Inverse Trigonometric Function JEE Advanced previous year questions with solutions are given on this page. As far as the JEE Advanced exam is concerned, the inverse trigonometric function has great importance. Trigonometric functions and their inverses, the principal value of inverse trigonometric functions, properties of inverse trigonometric functions, infinite series of inverse trigonometric functions, etc., are the important topics in this chapter. By solving these questions, students will further develop adept problem-solving and time management skills that will help them to perform excellently in the JEE Advanced exam.
Download Inverse Trigonometric Function Previous Year Solved Questions PDF
Question 1: If α = cos-1(⅗), β = tan-1(⅓), where 0< α, β<π/2, then α – β is
(a) tan-1 (9/5√10)
(b) cos-1(9/5√10)
(c) tan-1(9/14)
(d) sin-1(9/5√10)
Solution:
Given α = cos-1(⅗)
cos α = 3/5
sin α = ⅘
tan α = 4/3
Also β = tan-1(⅓)
So tan β = 1/3
tan (α-β) = (tan α – tan β)/(1 + tan α tan β)
= (4/3 – ⅓)/(1 + 4/9)
= 1/(13/9)
= 9/13
So (α-β) = tan-1 (9/13)
= sin-1(9/5√10)
Hence option d is the answer.
Question 2: The principal value of tan-1 (cot 43π/4) is
(a) -3π/4
(b) 3π/4
(c) -π/4
(d) π/4
Solution:
tan-1 (cot 43π/4) = tan-1[ cot (10π + 3π/4)]
= tan-1[ cot 3π/4] (since cot (2nπ + x = cot x)]
= tan-1 (tan (π/2 – 3π/4))
= (π/2 – 3π/4))
= (2π- 3π)/4
= -π/4
Hence option c is the answer.
Question 3: If α = 3 sin-1(6/11) and β = 3 cos-1(4/9), where the inverse trigonometric functions take only the principal values, then the correct option(s) is (are)
(a) cos β > 0
(b) sin β < 0
(c) cos(α + β) > 0
(d) cos α < 0
Solution:
α = 3 sin-1(6/11) > 3 sin-11/2 > π/2
=> α > π/2
So cos α < 0
β = 3 cos-1(4/9) > 3 cos-1(½) > π
=> β > π
So cos β < 0 and sin β < 0
Now α + β > 3π/2
cos (α + β) > 0
Hence option b, c and d are correct.
Question 4: The principal value of sin-1(sin 2π/3) is
(a) -2π/3
(b) 2π/3
(c) 4π/3
(d) none of these
Solution:
The principal value of sin-1(sin 2π/3) = sin-1(sin π – π/3)
= sin-1 sin π/3
= π/3
Hence option d is the answer.
Question 5: A possible value of tan (¼ sin-1 √63/8) is:
(a) 1/(2√2)
(b) 1/√7
(c) √7 – 1
(d) 2√2 – 1
Solution:
We have to find tan (¼ sin-1 √63/8).
Let sin-1(√63/8) = θ
sin θ = √63/8
cos θ = 1/8
2 cos2(θ/2) – 1 = 1/8
=> cos2 θ/2 = 9/16
cos θ/2 = 3/4
=> (1- tan2 θ/4 )/(1 + tan2 θ/4) = 3/4
tan θ/4 = 1/√7
Hence option b is the answer.
Question 6: If S is the sum of the first 10 terms of the series tan-1(⅓) + tan-1(1/7) + tan-1(1/13) + tan-1(1/21) + …., then tan S is equal to
(a) 5/6
(b) 5/11
(c) -6/5
(d) 10/11
Solution:
S = tan-1 (1/3) + tan-1 (1/7) + tan-1 (1/13) + ….upto 10 terms
= tan-1 (2-1)/(1 + 2.1) + tan-1(3-2)/(1 + 3.2) + tan-1(4-3)/(1 + 3.4) + ….+ tan-1(11 – 10)/(1 + 11.10)
= (tan-1 2 – tan-1 1) + (tan-1 3 – tan-1 2)+ (tan-1 4 – tan-1 3) + ….+ (tan-1 11 – tan-1 10)
= (tan-1 11 – tan-1 1)
= tan-1(5/6)
tan S = 5/6
Hence option a is the answer.
Question 7: The value of sin-1 (12/13) – sin-1(3/5) is equal to
(a) π – sin-1(63/65)
(b) π/2 – sin-1(56/65)
(c) π/2 – cos-1(56/65)
(d) π – cos-1(33/65)
Solution:
sin-1 (12/13) – sin-1(3/5)
= sin-1((12/13)×(⅘) – (⅗)×(5/13))
Since sin-1x – sin-1y = sin-1(x√(1-y2) – y√(1-x2))
= sin-1(33/65)
= cos-1(56/65)
= π/2 – sin-1(56/65)
Hence option b is the answer.
Question 8: Considering only the principal values of inverse functions, the set A = {x≥0: tan-1(2x) + tan-1(3x) = π/4}
(a) contains two elements
(b) contains more than two elements
(c) is a singleton set
(d) is an empty set
Solution:
Consider tan-1(2x) + tan-1(3x) = π/4
=> tan-1(5x/(1-6x2)) = π/4
=> 5x/(1-6x2) = 1
=> 5x = 1-6x2
=> 6x2 + 5x – 1 = 0
=> (6x – 1)(x + 1) = 0
=> x = 1/6 (since x ≥ 0, x = -1 is rejected)
So A is a singleton set.
Hence option c is the answer.
Question 9: The value of tan-1[(√(1+x2) + √(1-x2))/(√(1+x2) – √(1-x2))], |x| < ½, x ≠ 0, is equal to
(a) π/4 + ½ cos-1x2
(b) π/4 + cos-1x2
(c) π/4 – ½ cos-1x2
(d) π/4 – cos-1x2
Solution:
Let x2 = cos 2θ
=> θ = ½ cos-1 x2
=> tan-1[(√(1+x2) + √(1-x2))/(√(1+x2) – √(1-x2))] = [√(1 + cos 2θ) + √(1 -cos 2θ)]/[√(1 + cos 2θ) – √(1 -cos 2θ)]
= tan-1(1 + tan θ)/(1 – tan θ)
= tan-1(tan (π/4 + θ))
= π/4 + θ
= π/4 + ½ cos-1 x2
Hence option a is the answer.
Question 10: If f(x) = 2 tan-1x + sin-1(2x/(1+x2)), x>1, then f(5) is equal to
(a) tan-1 65/156
(b) π/2
(c) π
(d) 4 tan-1(5)
Solution:
Given f(x) = 2 tan-1x + sin-1(2x/(1+x2))
When x> 1, sin-1(2x/(1+x2)) = π – 2 tan-1x
So f(x) = 2 tan-1x + π – 2 tan-1x
= π
f(5) = π
Hence option c is the answer.
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