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All questions of Trigonometric Functions and Equations for Grade 12 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.
|
Shivani answered |
->>Sintheta=3/5....-> sectheta=1/costheta...-> tantheta=sintheta/costheta...by substituting in the given que u will get...1+sintheta/1-sintheta....at last by solving u will get ans 4...HOPE U GOT IT...
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
sec-1x + cosec-1x =- a)0
- b)π/2
- c)not defined
- d)π
Correct answer is option 'A'. Can you explain this answer?
sec-1x + cosec-1x =
a)
0
b)
π/2
c)
not defined
d)
π
|
Kumari Sakshi answered |
Are you sure the answer is 0 .according to me it should be pi/2
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
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
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
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)
is given by
sin2000+cos2000 is- a)Negative
- b)Positive
- c)Zero
- d)Zero or positive.
Correct answer is option 'A'. Can you explain this answer?
sin2000+cos2000 is
a)
Negative
b)
Positive
c)
Zero
d)
Zero or positive.
|
|
Mihir Joshi answered |
Because both sin 2000 and cos 2000 lies in 3rd quadrant. In 3rd quadrant values of both sine and cosine functions are negative.
- 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 :
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 |
- 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
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
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.
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
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