NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Mathematics (Maths) Class 12

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 ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
We know thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev[∵ tan- 1(- x) = - tan - 1 x]
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.2. EvaluateNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.3. Prove thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
L.H.S.NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
= cot [cot– 1 (7)] = 7  R.H.S.
Hence proved.

Q.4. Find the value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.5. Find the value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
We know thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.6. Show that NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Ans.
L.H.S. 2 tan– 1 (– 3) = – 2 tan– 1 (3)
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence proved.

Q.7. Find the real solutions of the equation
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ 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:
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ 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 
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.9. If 2 tan–1 (cos θ) = tan–1 (2 cosec θ), then show that θ =NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
where n is any integer.
Ans.
2 tan– 1(cos θ) = tan– 1(2 cosec θ)
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ 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° orNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence proved.

Q.10. Show that NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
L.H.S. NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
R.H.S. NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
L.H.S. = R.H.S.  Hence proved.


Q.11. Solve the following equationNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
Given thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Squaring both sides we get,
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Long Answer (L.A.)

Q.12. Prove thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
L.H.S.NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Put x2 = cos θ ∴ θ = cos– 1 x2
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
[Dividing the Nr. and Den. by cos θ/2]
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence proved.

Q.13. Find the simplified form of

NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
Given thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Put NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
= cos– 1 [cos (y – x)] = y – x
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.14. Prove thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
L.H.SNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Using sin– 1 x + sin– 1 y =NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevR.H.S. Hence proved.

Q.15. Show thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NowNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence proved.

Q.16. Prove thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.17. Find the value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.18. Show thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevand justify why the other value
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevis ignored?
Ans.
To prove thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev[∴ tan (tan– 1 θ) = θ]
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

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 ....
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ 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]
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Objective Type Questions

Q.20. Which of the following is the principal value branch of cos–1x?
(a)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(b) (0, π )
(c) [0, π]
(d)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
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)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(b)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(c)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(d)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans. (d)
Solution.
Principal value branch of cosec– 1 x is
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevas 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
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
∴ x = 1
Hence, the correct answer is (b).

Q.23. The value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

(a)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(b)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(c)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(d)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans. (d)
Solution.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
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–1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(a) [1, 2]
(b) [–1, 1]

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

Ans. (a)
Solution.
LetNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ 0 ≤ x - 1 ≤ 1 ⇒ 1 ≤ x ≤ 2 ⇒ x ∈ [1, 2]
Hence, the correct answer is (a).

Q.26. IfNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevthen x is equal to
(a) 1/5
(b) 2/5
(c) 0
(d) 1

Ans. (b)
Solution.
Given thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
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)]
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
= sin [sin– 1 (0.96)]
= 0.96
Hence, the correct answer is (c).

Q.28. The value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev is equal to
(a)2/π
(b)3π/2
(c)5π/2
(d)7π/2
Ans. (a)
Solution.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the correct answer is (a).

Q.29. The value of the expression NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(a) π/6
(b) 5π/6
(c) 7π/6
(d) 1
Ans. (b)
Solution.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the correct answer is (b).

Q.30. If tan–1 x + tan–1y =NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevthen 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 =NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the correct answer is (a).

Q.31. IfNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevwhere a, x ∈ ]0, 1,
then the value of x is
(a) 0
(b) a/2

(c) a
(d)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans. (d)
Solution.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ 4 tan– 1 a = 2 tan– 1 x ⇒ 2 tan– 1 a = tan– 1 x
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the correct answer is (d).

Q.32. The value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(a) 25/24
(b) 25/7
(c) 24/25
(d) 7/24
Ans. (d)
Solution.
We have,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
LetNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the correct answer is (d).

Q.33. The value of the expressionNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(a)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(b)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(c)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(d)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Ans. (b)
Solution.
We have,NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
LetNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the correct answer is (b).

Q.34. If | x | ≤  1, then 2 tan–1 x + sin–1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevis equal to
(a) 4 tan–1
(b) 0 
(c) 2/π
(d) π
Ans. (a)
Solution.
Here, we have 2 tan-1 sin -1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
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
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(a) 0 
(b) 1 
(c) 2 
(d) infinite
Ans. (d)
Solution.

NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Which does not satisfy for any value of x.
Hence, the correct answer is (d).

Q.37. If cos–1x > sin–1x, then
(a)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(b)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(c)NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
(d) x > 0
Ans. (c)
Solution.

 Here, given that cos– 1 x > sin– 1 x
⇒ sin [cos– 1 x] > x
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
We know that – 1 ≤ x ≤ 1
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the correct answer is (c).

Fill in the blanks 
Q.38. The principal value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevis______.
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, Principal value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.39. The value ofNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevis_____.
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the value of NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.40. If cos (tan–1 x + cot–1 √3 ) = 0, then value of x is_____.
Ans.
Given that
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the value of x is √3 .

Q.41. The set of values of NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev is_____.
Ans.
LetNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev⇒sec x =NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Since, the domain of sec– 1 x is R – {– 1, 1} andNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, sec-1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevhas no set of values.

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

Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence the principal value of tan - 1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Q.43. The value of cos–1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevis_____.
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the value of cos-1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

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

Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the value of cos (sin– 1 x + cos– 1 x) = 0.

Q.45. The value of expression tanNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
is______ .
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
Hence, the value of the given expression is 1.

Q.46. If y = 2 tan–1 x + sin–1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevfor all x, then____< y <____.
Ans.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ y = 2 tan– 1 x + 2 tan– 1 x
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ y = 4 tan– 1 x
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev⇒ – 2π < y < 2π
Hence, the value of y is (– 2π, 2π).

Q.47. The result tan–1x – tan–1y = tan–1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRevis 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-1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
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–1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
 is valid is 5.
Ans.
False.
Given thatNCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev
⇒ n > p ⇒ n > 3.14
Hence, the value of n is 4.

Q.55. The principal value of sin–1NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev

Ans.
True.
NCERT Exemplar- Inverse Trigonometric Functions Notes | EduRev 

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