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All questions of Inverse Trigonometric Functions for A Level 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
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 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 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
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
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 :-):-)^_^
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)
|
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)
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
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 +
is given by
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
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.
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
|
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
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
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)
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)
- 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
- 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.
|
Seblewongel Girma answered |
Sin(2(0.705π))=-24/25
- 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
are :
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.
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.
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) )
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)
|
Top Rankers answered |
(cosx−sinx)/(cosx+sinx) = (1−tanx)/(1+tanx)
tan(A−B) = (tanA−tanB)/(1 + tanAtanB)
= tan(π/4−x)
putting this value in question.
tan−1tan(π/4−x)
π/4−x.
tan(A−B) = (tanA−tanB)/(1 + tanAtanB)
= tan(π/4−x)
putting this value in question.
tan−1tan(π/4−x)
π/4−x.
Evaluate 
- a)
- b)
- c)
- d)
Correct answer is 'A'. Can you explain this answer?
Evaluate
a)
b)
c)
d)
|
Knowledge Hub 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)
If 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)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.
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.
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.
What is the maximum and minimum value of sin x +cos x?- a)0, -1
- b)1, 0
- c)-1, -√2
- d)√2, –√2
Correct answer is option 'D'. Can you explain this answer?
What is the maximum and minimum value of sin x +cos x?
a)
0, -1
b)
1, 0
c)
-1, -√2
d)
√2, –√2
|
|
Rajiv Reddy answered |
Let y= sin x + cos x
dy/dx=cos x- sin x
For maximum or minimum dy/dx=0
Setting cosx- sin x=0
We get cos x = sin x
x= π/4, 5π/4———-
Whether these correspond to maximum or minimum, can be found from the sign of second derivative.
d^2y/dx^2=-sin x - cos x=-1/√2–1/√2 (for x=π/4) which is negative. Hence x=π/4 corresponds to maximum.For x=5π/4
d^2y/dx^2=-(-1/√2)-(-1/√2)=2/√2 a positive quantity. Hence 5π/4 corresponds to minimum
Maximum value of the function
y= sin π/4 + cos π/4= 2/√2=√2
Minimum value is
Sin(5π/4)+cos (5π/4)=-2/√2=-√2
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
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.
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
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.
The number of solutions of the equation 
- a)One
- b)two
- c)more than one
- d)none of these.
Correct answer is option 'A'. Can you explain this answer?
The number of solutions of the equation
a)
One
b)
two
c)
more than one
d)
none of these.
|
Nilesh Goyal answered |
As no value of x in (0, 1) can satisfy the given equation.Thus, the given equation has only one solution.
2cos−1x = cos−1(2x2−1)holds true for all- a)|x| ≤ 1/2
- b)|x|⩽1
- c)0⩽x⩽1
- d)none of these.
Correct answer is option 'C'. Can you explain this answer?
2cos−1x = cos−1(2x2−1)holds true for all
a)
|x| ≤ 1/2
b)
|x|⩽1
c)
0⩽x⩽1
d)
none of these.
|
Akshita Nair answered |
This is true for all real values of x ∈ [0,1].
Find the value of sin-1(3/5) + sin-1(4/5) + cos-1(√3/2).- a)π/3
- b)2π/3
- c)4π/3
- d)π/4
Correct answer is option 'B'. Can you explain this answer?
Find the value of sin-1(3/5) + sin-1(4/5) + cos-1(√3/2).
a)
π/3
b)
2π/3
c)
4π/3
d)
π/4
|
Varun Kapoor answered |
Using the formula sin-1x + sin-1y
=
=
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'.
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