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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
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
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
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
|
|
Lekshmi Choudhury answered |
Because both sin 2000 and cos 2000 lies in 3rd quadrant. In 3rd quadrant values of sin and cos are 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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