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# NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

## JEE : NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

The document NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Q.1. Find the value of
Ans.
We know that

[âˆµ tan- 1(- x) = - tan - 1 x]

Hence,

Q.2. Evaluate
Ans.

Hence,

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

= cot [cotâ€“ 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 tanâ€“ 1 (â€“ 3) = â€“ 2 tanâ€“ 1 (3)

Hence proved.

Q.7. Find the real solutions of the equation

Ans.

â‡’ x2 + x = 0 â‡’ x(x + 1) = 0
â‡’ x = 0 or x + 1 = 0 â‡’ x = 0 or x = â€“ 1
Hence the real solutions are x = 0 and x = â€“ 1.
Alternate Method:

â‡’ x2 + x + 1 = 1 â‡’ x2 + x = 0
â‡’ x(x + 1) = 0 â‡’ x = 0 or x + 1 = 0
âˆ´ x = 0, x = â€“ 1

Q.8. Find the value of the expression

Ans.

Hence,

Q.9. If 2 tanâ€“1 (cos Î¸) = tanâ€“1 (2 cosec Î¸), then show that Î¸ =
where n is any integer.
Ans.
2 tanâ€“ 1(cos Î¸) = tanâ€“ 1(2 cosec Î¸)

â‡’ cos Î¸ sin Î¸ = sin2Î¸
â‡’ cos Î¸ sin Î¸ â€“ sin2Î¸ = 0 â‡’ sin Î¸(cos Î¸ - sin Î¸) = 0
â‡’ sin Î¸ = 0  or cos Î¸ - sin Î¸ = 0 â‡’ sin Î¸ = 0  or 1 - tan Î¸ = 0
â‡’ Î¸ = 0  or tan Î¸ = 1 â‡’ Î¸ = 0Â° 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

â‡’
Squaring both sides we get,

â‡’
Hence,

Q.12. Prove that
Ans.
L.H.S.
Put x2 = cos Î¸ âˆ´ Î¸ = cosâ€“ 1 x2

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

Hence proved.

Q.13. Find the simplified form of

Ans.
Given that
Put

= cosâ€“ 1 [cos (y â€“ x)] = y â€“ x

Q.14. Prove that
Ans.
L.H.S
Using sinâ€“ 1 x + sinâ€“ 1 y =

R.H.S. Hence proved.

Q.15. Show that

Ans.

Now
â‡’
â‡’

Hence proved.

Q.16. Prove that
Ans.

âˆ´
Hence,

Q.17. Find the value of
Ans.

Q.18. Show thatand justify why the other value
is ignored?
Ans.
To prove that
[âˆ´ tan (tanâ€“ 1 Î¸) = Î¸]

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
âˆ´ d = a2 â€“ a1 = a3 â€“ a2 = a4 â€“ a3 ....

â‡’ 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]
â‡’ tan [tan-1 an tan-1 a1]

Objective Type Questions

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

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

Q.22. If 3 tanâ€“1 x + cotâ€“1 x = Ï€, then x equals
(a) 0
(b) 1
(c) â€“ 1
(d) 1/2
Ans. (b)
Solution.
Given that 3 tanâ€“ 1 x + cotâ€“ 1 x = Î¸
â‡’ 2 tanâ€“ 1 x + tanâ€“ 1 x + cotâ€“ 1 x = Î¸

âˆ´ 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 cosâ€“1 (2x â€“ 1) is
(a) [0, 1]
(b) [â€“1, 1]

(c) ( â€“1, 1)
(d) [0, Ï€]

Ans. (a)
Solution.
The given function is cosâ€“ 1(2x â€“ 1)
Let f(x) = cosâ€“ 1(2x â€“ 1)
â€“ 1 â‰¤ 2x â€“ 1 â‰¤ 1 â‡’ - 1 + 1 â‰¤ 2x â‰¤ 1 + 1
0 â‰¤ 2x â‰¤ 2 â‡’ 0 â‰¤ x â‰¤ 1
âˆ´ 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) = sinâ€“1
(a) [1, 2]
(b) [â€“1, 1]

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

Ans. (a)
Solution.
Let

â‡’ 0 â‰¤ x - 1 â‰¤ 1 â‡’ 1 â‰¤ x â‰¤ 2 â‡’ x âˆˆ [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
â‡’
â‡’
â‡’
â‡’
Hence, the correct answer is (b).

Q.27. The value of sin (2 tanâ€“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 tanâ€“ 1 (0.75)]

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

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

Hence, the correct answer is (a).

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

Hence, the correct answer is (b).

Q.30. If tanâ€“1 x + tanâ€“1y =then cotâ€“1 x + cotâ€“1 y equals
(a)Ï€/ 5
(b)2Ï€/ 5
(c)3Ï€/5
(d) Ï€
Ans. (a)
Solution.
Given that tanâ€“ 1 x + tanâ€“ 1 y =

Hence, the correct answer is (a).

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

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

â‡’ 4 tanâ€“ 1 a = 2 tanâ€“ 1 x â‡’ 2 tanâ€“ 1 a = tanâ€“ 1 x
â‡’
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
âˆ´

Hence, the correct answer is (d).

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

Ans. (b)
Solution.
We have,
Let
â‡’

Hence, the correct answer is (b).

Q.34. If | x | â‰¤  1, then 2 tanâ€“1 x + sinâ€“1is equal to
(a) 4 tanâ€“1
(b) 0
(c) 2/Ï€
(d) Ï€
Ans. (a)
Solution.
Here, we have 2 tan-1 sin -1

Hence, the correct answer is (a).

Q.35. If  cosâ€“1 Î± + cosâ€“1 Î² + cosâ€“1 Î³ = 3Ï€, then Î± (Î² + Î³) + Î² (Î³ + Î±) + Î³ (Î± + Î²) equals
(a) 0
(b) 1
(c) 6
(d) 12
Ans. (c)
Solution.

We have cosâ€“1 Î± + cosâ€“1 Î² + cosâ€“1 Î³ = 3Ï€
â‡’ cosâ€“1 Î± + cosâ€“1 Î² + cosâ€“1 Î³ = Ï€ + Ï€ + Ï€
â‡’ cosâ€“1 Î± = Ï€, cosâ€“1 Î² = Ï€ and cosâ€“1 Î³ = Ï€
â‡’ Î± = cos Ï€, Î² = cos Ï€ and Î³ = cos Ï€
âˆ´ Î±  = â€“ 1, Î² = â€“ 1 and Î³ = â€“ 1
Which gives a = Î² = Î³ = â€“1
So Î± (Î² + Î³) + Î²( Î³+ Î±) + Î³(Î±  + Î²)
â‡’ (â€“ 1)(â€“ 1 â€“ 1) + (â€“ 1)(â€“ 1 â€“ 1) + (â€“ 1)(â€“ 1 â€“ 1)
â‡’ (â€“ 1)(â€“ 2) + (â€“ 1)(â€“ 2) + (â€“ 1)(â€“ 2) â‡’ 2 + 2 + 2 â‡’ 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 cosâ€“1x > sinâ€“1x, then
(a)
(b)
(c)
(d) x > 0
Ans. (c)
Solution.

Here, given that cosâ€“ 1 x > sinâ€“ 1 x
â‡’ sin [cosâ€“ 1 x] > x

We know that â€“ 1 â‰¤ x â‰¤ 1

Hence, the correct answer is (c).

Fill in the blanks
Q.38. The principal value ofis______.
Ans.

âˆ´
Hence, Principal value of

Q.39. The value ofis_____.
Ans.

Hence, the value of

Q.40. If cos (tanâ€“1 x + cotâ€“1 âˆš3 ) = 0, then value of x is_____.
Ans.
Given that

Hence, the value of x is âˆš3 .

Q.41. The set of values of  is_____.
Ans.
Letâ‡’sec x =
Since, the domain of secâ€“ 1 x is R â€“ {â€“ 1, 1} and
Hence, sec-1has no set of values.

Q.42. The principal value of tanâ€“1 âˆš3 is_____.

Ans.

Hence the principal value of tan - 1

Q.43. The value of cosâ€“1is_____.
Ans.

Hence, the value of cos-1

Q.44. The value of cos (sinâ€“1 x + cosâ€“1 x), |x| â‰¤ 1 is______ .

Ans.

Hence, the value of cos (sinâ€“ 1 x + cosâ€“ 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 tanâ€“1 x + sinâ€“1for all x, then____< y <____.
Ans.

â‡’ y = 2 tanâ€“ 1 x + 2 tanâ€“ 1 x

â‡’ y = 4 tanâ€“ 1 x

â‡’ â€“ 2Ï€ < y < 2Ï€
Hence, the value of y is (â€“ 2Ï€, 2Ï€).

Q.47. The result tanâ€“1x â€“ tanâ€“1y = tanâ€“1is true when value of xy is _____.
Ans.
The given result is true when xy > â€“ 1.

Q.48. The value of cotâ€“1 (â€“x) for all x âˆˆ R in terms of cotâ€“1x is _______.
Ans.
cotâ€“1(â€“ x) = Ï€ â€“ cotâ€“1 x, x âˆˆ R [âˆµ as cot-1 (- x) = Ï€ - 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 (cosâ€“1 x)2 is equal to sec2 x.
Ans.
False.
We know that cosâ€“1 x = sec-1
So (cosâ€“1 x)2 â‰  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 Î¸ 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 tanâ€“1
is valid is 5.
Ans.
False.
Given that
â‡’
â‡’ n > p â‡’ n > 3.14
Hence, the value of n is 4.

Q.55. The principal value of sinâ€“1

Ans.
True.

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## Mathematics (Maths) Class 12

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