Trigonometry is a measurement of a triangle and it is included with inverse functions. There are six basic inverse trigonometric functions: arcsine, arccosine, arctangent, arccotangent, arcsecant, and arccosecant. In this article, we will discuss the topic of inverse trigonometric functions along with JEE previous year problems. Students can make use of the solutions that we are offering and be one step ahead in the competition. JEE previous year questions with solutions are extremely helpful for students preparing for JEE and other competitive exams. All solutions are prepared by our subject experts to present factual information to the students related to problems covered in JEE previous exams. Students can download freely these solutions and start practising offline.
Properties of Inverse Trigonometric Functions
It’s important to learn the formulas well; the better you know the basic identities, the easier it will be to recognize the problem and solve them.
| Let’s have a look at some of the properties of inverse trigonometric functions:
1. sin(arcsin x) = x and arcsin(sin y) = y where -1 ≤ x ≤ 1 and -π/2 ≤ y ≤ π/2 2. tan(arctan x) = x and arctan(tan y) = y where -π/2 < y < π/2 3. sec(arcsec x) = x and arcsec(sec y) = y where, |x| ≥ 1 and 0 ≤ y < π/2 4. sin-1 (1/x) = csc-1 x 5. cos-1 (1/x) = sec-1 x 6. tan-1 (1/x) = cot-1 x 7. sin-1x + cos-1 x = π/2 8. tan-1x + cot-1x = π/2 9. sec-1 x + csc-1 x = π/2 |
JEE Main Past Year Questions With Solutions on Application of Integrations
Question 1: A value of x satisfying the equation sin[cot−1(1 + x)] = cos[tan-1x], is:
(a) 1/2
(b) -1
(c) 0
(d) -1/2
Answer: (d)
Solution:
Given equation is: sin[cot−1(1 + x)] = cos[tan-1x]
Let cot−1(1 + x)] = a => cot a = 1 + x …(1)
And tan-1x =b => x = tan b …(2)
Solve (1) for a in terms of sin function:
cot a = 1 + x
We know, cosec a = √(1+cot2a) = √(1 + (1 + x)2 = √(x2 + 2x + 2)
Also, sin a = 1/cosec a
⇒ sin a = 1/√(x2 + 2x + 2)
Or a = sin-1[1/ √(x2 + 2x + 2)]
Solve (2) for b in terms of cos function:
x = tan b
We know, sec b = √(1+tan2b) = √(1+x2)
Also, cos b = 1/sec b = 1/√(1+x2)
Or b = cos-1[1/√(1+x2)]
Given equation =>
sin(sin-1[1/ √(x2 + 2x + 2)]) = cos[cos-1(1/√(1+x2))]
⇒ 1/ √(x2 + 2x + 2) = 1/√(1+x2)
Squaring both sides and solving, we get
2x = -1
Or x = -1/2
Question 2: 1 + cot2 (sin-1 x) =
(a) 1/2x
(b) x2
(c) 1/x2
(d) 2/x
Solution:
Let sin-1 x = θ
So sin θ = x
1 + cot2 θ = cosec2 θ
= 1/x2
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 trig functions take only the principal values, then the right option is
(a) cosβ > 0
(b) cos(α + β) > 0
(c) sin β < 0
(d) cosα < 0
Answer: (b), (c) and (d)
Solution:
α = 3 sin-1 (6/11) > 3 sin-1 (6/12) > 3 sin-1 (1/2) > π / 2
β = 3 cos-1 (4/9) > 3 cos-1 (4/8) > 3 cos-1 (1/2) > π
α + β > 3π/2
sin β < 0,cos α < 0,cos (α + β) > 0
Question 4: If f'(x) = tan-1(sec x + tan x), -π/2 < x < π/2, and f(0) = 0 then f(1) is equal to
(a) (π+1)/4
(b) (π+2)/4
(c) ¼
(d) (π-1)/4
Answer: (a)
Solution:
f'(x) = tan-1(sec x + tan x) = tan-1(1/cosx + sin x/cos x) = tan-1[(1+sin x)/cos x]
f'(x) = π/4 + x/2
⇒ f(x) = (π/4) x + x2/4 + c
At f(0) = 0 => c = 0
At f(1) = π/4 + 1/4 = (π+1)/4
Question 5: The value of
|x| < 1/2, x ≠ 0, is equal to
(a) π/4 + (1/2) cos-1x2
(b) π/4 + cos-1x2
(c) π/4 – (1/2) cos-1x2
(d) π/4 – cos-1x2
Answer: (a)
Solution: Let x2 = cos 2A => A = (1/2) cos-1 (x2) …(1)
We know tan π/4 = 1; using in the above equation, we get
= tan-1(tan(π/4 + A))
= π/4 + A
Using (1)
= π/4 + (1/2) cos-1(x2)
Question 6:
(a) 23/25
(b) 25/23
(c) 23/24
(d) 24/23
Answer: (b)
Solution:
= 25/23
Question 7: Let f: [0, 4π] -> [0, π] be defined by f(x) = cos-1(cos x). The number of points x Є [0, 4π] satisfying the equation f(x) = (10-x)/10 is
(a) 1
(b) 2
(c) 3
(d) None of these
Answer: (c)
Solution:
Draw graph for f(x) = cos-1(cos x) and f(x) = (10-x)/10
Both equations intersect at three different points, so the number of solutions is 3.
Question 8: Let f(x) = x cos-1(sin(-|x|)), x Є (-π/2, π/2) then which of the following is true?
(a) f′(0) = -π/2
(b) f’ is decreasing in (-π/2, 0) and increasing in (0, π/2)
(c) f is not differentiable at x = 0
(d) f’ is increasing in (-π/2, 0) and decreasing in (0, π/2)
Answer: (b)
Solution:
f(x) = x cos-1(sin(-|x|)) (Given)
⇒ f(x) = x cos-1(-sin(|x|))
[As sine is an odd function]⇒ f(x) = x [π – cos-1(sin(|x|))
⇒ f(x) = x [π – (π/2 – sin-1(sin(|x|))
⇒ f(x) = x(π/2 + |x|)
Therefore, f'(x) is decreasing (-π/2, 0) and increasing in (0, π/2).
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JEE Advanced Maths Inverse Trigonometric Function Previous Year Questions with Solutions
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