value of the given expression is 1.

Q.46. If y = 2 tan–1 x + sin–1for all x, then____< y <____.
Ans.

⇒ y = 2 tan– 1 x + 2 tan– 1 x

⇒ y = 4 tan– 1 x

⇒ – 2π < y < 2π
Hence, the value of y is (– 2π, 2π).

Q.47. The result tan–1x – tan–1y = tan–1is true when value of xy is _____.
Ans.
The given result is true when xy > – 1.

Q.48. The value of cot–1 (–x) for all x ∈ R in terms of cot–1x is _______.
Ans.
cot–1(– x) = π – cot–1 x, x ∈ R [∵ as cot-1 (- x) = π - cot-1 x]

## State True or False

Q.49. All trigonometric functions have inverse over their respective domains.
Ans.
False.
We know that all inverse trigonometric functions are restricted over their domains.

Q.50. The value of the expression (cos–1 x)2 is equal to sec2 x.
Ans.
False.
We know that cos–1 x = sec-1
So (cos–1 x)2 ≠ sec2 x

Q.51. The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.
Ans.
True.
We know that all trigonometric functions are restricted over their domains to obtain their inverse functions.

Q.52. The least numerical value, either positive or negative of angle θ is called principal value of the inverse trigonometric function.
Ans.
True.

Q.53. The graph of inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging x and y axes.
Ans.
True.
We know that the domain and range are interchanged in the graph of inverse trigonometric functions to that of their corresponding trigonometric functions.

Q.54. The minimum value of n for which tan–1
is valid is 5.
Ans.
False.
Given that

⇒ n > p ⇒ n > 3.14
Hence, the value of n is 4.

Q.55. The principal value of sin–1

Ans.
True.

The document NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

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## FAQs on NCERT Exemplar: Inverse Trigonometric Functions - Mathematics (Maths) for JEE Main & Advanced

 1. What are the basic properties of inverse trigonometric functions?
Ans. The basic properties of inverse trigonometric functions include the principal value branch, the range of values they can take, and their relationship with the original trigonometric functions.
 2. How are inverse trigonometric functions used in solving trigonometric equations?
Ans. Inverse trigonometric functions are used to solve trigonometric equations by undoing the trigonometric operations applied to find the original angle.
 3. What are the restrictions on the domains of inverse trigonometric functions?
Ans. The domains of inverse trigonometric functions are restricted to ensure that they are one-to-one functions, which means that each input value corresponds to a unique output value.
 4. How do inverse trigonometric functions help in finding angles in trigonometry problems?
Ans. Inverse trigonometric functions help in finding angles in trigonometry problems by allowing us to find the angle that corresponds to a specific trigonometric ratio.
 5. Can inverse trigonometric functions be used to calculate the angles of a triangle?
Ans. Yes, inverse trigonometric functions can be used to calculate the angles of a triangle when the lengths of the sides are known, by using trigonometric ratios and their inverses to find the angles.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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