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NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced PDF Download

NCERT Exemplar: Inverse Trigonometric Functions


Short Answer Type Questions

Q.1. Find the value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
We know thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced[∵ tan- 1(- x) = - tan - 1 x]
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.2. EvaluateNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.3. Prove thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
L.H.S.NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
= cot [cot– 1 (7)] = 7  R.H.S.
Hence proved.

Q.4. Find the value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.5. Find the value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
We know thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.6. Show that NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Ans.
L.H.S. 2 tan– 1 (– 3) = – 2 tan– 1 (3)
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence proved.

Q.7. Find the real solutions of the equation
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.9. If 2 tan–1 (cos θ) = tan–1 (2 cosec θ), then show that θ =NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
where n is any integer.
Ans.
2 tan– 1(cos θ) = tan– 1(2 cosec θ)
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
Hence proved.

Q.10. Show that NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
L.H.S. NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
R.H.S. NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
L.H.S. = R.H.S.  Hence proved.


Q.11. Solve the following equationNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Squaring both sides we get,
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Long Answer Type Questions

Q.12. Prove thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
L.H.S.NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Put x2 = cos θ ∴ θ = cos– 1 x2
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
[Dividing the Nr. and Den. by cos θ/2]
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence proved.

Q.13. Find the simplified form of

NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Put NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
= cos– 1 [cos (y – x)] = y – x
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.14. Prove thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
L.H.SNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Using sin– 1 x + sin– 1 y =NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & AdvancedR.H.S. Hence proved.

Q.15. Show thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NowNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence proved.

Q.16. Prove thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.17. Find the value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.18. Show thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedand justify why the other value
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedis ignored?
Ans.
To prove thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced[∴ tan (tan– 1 θ) = θ]
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced

Objective Type Questions

Q.20. Which of the following is the principal value branch of cos–1x?
(a)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(b) (0, π )
(c) [0, π]
(d)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
(b)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(c)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(d)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans. (d)
Solution.
Principal value branch of cosec– 1 x is
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedas 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 | Mathematics (Maths) for JEE Main & Advanced
∴ x = 1
Hence, the correct answer is (b).

Q.23. The value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

(a)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(b)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(c)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(d)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans. (d)
Solution.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
(a) [1, 2]
(b) [–1, 1]

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

Ans. (a)
Solution.
LetNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
⇒ 0 ≤ x - 1 ≤ 1 ⇒ 1 ≤ x ≤ 2 ⇒ x ∈ [1, 2]
Hence, the correct answer is (a).

Q.26. IfNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedthen x is equal to
(a) 1/5
(b) 2/5
(c) 0
(d) 1

Ans. (b)
Solution.
Given thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
= sin [sin– 1 (0.96)]
= 0.96
Hence, the correct answer is (c).

Q.28. The value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced is equal to
(a)2/π
(b)3π/2
(c)5π/2
(d)7π/2
Ans. (a)
Solution.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct answer is (a).

Q.29. The value of the expression NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(a) π/6
(b) 5π/6
(c) 7π/6
(d) 1
Ans. (b)
Solution.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct answer is (b).

Q.30. If tan–1 x + tan–1y =NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedthen 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 | Mathematics (Maths) for JEE Main & Advanced

NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct answer is (a).

Q.31. IfNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedwhere a, x ∈ ]0, 1,
then the value of x is
(a) 0
(b) a/2

(c) a
(d)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans. (d)
Solution.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
⇒ 4 tan– 1 a = 2 tan– 1 x ⇒ 2 tan– 1 a = tan– 1 x
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct answer is (d).

Q.32. The value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(a) 25/24
(b) 25/7
(c) 24/25
(d) 7/24
Ans. (d)
Solution.
We have,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
LetNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct answer is (d).

Q.33. The value of the expressionNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(a)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(b)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(c)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(d)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Ans. (b)
Solution.
We have,NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
LetNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct answer is (b).

Q.34. If | x | ≤  1, then 2 tan–1 x + sin–1NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedis 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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
(a) 0 
(b) 1 
(c) 2 
(d) infinite
Ans. (d)
Solution.

NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
(b)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(c)NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
(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 | Mathematics (Maths) for JEE Main & Advanced
We know that – 1 ≤ x ≤ 1
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct answer is (c).

Fill in the blanks 

Q.38. The principal value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedis______.
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, Principal value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.39. The value ofNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedis_____.
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.40. If cos (tan–1 x + cot–1 √3 ) = 0, then value of x is_____.
Ans.
Given that
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of x is √3 .

Q.41. The set of values of NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced is_____.
Ans.
LetNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced⇒sec x =NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Since, the domain of sec– 1 x is R – {– 1, 1} andNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, sec-1NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedhas no set of values.

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

Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence the principal value of tan - 1NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Q.43. The value of cos–1NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedis_____.
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of cos-1NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

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

Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of cos (sin– 1 x + cos– 1 x) = 0.

Q.45. The value of expression tanNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
is______ .
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the given expression is 1.

Q.46. If y = 2 tan–1 x + sin–1NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advancedfor all x, then____< y <____.
Ans.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
⇒ y = 2 tan– 1 x + 2 tan– 1 x
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
⇒ y = 4 tan– 1 x
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced⇒ – 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 | Mathematics (Maths) for JEE Main & Advancedis 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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
 is valid is 5.
Ans.
False.
Given thatNCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced
⇒ n > p ⇒ n > 3.14
Hence, the value of n is 4.

Q.55. The principal value of sin–1NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced

Ans.
True.
NCERT Exemplar: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced 

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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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.
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