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All questions of Chapter 2 - Inverse Trigonometric Functions for JEE Exam
Can you explain the answer of this question below:
- A:
4
- B:
1/4
- C:
2
- D:
none of these.
The answer is a.
4
1/4
2
none of these.
|
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
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
Can you explain the answer of this question below:Value of 
is
- A:
π/2
- B:
1/x
- C:
x
- D:
π
The answer is a.
Value of is
π/2
1/x
x
π
|
Case Studies answered |
Tan-1(1/x) = cot-1(x) and tan-1(x) + cot-1(x) =90degree
Evaluate sin(3 sin–10.4)a)0.56b)0.31c)0.64d)0.9Correct 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
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]
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
Value of 
- a)
- b)
- c)-1
- d)
Correct answer is option 'A'. Can you explain this answer?
Value of
a)
b)
c)
-1
d)
|
Aravind Rane answered |
Cot-1[sin(-90)] =cot-1(-1) =π-cot-1(1) =π-π/4 =3π/4
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?
The maximum value of sin x + cos x is
a)
1
b)
2
c)
√2
d)
|
Shreya Gupta answered |
sinx + cosx=sinx + sin(90-x)=2sin{(x+90-x)/2}cos{(x-90+x)/2}using the formula
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?
The simplest form of for x > 0 is …
for x > 0 is …a)
x
b)
-x/2
c)
2x
d)
x/2
|
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
= tan-1{(2sin2 x/2) / (2cos2 x/2)}½
= tan-1{(2sin2 x/2) / (2cos2 x/2)}
= tan-1(tan x/2)
= x/2
is given by
Evaluate :cos (tan–1 x)- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Evaluate :cos (tan–1 x)
a)
b)
c)
d)
|
Sushil Kumar 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)
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?
a)
4
b)
1/4
c)
2
d)
none of these.
|
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
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
The value ofcos150−sin150 is- a)
- b)
- c)0
- d)
Correct answer is option 'D'. Can you explain this answer?
The value ofcos150−sin150 is
a)
b)
c)
0
d)
|
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 :-):-)^_^
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?
If ab + bc + ca = 0, then find 1/a2-bc + 1/b2 – ca + 1/c2- ab
a)
π
b)
0
c)
-1
d)
2π
|
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 +
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.
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.
If sin A + cos A = 1, then sin 2A is equal to- a)2
- b)1/2
- c)0
- d)1
Correct answer is option 'C'. Can you explain this answer?
If sin A + cos A = 1, then sin 2A is equal to
a)
2
b)
1/2
c)
0
d)
1
|
|
Suresh Iyer answered |
(Sin A + Cos A)2 = sin2A + cos2A + 2 sinAcosA
1 = 1 + Sin 2A
so, Sin 2A = 0
Hence A = 0
Evaluate: sin (2 sin–10.6)- a)0.6
- b)0.66
- c)0.36
- d)0.96
Correct answer is option 'D'. Can you explain this answer?
Evaluate: sin (2 sin–10.6)
a)
0.6
b)
0.66
c)
0.36
d)
0.96
|
Nikita Singh answered |
Let, sin-1(0.6) = A…………(1)
( Since, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) )
( Since, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) )When x = π/2, then tan x, is- a)∞ or −∞
- b)−∞−∞
- c)∞
- d)not defined.
Correct answer is option 'D'. Can you explain this answer?
When x = π/2, then tan x, is
a)
∞ or −∞
b)
−∞−∞
c)
∞
d)
not defined.
|
Poulomi Datta answered |
tan π/2 = n.d.i.e. not defined.
cos2θ is not equal to- a)1−2sin2θ
- b)2cos2θ−1
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
cos2θ is not equal to
a)
1−2sin2θ
b)
2cos2θ−1
c)
d)
|
Priyanka Sharma answered |
cos 2θ is equal to cos2θ - sin2θ = 2cos2θ - 1
- a)±√3.
- b)0
- c)
- d)1
Correct answer is option 'D'. Can you explain this answer?
a)
±√3.
b)
0
c)
d)
1
|
Devendra Singh answered |
Apply apply formula of Sin inverse X + Cos inverse X and the question will be solved
- a)7/25
- b)-24/25
- c)24/25
- d)none of these.
Correct answer is option 'B'. Can you explain this answer?
a)
7/25
b)
-24/25
c)
24/25
d)
none of these.
|
Om Desai answered |
Put 
therefore the given expressionis sin2θ = 2sinθcosθ
therefore the given expressionis sin2θ = 2sinθcosθIf ƒ(x) = √(2tan(x)),then f-1(√(2)) =- a)0
- b)1
- c)√(2)
- d)2
Correct answer is option 'B'. Can you explain this answer?
If ƒ(x) = √(2tan(x)),then f-1(√(2)) =
a)
0
b)
1
c)
√(2)
d)
2
|
|
Siddharth Chakraborty answered |
If what? Please provide more information for me to assist you better.
- a)
- b)1
- c)0
- d)none of these.
Correct answer is option 'B'. Can you explain this answer?
a)
b)
1
c)
0
d)
none of these.
|
|
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
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
If and x + y + z = xyz, then a value of tan−1x+tan−1y+tan−1z is- a)3π/2
- b)π/2
- c)π
- d)none of these.
Correct answer is option 'C'. Can you explain this answer?
If and x + y + z = xyz, then a value of tan−1x+tan−1y+tan−1z is
a)
3π/2
b)
π/2
c)
π
d)
none of these.
|
|
Ruchi Kumar answered |
This question is incomplete and cannot be answered. Please provide more information.
The principal value of cos–1 (cos 5) is- a)5
- b)π – 5
- c)5 – π
- d)2π – 5
Correct answer is option 'D'. Can you explain this answer?
The principal value of cos–1 (cos 5) is
a)
5
b)
π – 5
c)
5 – π
d)
2π – 5
|
Madhurima Patel answered |
Sorry, your question is incomplete. Please provide more information.
Evaluate 
- a)
- b)
- c)
- d)
Correct answer is 'A'. Can you explain this answer?
Evaluate
a)
b)
c)
d)
|
Ritu Singh answered |
Correct Answer :- a
Explanation : cos-1(12/13) + sin-1 (3/5)
⇒ sin-1 5/13 + sin-1 3/5
Using the formula, sin-1x + sin-1y = sin-1( x√1-y² + y√1-x² )
⇒ sin-1 ( 5/13√1-9/25 + 3√1-25/169)
⇒ sin-1 ( 5/13 × 4/5 + 3/5 × 12/13)
⇒ sin-1 (30 + 36 / 65)
⇒ sin-1 (56/ 65)
- a)4
- b)1/4
- c)2
- d)none of these.
Correct answer is 'A'. Can you explain this answer?
a)
4
b)
1/4
c)
2
d)
none of these.
|
Ishita Deshpande answered |
SinQ is 3/5.
on simplifying:
(secQ+tanQ)/(secQ-tanQ)
We get...(1+sinQ)/(1-sinQ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
on simplifying:
(secQ+tanQ)/(secQ-tanQ)
We get...(1+sinQ)/(1-sinQ)
=(1+3/5)/(1-3/5)
=(8/2)
=4
If ƒ(x) = tan(x), then f-1(1/√(3)) =- a)π/6
- b)π/4
- c)π/3
- d)π
Correct answer is option 'A'. Can you explain this answer?
If ƒ(x) = tan(x), then f-1(1/√(3)) =
a)
π/6
b)
π/4
c)
π/3
d)
π
|
|
Sanjana Mishra answered |
Please provide more context or ask a specific question.
The simplest form of 
- a)
- b)
- c)2x
- d)
Correct answer is option 'A'. Can you explain this answer?
The simplest form of
a)
b)
c)
2x
d)
|
Seelam Yogitha answered |
Divide inner functions(both nr n Dr) with cosx,, tan^-(1-tanx/1+tanx)=tan^-(tan(π/4 -X) )=π/4 -x
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?
Evaluate sin(3 sin–1 0.4)
a)
0.9
b)
0.31
c)
0.64
d)
0.56
|
Mira Sharma answered |
sin(3 sin^−1 0.4)
= sin[sin^−1(3 x 0.4−4 x 0.4^2)] [∵3sin^−1 x = sin^−1(3x − 4x^3)]
= sin[sin^−1(1.2 − 4 x 0.16)]
= sin[sin^−1(1.2 − 0.64)]
= sin[sin^−1(0.56)] = 0.56
is equal to- a)
- b)
- c)
- d)
Correct answer is 'C'. Can you explain this answer?
is equal toa)
b)
c)
d)
|
|
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)
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)
are :
- a)4/3
- b)-3/8
- c)4/3 or -3/8
- d)none of these.
Correct answer is option 'B'. Can you explain this answer?
a)
4/3
b)
-3/8
c)
4/3 or -3/8
d)
none of these.
|
Chirag Verma answered |
The correct option is B.
cot (cos−1x) is equal to- a)
- b)
- c)
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
cot (cos−1x) is equal to
a)
b)
c)
d)
none of these
|
Athul Patel answered |
Put,
cos-1x = θ ⇒ x = cos θ ⇒ cos θ
cos-1x = θ ⇒ x = cos θ ⇒ cos θ
Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion (A): Range of cot–1 x is (0, π)
Reason (R): Domain of tan–1 x is R.- a)Both A and R are true and R is the correct explanation of A
- b)Both A and R are true but R is NOT the correct explanation of A
- c)A is true but R is false
- d)A is false and R is True
Correct answer is option 'B'. Can you explain this answer?
Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion (A): Range of cot–1 x is (0, π)
Reason (R): Domain of tan–1 x is R.
Assertion (A): Range of cot–1 x is (0, π)
Reason (R): Domain of tan–1 x is R.
a)
Both A and R are true and R is the correct explanation of A
b)
Both A and R are true but R is NOT the correct explanation of A
c)
A is true but R is false
d)
A is false and R is True
|
Geetika Shah answered |
Value of 
- a)
- b)tan-1 3 + tan-1 2x.
- c)
- d)tan-1 6 + tan-1 x.
Correct answer is option 'A'. Can you explain this answer?
Value of
a)
b)
tan-1 3 + tan-1 2x.
c)
d)
tan-1 6 + tan-1 x.
|
Tejas Verma answered |
tan-1[(3+2x)/(2-3x)]
=> tan-1[2(3/2 + x)/2(1-3/2x)]
=> tan-1[(3/2 + x)/(1-3/2x)]
=> tan-1(3/2) + tan-1(x)
=> tan-1[2(3/2 + x)/2(1-3/2x)]
=> tan-1[(3/2 + x)/(1-3/2x)]
=> tan-1(3/2) + tan-1(x)
sin (200)0 + cos (200)0 is- a)Zero
- b)Positive
- c)Zero or positive.
- d)Negative
Correct answer is option 'D'. Can you explain this answer?
sin (200)0 + cos (200)0 is
a)
Zero
b)
Positive
c)
Zero or positive.
d)
Negative
|
|
Dipika Dey answered |
To determine the sign of sin(200)0 and cos(200)0, we need to recall the unit circle and the values of sine and cosine for angles in the different quadrants.
The unit circle is a circle with a radius of 1 unit, centered at the origin (0,0) on a coordinate plane. It helps us understand the relationship between angles and the coordinates on the circle.
The circle is divided into four quadrants: Quadrant I (0°-90°), Quadrant II (90°-180°), Quadrant III (180°-270°), and Quadrant IV (270°-360°).
In Quadrant I, both the x-coordinate (cosine) and the y-coordinate (sine) are positive. In Quadrant II, the x-coordinate (cosine) is negative, but the y-coordinate (sine) is positive. In Quadrant III, both the x-coordinate (cosine) and the y-coordinate (sine) are negative. In Quadrant IV, the x-coordinate (cosine) is positive, but the y-coordinate (sine) is negative.
Now let's consider the given angle, 200°.
Since 200° is greater than 180°, it lies in Quadrant III.
- The cosine of 200° will be negative because it is in Quadrant III.
- The sine of 200° will also be negative because it is in Quadrant III.
Hence, sin(200)0 and cos(200)0 are both negative.
Therefore, the correct answer is option 'D' - Negative.
The unit circle is a circle with a radius of 1 unit, centered at the origin (0,0) on a coordinate plane. It helps us understand the relationship between angles and the coordinates on the circle.
The circle is divided into four quadrants: Quadrant I (0°-90°), Quadrant II (90°-180°), Quadrant III (180°-270°), and Quadrant IV (270°-360°).
In Quadrant I, both the x-coordinate (cosine) and the y-coordinate (sine) are positive. In Quadrant II, the x-coordinate (cosine) is negative, but the y-coordinate (sine) is positive. In Quadrant III, both the x-coordinate (cosine) and the y-coordinate (sine) are negative. In Quadrant IV, the x-coordinate (cosine) is positive, but the y-coordinate (sine) is negative.
Now let's consider the given angle, 200°.
Since 200° is greater than 180°, it lies in Quadrant III.
- The cosine of 200° will be negative because it is in Quadrant III.
- The sine of 200° will also be negative because it is in Quadrant III.
Hence, sin(200)0 and cos(200)0 are both negative.
Therefore, the correct answer is option 'D' - Negative.
Direction: Read the following text and answer the following questions on the basis of the same:The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.Principal value of sin -1(1) + sin-1(1/√2) is- a)2π
- b)π
- c)3π/4
- d)π/3
Correct answer is option 'C'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.
Principal value of sin -1(1) + sin-1(1/√2) is
a)
2π
b)
π
c)
3π/4
d)
π/3
|
Varun Kapoor answered |
Sin-1(1) + sin-1(1/√2) = π//2 + π/4
= 3π/4
If sin A + cos A = 1, then sin 2A is equal to- a)2
- b)0
- c)1/2
- d)1
Correct answer is option 'B'. Can you explain this answer?
If sin A + cos A = 1, then sin 2A is equal to
a)
2
b)
0
c)
1/2
d)
1
|
Anjali Sen answered |
Understanding the Given Equation
We start with the equation:
- sinA + cosA = 1
This can be analyzed using the Pythagorean identity for sine and cosine.
Squaring Both Sides
If we square both sides:
- (sinA + cosA)^2 = 1^2
- sin²A + 2sinAcosA + cos²A = 1
Using the identity sin²A + cos²A = 1, we replace it in the equation:
- 1 + 2sinAcosA = 1
This simplifies to:
- 2sinAcosA = 0
Finding Values of Sine and Cosine
From 2sinAcosA = 0, we can conclude:
- sinA = 0 or cosA = 0
This means A can be:
- A = nπ (where n is an integer) for sinA = 0.
- A = π/2 + nπ for cosA = 0.
Calculating Sin 2A
We know that:
- sin 2A = 2sinAcosA
Given that either sinA = 0 or cosA = 0, we have:
- If sinA = 0, then sin 2A = 2 * 0 * cosA = 0.
- If cosA = 0, then sin 2A = 2 * sinA * 0 = 0.
In both scenarios, we find that:
- sin 2A = 0
Conclusion
Thus, the value of sin 2A is:
- 0
Therefore, the correct answer is option 'B'.
We start with the equation:
- sinA + cosA = 1
This can be analyzed using the Pythagorean identity for sine and cosine.
Squaring Both Sides
If we square both sides:
- (sinA + cosA)^2 = 1^2
- sin²A + 2sinAcosA + cos²A = 1
Using the identity sin²A + cos²A = 1, we replace it in the equation:
- 1 + 2sinAcosA = 1
This simplifies to:
- 2sinAcosA = 0
Finding Values of Sine and Cosine
From 2sinAcosA = 0, we can conclude:
- sinA = 0 or cosA = 0
This means A can be:
- A = nπ (where n is an integer) for sinA = 0.
- A = π/2 + nπ for cosA = 0.
Calculating Sin 2A
We know that:
- sin 2A = 2sinAcosA
Given that either sinA = 0 or cosA = 0, we have:
- If sinA = 0, then sin 2A = 2 * 0 * cosA = 0.
- If cosA = 0, then sin 2A = 2 * sinA * 0 = 0.
In both scenarios, we find that:
- sin 2A = 0
Conclusion
Thus, the value of sin 2A is:
- 0
Therefore, the correct answer is option 'B'.
Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.Assertion (A): Principal value of cos–1(1) is πReason (R): Value of cos 0° is 1- a)Both A and R are true and R is the correct explanation of A
- b)Both A and R are true but R is NOT the correct explanation of A
- c)A is true but R is false
- d)A is false and R is True
Correct answer is option 'D'. Can you explain this answer?
Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion (A): Principal value of cos–1(1) is π
Reason (R): Value of cos 0° is 1
a)
Both A and R are true and R is the correct explanation of A
b)
Both A and R are true but R is NOT the correct explanation of A
c)
A is true but R is false
d)
A is false and R is True
|
|
Nandini Iyer answered |
In case of Assertion
cos-1(1) = y
Cos y= 1
Cos y = cos 00 [∴ cos 00 = 1]
∴ y = 0
⇒ Principal value of cos–1 (1) is 0
Hence Assertion is in correct.
Reason is correct.
If 2tan−1(cos x) = tan−1(2cosecx) , then x =- a)
- b)
- c)
- d)none of these.
Correct answer is option 'B'. Can you explain this answer?
If 2tan−1(cos x) = tan−1(2cosecx) , then x =
a)
b)
c)
d)
none of these.
|
Sagar Mukherjee answered |
If 2 tan-1 (cos x) = tan -1(2 cosec x),
2tan-1(cos x) = tan-1 (2 cosec x)
= tan-1(2 cosec x)
= cot x cosec x = cosec x = x = π/4
If cos(-1)x + cos(-1)y = 2π, then the value of sin(-1)x + sin(-1)y is - a)−π
- b)π
- c)0
- d)none of these.
Correct answer is option 'A'. Can you explain this answer?
If cos(-1)x + cos(-1)y = 2π, then the value of sin(-1)x + sin(-1)y is
a)
−π
b)
π
c)
0
d)
none of these.
|
Rishika Mishra answered |
If cos(-1)x + cos(-1) y = 2π, then the value of sin(-1)x + sin(-1)y = π−2π = −π.
Direction: Read the following text and answer the following questions on the basis of the same:The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.Principal value of cot-1(√3) is :- a)π/3
- b)π
- c)π/6
- d)π/2
Correct answer is option 'C'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
The value of an inverse trigonometric functions which lies in the range of Principal branch is called the principal value of that inverse trigonometric functions.
Principal value of cot-1(√3) is :
a)
π/3
b)
π
c)
π/6
d)
π/2
|
Disha Reddy answered |
Understanding Principal Values of Inverse Trigonometric Functions
The principal value of an inverse trigonometric function is the value that lies within a specific range, typically chosen to make the function single-valued. For cotangent, the principal value is defined in the range (0, π).
Evaluating cot-1(√3)
To find the principal value of cot-1(√3), we need to determine the angle θ such that:
- cot(θ) = √3
Identifying the Angle
- The cotangent function is defined as the ratio of cosine to sine: cot(θ) = cos(θ)/sin(θ).
- We know that cot(π/6) = √3. This is because:
- In a 30-60-90 triangle, the adjacent side (cos) is √3, and the opposite side (sin) is 1.
Conclusion
- Therefore, when θ = π/6, cot(θ) yields √3.
- Since π/6 lies within the principal value range of (0, π), the answer for cot-1(√3) is indeed:
Correct Answer: Option C - π/6
The principal value of an inverse trigonometric function is the value that lies within a specific range, typically chosen to make the function single-valued. For cotangent, the principal value is defined in the range (0, π).
Evaluating cot-1(√3)
To find the principal value of cot-1(√3), we need to determine the angle θ such that:
- cot(θ) = √3
Identifying the Angle
- The cotangent function is defined as the ratio of cosine to sine: cot(θ) = cos(θ)/sin(θ).
- We know that cot(π/6) = √3. This is because:
- In a 30-60-90 triangle, the adjacent side (cos) is √3, and the opposite side (sin) is 1.
Conclusion
- Therefore, when θ = π/6, cot(θ) yields √3.
- Since π/6 lies within the principal value range of (0, π), the answer for cot-1(√3) is indeed:
Correct Answer: Option C - π/6
Chapter doubts & questions for Chapter 2 - Inverse Trigonometric Functions - Mathematics CUET UG Mock Test Series 2025 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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