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All questions of Inverse Trigonometric Functions for JEE Exam

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Can you explain the answer of this question below:

  • A:

    4

  • B:

    1/4

  • C:

    2

  • D:

    none of these.

The answer is a.

Divey Sethi answered
sin θ is 3/5.
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4

 If xy + yz + zx = 1, then, tan–1x + tan–1y + tan–1z =
  • a)
    π
  • b)
    π/2
  • c)
    1
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered
If xy + yz + zx = 1,
Then, 1 – xy – yz – zx = 0 ----- eqn1 
tan-1x + tan-1y = tan-1((x+y)/(1-xy))
tan-1x + tan-1y + tan-1z = tan-1((x+y)/(1-xy)) + tan-1z
= tan-1( ( ((x+y)/(1-xy)) + z ) / (1 – ((x+y)*z)/(1+xy)))
= tan-1((x+y+z-xyz)/(1-xy-yz-zx))
= tan-1((x+y+z-xyz)/0)  [From eqn1]
= π/2

The simplest form of for x > 0 is …​
  • a)
    x
  • b)
    -x/2
  • c)
    2x
  • d)
    x/2
Correct answer is option 'D'. Can you explain this answer?

Mira Joshi answered
tan-1(1-cosx/1+cosx)½
= tan-1{(2sin2 x/2) / (2cos2 x/2)}½
= tan-1{(2sin2 x/2) / (2cos2 x/2)}
= tan-1(tan x/2)
= x/2

Evaluate sin(3 sin–10.4)
​a)0.56
b)0.31
c)0.64
d)0.9
Correct answer is 'D'. Can you explain this answer?

Subhankar Choudhary answered
3sin^-1(x) = sin^-1(3x - 4x^3) when -1/2<=x<=1/2
Definitely 0.4 comes in this range of x and so
3sin^-1(0.4) = sin^-1[3*0.4 - 4*0.4^3]
3sin^-1(0.4) = sin^-1[1.2 - 4*0.064]
3sin^-1(0.4) = sin^-1[1.2 - 0.256]
3sin^-1(0.4) = sin^-1[0.944]
Finally , sin(3sin^-1(0.4)) = sin{sin^-1(0.944)} = 0.944

The maximum value of sin x + cos x is
  • a)
    1
  • b)
    2
  • c)
    √2
  • d)
Correct answer is option 'C'. Can you explain this answer?

Shreya Gupta answered
sinx + cosx=sinx + sin(90-x)=2sin{(x+90-x)/2}cos{(x-90+x)/2}using the formula 

The value of  tan–1(1) + cos–1(–1/2) + sin–1(–1/2) is equal to -
  • a)
    3π/4
  • b)
    5π/12
  • c)
    π/4
  • d)
    13π/12
Correct answer is option 'C'. Can you explain this answer?

Bs Academy answered
Since, here we are considering only principle solutions
tan-1(1) = π/4
cos-1(-1/2) = 2π/3
sin-1(-1/2) = -π/6
Sol : π/4 +2π/3 -π/6
: 3π/4

Can you explain the answer of this question below:

Domain of f(x) = cos–1 x + cot–1 x + cosec–1 x is

  • A:

    [–1, 1]

  • B:

    R

  • C:

  • D:

    {–1, 1}

The answer is D.

Suresh Iyer answered
Domain of cos-1x = [-1 ,1] 
Domain of cot-1x = R
Domain of cosec-1x = R - (-1,1)
So taking intersection of domains of all three we have  only {-1 ,1}
As cos-1x is confined between [-1,1] and cosec-1x is not defined between (-1,1), So only -1 and 1 are left.

The value ofcos150−sin150 is
  • a)
  • b)
  • c)
    0
  • d)
     
Correct answer is option 'D'. Can you explain this answer?

Poojan Angiras answered
Heyy!!! write cos 15 and sin 15 as cos(60-45) and sin(60-45) respectively.And then apply formula of cos(a+b) and sin(a+b).Proceed as the question u will get correct answer :-):-)^_^

Evaluate :cos (tan–1 x)
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Preeti Iyer answered
tan−1 x = θ , so that  x=tanθ . We need to determine  cosθ .
sec2θ = 1 + tan2θ = 1 + x2 
∴s ecθ = ±√(1+x2) 
Then, cos(tan−1x) = cosθ=1/secθ = ±1/√(1+x2)

  • a)
    4
  • b)
    1/4
  • c)
    2
  • d)
    none of these.
Correct answer is option 'A'. Can you explain this answer?

Krishna Iyer answered
sin θ is 3/5.
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4

If ab + bc + ca = 0, then find 1/a2-bc + 1/b2 – ca + 1/c2- ab
  • a)
    π
  • b)
    0
  • c)
    -1
  • d)
    2π
Correct answer is option 'B'. Can you explain this answer?

Preeti Iyer answered
Given ab+bc+ca=0 and asked to find value 1/(a2-bc ) + 1/(c2-ab) + 1/(a2-bc)Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + 

If x< 0 then value of tan-1(x) + tan-1  is equal to 
  • a)
    π/2
  • b)
    –π/2
  • c)
    0
  • d)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Lavanya Menon answered
 Let y = tan-1 x
x = tan y
⇒ 1/x = 1/tan y
⇒ 1/x = cot y
⇒ 1/x = tan(π/2 - y)
⇒ π/2 - y = tan-1 (1/x)
As we know that y = tan-1 x
π/2 = tan-1(x) + tan-1(1/x)

Range of f(x) = sin–1 x + tan–1 x + sec–1 x is
  • a)
  • b)
  • c)
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Tanuja Kapoor answered
The function f is defined for x = ±1 
Now, f(1) = sin-1(1) + sec-1(1) + tan-1(1) 
= π/2 + 0 + π/4 = 3π/4 
Also, f(-1) = sin-1(-1) + sec-1(-1) + tan-1(-1) 
= - π/2 + π - 0 - π/4 
= 3π/4
Range = { π/4, 3π/4}

The number of solutions of the equation sin-1 x - cos-1 x = sin-1(1/2) is
a) 3
b) 1
c) 2
d) infinite.
Correct answer is option 'B'. Can you explain this answer?

Raghavendra Ghoshal answered
Solution:

Given equation is sin^-1(x) - cos^-1(x) = sin^-1(1/2)

We know that sin(x) + cos(x) = √2 cos(x - π/4)

So, sin^-1(x) - cos^-1(x) = π/2 - sin^-1(√2x)

Therefore, the given equation becomes π/2 - sin^-1(√2x) = sin^-1(1/2)

sin(sin^-1(x)) = xsin(π/2 - sin^-1(√2x)) = √[1 - 2x^2]

√[1 - 2x^2] = 1/2

2x^2 = 3/4

x = ±√3/2

Therefore, the given equation has only 1 solution, which is x = √3/2 or x = -√3/2.

Hence, the correct answer is option B.

  • a)
     tan-1 (9/5√10)
  • b)
    cos-1(9/5√10)
  • c)
    tan-1(9/14)
  • d)
    sin-1(9/5√10)
Correct answer is option 'D'. Can you explain this answer?

Defence Exams answered
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.

The complete solution set of the inequality [cot–1 x]2  – 6[cot–1 x] + 9 ≤0 is (where [ * ] denotes the greatest integer function)
  • a)
    (– ∞, cot 3]
  • b)
     [cot 3, cot 2]
  • c)
     [cot 3, ∞)
  • d)
    None of these
Correct answer is 'A'. Can you explain this answer?

Janani Roy answered
Since $\cot x>0$ for $0<><\pi $,="" the="" given="" inequality="" is="" equivalent="" to=""><\frac{\pi}{4}$, or=""><\frac{\pi}{2}$. thus,="" the="" solution="" set="" is=""><><\frac{\pi}{2}.$$>

The value of   , if ∠C = 900 in triangle ABC, is 
  • a)
    π/4
  • b)
    π/3
  • c)
    π/2
  • d)
    π
Correct answer is option 'A'. Can you explain this answer?

Swati Verma answered
a2 + b2 = c2
tan-1 A + tan-1 B = tan(A + B)/(1 - AB)
=> tan-1 (a/(b+c)) + tan-1 (b/(c+a))
tan-1 [(a/(b+c) + b/(c+a))/(1 - ab/(b+c)(c+a))]
=> tan-1(ac + a2 + b2 + bc)/(bc + ab + c2 + ac - ab)
= tan-1 (ac + bc + a2 + b2)/(ac + bc + c2)
= tan-1 (ac + bc + c2)/(ac + bc + c2)  As we know that {a2 + b2 = c2}
= tan-1(1)
= π/4

tan–1 a + tan–1b, where a > 0, b > 0, ab > 1, is equal to
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
tan-1 a + tan-1 b = tan-1[(a+b)/(1 - ab)]    (ab>1)
tan-1[-(a+b)/(1 - ab)] 
π - tan-1[(a+b)/(1 - ab)]

The number of real solution of 
  • a)
    zero
  • b)
    one
  • c)
    two
  • d)
    infinite 
Correct answer is option 'C'. Can you explain this answer?

Pallavi Desai answered
From function it is clear that
(1) x(x + 1) > 0    ∴ Domain of square root function.
(2) x2 + x + 1>0   ∴ Domain of square root function. 
Domain of sin-1 function. From (2) and (3)

Evaluate sin(3 sin–1 0.4)​
  • a)
    0.9
  • b)
    0.31
  • c)
    0.64
  • d)
    0.56
Correct answer is option 'D'. Can you explain this answer?

Mohit Rajpoot answered
3sin^-1(x) = sin^-1(3x - 4x^3) when -1/2<=x<=1/2
Definitely 0.4 comes in this range of x and so
3sin^-1(0.4) = sin^-1[3*0.4 - 4*0.4^3]
3sin^-1(0.4) = sin^-1[1.2 - 4*0.064]
3sin^-1(0.4) = sin^-1[1.2 - 0.256]
3sin^-1(0.4) = sin^-1[0.944]
Finally , sin(3sin^-1(0.4)) = sin{sin^-1(0.944)} = 0.944

 is equal to
  • a)
  • b)
  • c)
  • d)
Correct answer is 'C'. Can you explain this answer?

Nikita Singh answered
Let y = tan−1[(a−x)/(a+x)]1/2
put x = a cos θ
Now, y = tan−1[(a − a cos θ)/(a + a cos θ)]1/2
⇒ y = tan−1((1 − cos θ)/(1 + cos θ))1/2
⇒ y = tan−1[[(1−cos θ)(1−cos θ)]/(1+cos θ)(1−cos θ)]1/2
⇒ y = tan−1[(1−cos θ)2/1 − cos2θ]1/2
⇒ y = tan−1[(1 − cos θ)/sin θ]
⇒ y = tan−1[2 sin2(θ/2)/2 sin(θ/2) . cos(θ/2)]
⇒ y = tan−1[tan(θ/2)]
⇒ y = θ/2
⇒ y = 1/2 cos−1(x/a)

  • a)
  • b)
    1
  • c)
    0
  • d)
    none of these.
Correct answer is option 'B'. Can you explain this answer?

Nikita Singh answered
sin-1√3/5 = A
Sin A = √3/5 , cos A = √22/5 
Therefore Cos-1√3/5 = B
Cos B = √3/5 , sin B = √22/5  
sin(A+B) = sinA cosB + cosA sinB
= √3/5 * √3/5 + √22/5 * √22/5
= 3/25 * 22/25
= 25/25 
= 1

The value of sin-1[cos{cos-1 (cos x) + sin-1(sin x)}], where x   
  • a)
    π/2
  • b)
    π/4
  • c)
    –π/4
  • d)
    –π/2
Correct answer is option 'D'. Can you explain this answer?

Riya Banerjee answered
x implies(pi/2, pi)
= sin-1(cos(cos-1(cosx) + sin-1(sinx)))
= sin-1(cos(x + π - x)
= sin-1(cos π)
= sin-1(-1)
= -π/2

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