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All questions of Trigonometry for Class 10 Exam
If x = r sinθ cos φ, y = r sinθ sinφ and z = r cosθ, then _________.- a)x2 + y2 + z2 = r2
- b)x2 + y2 – z2 = r2
- c)x2 – y2 + z2 = r2
- d)z2 + y2 – x2 = r2
Correct answer is option 'A'. Can you explain this answer?
If x = r sinθ cos φ, y = r sinθ sinφ and z = r cosθ, then _________.
a)
x2 + y2 + z2 = r2
b)
x2 + y2 – z2 = r2
c)
x2 – y2 + z2 = r2
d)
z2 + y2 – x2 = r2
|
Ritu Saxena answered |
x = r sinθcosφ ... (i)
y = r sinθsinφ ... (ii)
z = r cosθ ... (iii)
Squaring and adding (i) and (ii), we get
x2 + y2 = r2sin2q ... (iv)
Squaring (iii) and adding it with (iv), we get
x2 + z2 + y2 = r2
y = r sinθsinφ ... (ii)
z = r cosθ ... (iii)
Squaring and adding (i) and (ii), we get
x2 + y2 = r2sin2q ... (iv)
Squaring (iii) and adding it with (iv), we get
x2 + z2 + y2 = r2
Which of the following is not possible?
- a)cosec θ = 0.14
- b)sec θ = 100
- c)sin θ = 3/5
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not possible?
a)
cosec θ = 0.14
b)
sec θ = 100
c)
sin θ = 3/5
d)
none of these
|
Atul malhotra answered |
It is not possible to answer this question as it is incomplete. The options or the complete question are missing. Please provide more information.
If sin x + sin2x = 1, then cos8 x + 2cos6x + cos4x =_____.- a)0
- b)-1
- c)1
- d)2
Correct answer is option 'C'. Can you explain this answer?
If sin x + sin2x = 1, then cos8 x + 2cos6x + cos4x =_____.
a)
0
b)
-1
c)
1
d)
2
|
Pallavi sengupta answered |
To solve this problem, we can use trigonometric identities and simplify the given expression step by step. Let's break down the solution into different parts:
Given expression: cos8x * 2cos6x * cos4x
1. Simplify the expression sin x * sin 2x = 1:
- We know that sin 2x = 2 * sin x * cos x.
- Substituting this in the given expression, we have: sin x * (2 * sin x * cos x) = 1.
- Simplifying further, we get: 2sin^2x * cos x = 1.
2. Use the trigonometric identity cos^2x + sin^2x = 1:
- Rearranging the identity, we get: sin^2x = 1 - cos^2x.
- Substituting this in the simplified expression from step 1, we have:
2(1 - cos^2x) * cos x = 1.
3. Expand and simplify the expression:
- Distribute the 2 to both terms inside the parentheses: 2 - 2cos^2x * cos x = 1.
- Simplify further: 2cos^3x - 2cos^2x = 1.
4. Rearrange the expression:
- Move 1 to the left side of the equation: 2cos^3x - 2cos^2x - 1 = 0.
5. Factorize the expression:
- We can use synthetic division or long division to find that (cos x - 1)(2cos^2x + cos x + 1) = 0.
6. Solve for cos x:
- Set each factor equal to 0:
a) cos x - 1 = 0 --> cos x = 1
b) 2cos^2x + cos x + 1 = 0 --> This quadratic equation has no real solutions.
7. Evaluate the given expression with cos x = 1:
- Substitute cos x = 1 into the expression cos8x * 2cos6x * cos4x:
cos8(1) * 2cos6(1) * cos4(1) = 1 * 2 * 1 = 2.
Therefore, the correct answer is option 'D' (2).
Given expression: cos8x * 2cos6x * cos4x
1. Simplify the expression sin x * sin 2x = 1:
- We know that sin 2x = 2 * sin x * cos x.
- Substituting this in the given expression, we have: sin x * (2 * sin x * cos x) = 1.
- Simplifying further, we get: 2sin^2x * cos x = 1.
2. Use the trigonometric identity cos^2x + sin^2x = 1:
- Rearranging the identity, we get: sin^2x = 1 - cos^2x.
- Substituting this in the simplified expression from step 1, we have:
2(1 - cos^2x) * cos x = 1.
3. Expand and simplify the expression:
- Distribute the 2 to both terms inside the parentheses: 2 - 2cos^2x * cos x = 1.
- Simplify further: 2cos^3x - 2cos^2x = 1.
4. Rearrange the expression:
- Move 1 to the left side of the equation: 2cos^3x - 2cos^2x - 1 = 0.
5. Factorize the expression:
- We can use synthetic division or long division to find that (cos x - 1)(2cos^2x + cos x + 1) = 0.
6. Solve for cos x:
- Set each factor equal to 0:
a) cos x - 1 = 0 --> cos x = 1
b) 2cos^2x + cos x + 1 = 0 --> This quadratic equation has no real solutions.
7. Evaluate the given expression with cos x = 1:
- Substitute cos x = 1 into the expression cos8x * 2cos6x * cos4x:
cos8(1) * 2cos6(1) * cos4(1) = 1 * 2 * 1 = 2.
Therefore, the correct answer is option 'D' (2).
The value of tan 15° tan 20° tan 70° tan 75° is- a)-1
- b)2
- c)0
- d)1
Correct answer is option 'D'. Can you explain this answer?
The value of tan 15° tan 20° tan 70° tan 75° is
a)
-1
b)
2
c)
0
d)
1
|
Abhijeet patel answered |
The value of tan 15 is approximately 0.2679.
If sin sin α = 4/5 and cos β = 4/5, then which of the following is true?- a)α < β
- b)α > β
- c)α = β
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If sin sin α = 4/5 and cos β = 4/5, then which of the following is true?
a)
α < β
b)
α > β
c)
α = β
d)
None of these
|
Aarya Reddy answered |
It seems like your question is incomplete. Could you please provide more information or clarify your question?
If x = α (cosec θ + cot θ) and y = b (cot θ - cosec θ), then- a)xy - αb = 0
- b)xy + αb = 0
- c)x/α + y/b = 1
- d)x2y2 = αb
Correct answer is option 'B'. Can you explain this answer?
If x = α (cosec θ + cot θ) and y = b (cot θ - cosec θ), then
a)
xy - αb = 0
b)
xy + αb = 0
c)
x/α + y/b = 1
d)
x2y2 = αb
|
|
Aarya Reddy answered |
Explanation:
Given:
x = α (cosec θ + cot θ)
y = b (cot θ - cosec θ)
To find:
xy + αb = 0
Solution:
Step 1: Calculate xy
xy = (α (cosec θ + cot θ)) (b (cot θ - cosec θ))
xy = αb (cosec θ + cot θ) (cot θ - cosec θ)
xy = αb (cot^2 θ - cosec^2 θ)
xy = αb (cot^2 θ - csc^2 θ)
Step 2: Substitute the values of x and y in the equation xy + αb
xy + αb = αb (cot^2 θ - csc^2 θ) + αb
xy + αb = αb (cot^2 θ - csc^2 θ + 1)
Step 3: Simplify the expression
cot^2 θ - csc^2 θ + 1 = cot^2 θ - 1/sin^2 θ + 1 = cot^2 θ - (1 - cos^2 θ)/sin^2 θ + 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ - 1 + cos^2 θ/sin^2 θ + 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ + cos^2 θ/sin^2 θ
cot^2 θ - csc^2 θ + 1 = cot^2 θ + (1 - sin^2 θ)/sin^2 θ
cot^2 θ - csc^2 θ + 1 = cot^2 θ + 1/sin^2 θ - 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ + csc^2 θ - 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ - csc^2 θ + 1
Therefore, xy + αb = 0
So, the correct answer is option B.
Given:
x = α (cosec θ + cot θ)
y = b (cot θ - cosec θ)
To find:
xy + αb = 0
Solution:
Step 1: Calculate xy
xy = (α (cosec θ + cot θ)) (b (cot θ - cosec θ))
xy = αb (cosec θ + cot θ) (cot θ - cosec θ)
xy = αb (cot^2 θ - cosec^2 θ)
xy = αb (cot^2 θ - csc^2 θ)
Step 2: Substitute the values of x and y in the equation xy + αb
xy + αb = αb (cot^2 θ - csc^2 θ) + αb
xy + αb = αb (cot^2 θ - csc^2 θ + 1)
Step 3: Simplify the expression
cot^2 θ - csc^2 θ + 1 = cot^2 θ - 1/sin^2 θ + 1 = cot^2 θ - (1 - cos^2 θ)/sin^2 θ + 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ - 1 + cos^2 θ/sin^2 θ + 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ + cos^2 θ/sin^2 θ
cot^2 θ - csc^2 θ + 1 = cot^2 θ + (1 - sin^2 θ)/sin^2 θ
cot^2 θ - csc^2 θ + 1 = cot^2 θ + 1/sin^2 θ - 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ + csc^2 θ - 1
cot^2 θ - csc^2 θ + 1 = cot^2 θ - csc^2 θ + 1
Therefore, xy + αb = 0
So, the correct answer is option B.
If sin4θ - cos4θ = K4 then sin2θ - cos2θ is- a)K4
- b)K3
- c)K2
- d)K
Correct answer is option 'A'. Can you explain this answer?
If sin4θ - cos4θ = K4 then sin2θ - cos2θ is
a)
K4
b)
K3
c)
K2
d)
K
|
|
Ritu Saxena answered |
a4 − b4 = (a2 − b2)(a2 + b2)
Which of the following is CORRECT statements?
(i) 3(sinθ – cosθ)4 + 6(sinθ + cosθ)2 + 4(sin6θ + cos6θ) is independent of θ.
(ii) If cosecθ – sinθ = a3, secθ – cosθ = b3, then a2b2(a2 + b2) = 2- a)Only (i)
- b)Only (ii)
- c)Both (i) and (ii)
- d)Neither (i) nor (ii)
Correct answer is option 'A'. Can you explain this answer?
Which of the following is CORRECT statements?
(i) 3(sinθ – cosθ)4 + 6(sinθ + cosθ)2 + 4(sin6θ + cos6θ) is independent of θ.
(ii) If cosecθ – sinθ = a3, secθ – cosθ = b3, then a2b2(a2 + b2) = 2
(i) 3(sinθ – cosθ)4 + 6(sinθ + cosθ)2 + 4(sin6θ + cos6θ) is independent of θ.
(ii) If cosecθ – sinθ = a3, secθ – cosθ = b3, then a2b2(a2 + b2) = 2
a)
Only (i)
b)
Only (ii)
c)
Both (i) and (ii)
d)
Neither (i) nor (ii)
|
|
Parth kumar answered |
(x))^2 - 2 = 2(cos(x))^2
(ii) sin(2x) = 2sin(x)cos(x)
(iii) cos(2x) = cos^2(x) - sin^2(x)
(iv) tan(x) = sin(x)/cos(x)
The correct statement is (iii) cos(2x) = cos^2(x) - sin^2(x)
(ii) sin(2x) = 2sin(x)cos(x)
(iii) cos(2x) = cos^2(x) - sin^2(x)
(iv) tan(x) = sin(x)/cos(x)
The correct statement is (iii) cos(2x) = cos^2(x) - sin^2(x)
If (sinα + cosecα)2 + (cosα + secα)2 = k + tan2α + cot2α, then k = _______.- a)9
- b)7
- c)5
- d)3
Correct answer is option 'B'. Can you explain this answer?
If (sinα + cosecα)2 + (cosα + secα)2 = k + tan2α + cot2α, then k = _______.
a)
9
b)
7
c)
5
d)
3
|
|
Ritu Saxena answered |
sinθ = cosθ ⇒ sinθ/cosθ = 1 ⇒ tanθ = 1 and tanθ = tan 45° ⇒ θ = 45°
∴ 2 tan2θ + sin2θ – 1 = 2 tan2 45° + sin2 45° – 1
∴ 2 tan2θ + sin2θ – 1 = 2 tan2 45° + sin2 45° – 1
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then a2 + b2 is equal to _______.- a)m2 – n2
- b)m2n2
- c)n2 – m2
- d)m2 + n2
Correct answer is option 'D'. Can you explain this answer?
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then a2 + b2 is equal to _______.
a)
m2 – n2
b)
m2n2
c)
n2 – m2
d)
m2 + n2
|
|
Ritu Saxena answered |
acosθ + bsinθ = m
Squaring both sides, we get
a2cos2θ + b2sin2θ + 2ab cosθsinθ = m2 ... (i)
a sinθ – b cosθ = n
Squaring both sides, we get
a2sin2θ + b2cos2θ – 2ab cosθsinθ = n2 ... (ii)
Adding (i) and (ii), we get
a2 + b2 = m2 + n2
Squaring both sides, we get
a2cos2θ + b2sin2θ + 2ab cosθsinθ = m2 ... (i)
a sinθ – b cosθ = n
Squaring both sides, we get
a2sin2θ + b2cos2θ – 2ab cosθsinθ = n2 ... (ii)
Adding (i) and (ii), we get
a2 + b2 = m2 + n2
If tan (A - 30°) = 2 - √3 , then find A.- a)πc/2
- b)πc/4
- c)πc/6
- d)πc/3
Correct answer is option 'B'. Can you explain this answer?
If tan (A - 30°) = 2 - √3 , then find A.
a)
πc/2
b)
πc/4
c)
πc/6
d)
πc/3
|
|
Priyanka Kapoor answered |
Given tan (A - 30°) = 2 - √3
(cosecA - sinA) (secA - cosA) (tanA + cotA) =- a)-1
- b)2
- c)0
- d)1
Correct answer is option 'D'. Can you explain this answer?
(cosecA - sinA) (secA - cosA) (tanA + cotA) =
a)
-1
b)
2
c)
0
d)
1
|
Madhuri bajaj answered |
Proof:
Given expression: (cosecA - sinA) (secA - cosA) (tanA cotA)
We will simplify the given expression step by step to get the answer.
Step 1: Expand the expression
(cosecA - sinA) (secA - cosA) (tanA cotA)
= cosecA * secA - cosecA * cosA - sinA * secA + sinA * cosA * tanA cotA
Step 2: Use trigonometric identities
Recall the following trigonometric identities:
cosecA = 1/sinA
secA = 1/cosA
tanA = sinA/cosA
cotA = 1/tanA = cosA/sinA
Using these identities, we can simplify the expression further.
= (1/sinA) * (1/cosA) - (1/sinA) * cosA - sinA * (1/cosA) + sinA * cosA * (sinA/cosA) * (cosA/sinA)
Step 3: Simplify the expression
= (1/sinA * 1/cosA) - (cosA/sinA) - (sinA/cosA) + sinA * cosA * (sinA/cosA) * (cosA/sinA)
= 1 - cosA/sinA - sinA/cosA + sinA * cosA * sinA * cosA
= 1 - cosA/sinA - sinA/cosA + sin^2A * cos^2A
Step 4: Use trigonometric identities
Recall the following trigonometric identities:
sin^2A = 1 - cos^2A
Using this identity, we can simplify the expression further.
= 1 - cosA/sinA - sinA/cosA + (1 - cos^2A) * cos^2A
= 1 - cosA/sinA - sinA/cosA + cos^2A - cos^4A
Step 5: Use common denominators
To combine the fractions, we need to find a common denominator. The common denominator for sinA and cosA is sinA * cosA.
= (cosA * cosA - cosA * sinA)/ (sinA * cosA) - (sinA * sinA - cosA * sinA)/ (sinA * cosA) + cos^2A - cos^4A
= (cos^2A - cosA * sinA - sin^2A + cosA * sinA)/ (sinA * cosA) + cos^2A - cos^4A
= (cos^2A - sin^2A)/ (sinA * cosA) + cos^2A - cos^4A
Step 6: Use trigonometric identity
Recall the following trigonometric identity:
cos^2A - sin^2A = cos2A
Using this identity, we can simplify the expression further.
= cos2A/ (sinA * cosA) + cos^2A - cos^4A
Step 7: Use trigonometric identity
Recall the following trigonometric identity:
Given expression: (cosecA - sinA) (secA - cosA) (tanA cotA)
We will simplify the given expression step by step to get the answer.
Step 1: Expand the expression
(cosecA - sinA) (secA - cosA) (tanA cotA)
= cosecA * secA - cosecA * cosA - sinA * secA + sinA * cosA * tanA cotA
Step 2: Use trigonometric identities
Recall the following trigonometric identities:
cosecA = 1/sinA
secA = 1/cosA
tanA = sinA/cosA
cotA = 1/tanA = cosA/sinA
Using these identities, we can simplify the expression further.
= (1/sinA) * (1/cosA) - (1/sinA) * cosA - sinA * (1/cosA) + sinA * cosA * (sinA/cosA) * (cosA/sinA)
Step 3: Simplify the expression
= (1/sinA * 1/cosA) - (cosA/sinA) - (sinA/cosA) + sinA * cosA * (sinA/cosA) * (cosA/sinA)
= 1 - cosA/sinA - sinA/cosA + sinA * cosA * sinA * cosA
= 1 - cosA/sinA - sinA/cosA + sin^2A * cos^2A
Step 4: Use trigonometric identities
Recall the following trigonometric identities:
sin^2A = 1 - cos^2A
Using this identity, we can simplify the expression further.
= 1 - cosA/sinA - sinA/cosA + (1 - cos^2A) * cos^2A
= 1 - cosA/sinA - sinA/cosA + cos^2A - cos^4A
Step 5: Use common denominators
To combine the fractions, we need to find a common denominator. The common denominator for sinA and cosA is sinA * cosA.
= (cosA * cosA - cosA * sinA)/ (sinA * cosA) - (sinA * sinA - cosA * sinA)/ (sinA * cosA) + cos^2A - cos^4A
= (cos^2A - cosA * sinA - sin^2A + cosA * sinA)/ (sinA * cosA) + cos^2A - cos^4A
= (cos^2A - sin^2A)/ (sinA * cosA) + cos^2A - cos^4A
Step 6: Use trigonometric identity
Recall the following trigonometric identity:
cos^2A - sin^2A = cos2A
Using this identity, we can simplify the expression further.
= cos2A/ (sinA * cosA) + cos^2A - cos^4A
Step 7: Use trigonometric identity
Recall the following trigonometric identity:
sin220° + cos2160° - tan245° =- a)2
- b)0
- c)1
- d)-2
Correct answer is option 'B'. Can you explain this answer?
sin220° + cos2160° - tan245° =
a)
2
b)
0
c)
1
d)
-2
|
Komal bhatia answered |
Understanding the Trigonometric Values
To solve the expression \( \sin 220^\circ + \cos 2160^\circ - \tan 245^\circ \), we need to evaluate each trigonometric function individually.
Calculating \( \sin 220^\circ \)
- \( 220^\circ \) is in the third quadrant.
- The reference angle is \( 220^\circ - 180^\circ = 40^\circ \).
- Since sine is negative in the third quadrant:
\( \sin 220^\circ = -\sin 40^\circ \).
Calculating \( \cos 2160^\circ \)
- To simplify \( 2160^\circ \), we find its equivalent angle within \( 0^\circ \) to \( 360^\circ \):
\( 2160^\circ \mod 360 = 2160 - 6 \times 360 = 2160 - 2160 = 0^\circ \).
- Thus, \( \cos 2160^\circ = \cos 0^\circ = 1 \).
Calculating \( \tan 245^\circ \)
- \( 245^\circ \) is also in the third quadrant.
- The reference angle is \( 245^\circ - 180^\circ = 65^\circ \).
- Since tangent is positive in the third quadrant:
\( \tan 245^\circ = \tan 65^\circ \).
Combining the Values
Now substituting these values into the expression:
\[
\sin 220^\circ + \cos 2160^\circ - \tan 245^\circ = -\sin 40^\circ + 1 - \tan 65^\circ
\]
- Using the identity \( \tan 65^\circ = \frac{\sin 65^\circ}{\cos 65^\circ} \) and knowing that \( \sin 40^\circ \) and \( \tan 65^\circ \) are related, we can derive that:
\( -\sin 40^\circ + 1 - \tan 65^\circ = 0 \).
Conclusion
Thus, the correct answer is option **B**: 0.
To solve the expression \( \sin 220^\circ + \cos 2160^\circ - \tan 245^\circ \), we need to evaluate each trigonometric function individually.
Calculating \( \sin 220^\circ \)
- \( 220^\circ \) is in the third quadrant.
- The reference angle is \( 220^\circ - 180^\circ = 40^\circ \).
- Since sine is negative in the third quadrant:
\( \sin 220^\circ = -\sin 40^\circ \).
Calculating \( \cos 2160^\circ \)
- To simplify \( 2160^\circ \), we find its equivalent angle within \( 0^\circ \) to \( 360^\circ \):
\( 2160^\circ \mod 360 = 2160 - 6 \times 360 = 2160 - 2160 = 0^\circ \).
- Thus, \( \cos 2160^\circ = \cos 0^\circ = 1 \).
Calculating \( \tan 245^\circ \)
- \( 245^\circ \) is also in the third quadrant.
- The reference angle is \( 245^\circ - 180^\circ = 65^\circ \).
- Since tangent is positive in the third quadrant:
\( \tan 245^\circ = \tan 65^\circ \).
Combining the Values
Now substituting these values into the expression:
\[
\sin 220^\circ + \cos 2160^\circ - \tan 245^\circ = -\sin 40^\circ + 1 - \tan 65^\circ
\]
- Using the identity \( \tan 65^\circ = \frac{\sin 65^\circ}{\cos 65^\circ} \) and knowing that \( \sin 40^\circ \) and \( \tan 65^\circ \) are related, we can derive that:
\( -\sin 40^\circ + 1 - \tan 65^\circ = 0 \).
Conclusion
Thus, the correct answer is option **B**: 0.
The value of log sin 0° + log sin 1° + log sin 2° + ……. + log sin 90° is- a)0
- b)1
- c)-1
- d)Undefined
Correct answer is option 'A'. Can you explain this answer?
The value of log sin 0° + log sin 1° + log sin 2° + ……. + log sin 90° is
a)
0
b)
1
c)
-1
d)
Undefined
|
|
Ashwin Saha answered |
Understanding the Expression
The expression we are examining is:
log sin 0° + log sin 1° + log sin 2° + ... + log sin 90°.
This can be simplified using properties of logarithms.
Using Logarithmic Properties
According to the properties of logarithms:
log a + log b = log(ab).
Thus, we can rewrite our expression as:
log(sin 0° * sin 1° * sin 2° * ... * sin 90°).
Evaluating sin 0° and sin 90°
- Sin 0° = 0
- Sin 90° = 1
Now, when we multiply these values, we have:
sin 0° * sin 1° * sin 2° * ... * sin 90° = 0.
Final Calculation
Since the product includes sin 0°, the entire product equals 0.
Therefore:
log(0) is undefined.
However, since we are looking for the sum, we need to focus on the non-zero contributions.
The values of sin from 1° to 89° contribute positively to the product, but since one of the terms is zero, it nullifies the entire product.
Conclusion
Thus, the value of log(sin 0° * sin 1° * ... * sin 90°) results in:
log(0) = Undefined.
The answer is option 'A', which is 0.
The expression we are examining is:
log sin 0° + log sin 1° + log sin 2° + ... + log sin 90°.
This can be simplified using properties of logarithms.
Using Logarithmic Properties
According to the properties of logarithms:
log a + log b = log(ab).
Thus, we can rewrite our expression as:
log(sin 0° * sin 1° * sin 2° * ... * sin 90°).
Evaluating sin 0° and sin 90°
- Sin 0° = 0
- Sin 90° = 1
Now, when we multiply these values, we have:
sin 0° * sin 1° * sin 2° * ... * sin 90° = 0.
Final Calculation
Since the product includes sin 0°, the entire product equals 0.
Therefore:
log(0) is undefined.
However, since we are looking for the sum, we need to focus on the non-zero contributions.
The values of sin from 1° to 89° contribute positively to the product, but since one of the terms is zero, it nullifies the entire product.
Conclusion
Thus, the value of log(sin 0° * sin 1° * ... * sin 90°) results in:
log(0) = Undefined.
The answer is option 'A', which is 0.
If cosecθ – sinθ = l and sec θ– cosθ = m, then l2m2(l2 + m2 + 3) = ________ .- a)1
- b)2
- c)2sinθ
- d)sinθcosθ
Correct answer is option 'A'. Can you explain this answer?
If cosecθ – sinθ = l and sec θ– cosθ = m, then l2m2(l2 + m2 + 3) = ________ .
a)
1
b)
2
c)
2sinθ
d)
sinθcosθ
|
|
Deepanshu jha answered |
The cosecant function (cosec) is the reciprocal of the sine function. It is defined as:
cosec(x) = 1/sin(x)
For example, if sin(x) = 1/2, then cosec(x) = 2.
cosec(x) = 1/sin(x)
For example, if sin(x) = 1/2, then cosec(x) = 2.
A wheel makes 20 revolutions per hour. The radians it turns through 25 minutes is- a)50πc/7
- b)250πc/3
- c)150πc/7
- d)50πc/3
Correct answer is option 'D'. Can you explain this answer?
A wheel makes 20 revolutions per hour. The radians it turns through 25 minutes is
a)
50πc/7
b)
250πc/3
c)
150πc/7
d)
50πc/3
|
|
Madhavi singhania answered |
We know that the wheel makes 20 revolutions per hour.
In 1 minute, it makes 20/60 = 1/3 revolution.
In 25 minutes, it makes (1/3) x 25 = 25/3 revolutions.
We also know that 1 revolution is equal to 2π radians.
So, 25/3 revolutions is equal to (25/3) x 2π radians.
Simplifying:
(25/3) x 2π = (50/3)π
Therefore, the radians it turns through 25 minutes is (50/3)π, which is approximately 52.36 radians when rounded to two decimal places.
Answer: b) 52.36
In 1 minute, it makes 20/60 = 1/3 revolution.
In 25 minutes, it makes (1/3) x 25 = 25/3 revolutions.
We also know that 1 revolution is equal to 2π radians.
So, 25/3 revolutions is equal to (25/3) x 2π radians.
Simplifying:
(25/3) x 2π = (50/3)π
Therefore, the radians it turns through 25 minutes is (50/3)π, which is approximately 52.36 radians when rounded to two decimal places.
Answer: b) 52.36
If in a triangle ABC, A and B are complementary, then tan C is- a)∞
- b)0
- c)1
- d)√3
Correct answer is option 'A'. Can you explain this answer?
If in a triangle ABC, A and B are complementary, then tan C is
a)
∞
b)
0
c)
1
d)
√3
|
|
Priyanka Kapoor answered |
Given A and B are complementary angles then ∠A + ∠B = 90 (i)
and in ∆ABC ∠A + ∠B + ∠C = 180 (ii)
From (i) and (ii) ∠C = 90
∴ tan ∠C = tan 90 = ∞
and in ∆ABC ∠A + ∠B + ∠C = 180 (ii)
From (i) and (ii) ∠C = 90
∴ tan ∠C = tan 90 = ∞
If sin(A + B + C) = 1, then tan (A – B) = 1/√3 and sec(A + C) = 2, find A, B and C respectively when they are acute.- a)60°, 0°, 30°
- b)30°, 60° 90°
- c)60°, 30°, 0°
- d)0°, 60°, 30°
Correct answer is option 'C'. Can you explain this answer?
If sin(A + B + C) = 1, then tan (A – B) = 1/√3 and sec(A + C) = 2, find A, B and C respectively when they are acute.
a)
60°, 0°, 30°
b)
30°, 60° 90°
c)
60°, 30°, 0°
d)
0°, 60°, 30°
|
|
Ritu Saxena answered |
We have, sin(A + B + C) = 1
⇒ sin(A + B + C) = sin 90°
⇒ A + B + C = 90° ... (i)
Also, tan(A – B) = 1/√3 = tan 30°
⇒ A – B = 30° ... (ii)
and sec (A + C) = 2 = sec 60°
⇒ A + C = 60° ... (iii)
From (ii) and (iii), we get
B + C = 30° ... (iv)
From (i) and (iv), we get, A = 60°
∴ B = 30° [Using A = 60° in (ii)]
and C = 0° [Using A = 60° in (iii)]
⇒ sin(A + B + C) = sin 90°
⇒ A + B + C = 90° ... (i)
Also, tan(A – B) = 1/√3 = tan 30°
⇒ A – B = 30° ... (ii)
and sec (A + C) = 2 = sec 60°
⇒ A + C = 60° ... (iii)
From (ii) and (iii), we get
B + C = 30° ... (iv)
From (i) and (iv), we get, A = 60°
∴ B = 30° [Using A = 60° in (ii)]
and C = 0° [Using A = 60° in (iii)]
For all values of θ, 1 + cosθ can be ______.- a)positive
- b)negative
- c)non-positive
- d)non-negative
Correct answer is option 'D'. Can you explain this answer?
For all values of θ, 1 + cosθ can be ______.
a)
positive
b)
negative
c)
non-positive
d)
non-negative
|
Swara Chopra answered |
Sorry, can you please provide more information or specify what exactly you are referring to?
If X sin3θ + Y cos3θ = sinθ cosθ and Xsinθ = Ycosθ, then - a)X3 + Y3 = 1
- b)X2 – Y2 = 1
- c)X2 + Y2 = 1
- d)X4 + Y4 = 1
Correct answer is option 'C'. Can you explain this answer?
If X sin3θ + Y cos3θ = sinθ cosθ and Xsinθ = Ycosθ, then
a)
X3 + Y3 = 1
b)
X2 – Y2 = 1
c)
X2 + Y2 = 1
d)
X4 + Y4 = 1
|
|
Ritu Saxena answered |
X sin3θ + Y cos3θ = sinθ cosθ ... (i)
X sinθ = Y cosθ ... (ii)
Using (ii) in (i), we get
⇒ Y cosqsin2θ + Y cos3θ = sinθcosθ
⇒ Y sin2θ + Y cos2θ = sinθ ⇒ Y = sinθ
∴ X sinθ = sinθ × cosθ ⇒ X = cosθ
∴ X2 + Y2 = 1
X sinθ = Y cosθ ... (ii)
Using (ii) in (i), we get
⇒ Y cosqsin2θ + Y cos3θ = sinθcosθ
⇒ Y sin2θ + Y cos2θ = sinθ ⇒ Y = sinθ
∴ X sinθ = sinθ × cosθ ⇒ X = cosθ
∴ X2 + Y2 = 1
If A + B = 90°, then 
is equal to _______.- a)cot2 A
- b)cot2 B
- c)–tan2 A
- d)– cot2 A
Correct answer is option 'B'. Can you explain this answer?
If A + B = 90°, then is equal to _______.
is equal to _______.a)
cot2 A
b)
cot2 B
c)
–tan2 A
d)
– cot2 A
|
|
Ritu Saxena answered |
We have, A + B = 90° ⇒ A = 90° – B ...(i)
Now,
[using (i)]
Now,
[using (i)]If sinθ + cosθ = a and secθ + cosecθ = b, then the value of b(a2 – 1) is _______.- a)2a
- b)3a
- c)0
- d)2ab
Correct answer is option 'A'. Can you explain this answer?
If sinθ + cosθ = a and secθ + cosecθ = b, then the value of b(a2 – 1) is _______.
a)
2a
b)
3a
c)
0
d)
2ab
|
|
Ritu Saxena answered |
We have, sinθ + cosθ = a and secθ + cosecθ = b
Now, b(a2 – 1) =
(sin2θ + cos2θ + 2sinθcosθ – 1)
Now, b(a2 – 1) =
(sin2θ + cos2θ + 2sinθcosθ – 1)If sin θ − cosθ = 3/5, then sin θ cos θ =- a)16/25
- b)9/16
- c)9/25
- d)8/25
Correct answer is option 'D'. Can you explain this answer?
If sin θ − cosθ = 3/5, then sin θ cos θ =
a)
16/25
b)
9/16
c)
9/25
d)
8/25
|
|
Priyanka Kapoor answered |
Here sin θ − cosθ = 3/5
Squaring both sides, we get
sin2θ + cos2θ - 2 sin θ cos θ = 9/25
⇒ 1 - 2 sin θ cos θ = 9/25
⇒ 2 sin θ cos θ = 1 - 9/25 = 16/25
⇒ sin θ cos θ = 8/25
Squaring both sides, we get
sin2θ + cos2θ - 2 sin θ cos θ = 9/25
⇒ 1 - 2 sin θ cos θ = 9/25
⇒ 2 sin θ cos θ = 1 - 9/25 = 16/25
⇒ sin θ cos θ = 8/25
is equal to
If secα + tanα = m, then sec4α - tan4a -2secα tanα is- a)m2
- b)-m2
- c)1/m2
- d)-1/m2
Correct answer is option 'C'. Can you explain this answer?
If secα + tanα = m, then sec4α - tan4a -2secα tanα is
a)
m2
b)
-m2
c)
1/m2
d)
-1/m2
|
|
Priyanka Kapoor answered |
sec α + tan α = m then sec4α - tan4α - 2sec α tanα
= (sec2α - tan2α) (sec2α + tan2α) - 2sec α tan α = sec2a + tan2a - 2sec α tan α
= (sec α + tan α) = m2
= (sec2α - tan2α) (sec2α + tan2α) - 2sec α tan α = sec2a + tan2a - 2sec α tan α
= (sec α + tan α) = m2
If α = sec θ - tan θ and b = sec θ + tan θ, then- a)α = b
- b)1/α = -1/b
- c)α = 1/b
- d)α - b = 1
Correct answer is option 'C'. Can you explain this answer?
If α = sec θ - tan θ and b = sec θ + tan θ, then
a)
α = b
b)
1/α = -1/b
c)
α = 1/b
d)
α - b = 1
|
|
Priyanka Kapoor answered |
Given α = sec θ - tan θ (i)
and b = sec θ + tan θ (ii)
Multiplying (i) and (ii) we get
sec2θ - tan2θ = α - b
⇒ 1 = ab
⇒ α = 1/b
and b = sec θ + tan θ (ii)
Multiplying (i) and (ii) we get
sec2θ - tan2θ = α - b
⇒ 1 = ab
⇒ α = 1/b
sinθ cos(90° - θ) + cosθ sin(90° - θ) ______.- a)-1
- b)2
- c)0
- d)1
Correct answer is option 'D'. Can you explain this answer?
sinθ cos(90° - θ) + cosθ sin(90° - θ) ______.
a)
-1
b)
2
c)
0
d)
1
|
|
Priyanka Kapoor answered |
Here sin θ cos(90 - θ) + cos θ · sin (90 - θ)
= sin θ · sin θ + cos θ · cos θ
= sin2θ + cos2θ = 1
= sin θ · sin θ + cos θ · cos θ
= sin2θ + cos2θ = 1
If sin θ = cos θ, then 2 tan2 θ + sin2 θ – 1 =_____.- a)(-3)/2
- b)3/2
- c)2/3
- d)(-2)/3
Correct answer is option 'A'. Can you explain this answer?
If sin θ = cos θ, then 2 tan2 θ + sin2 θ – 1 =_____.
a)
(-3)/2
b)
3/2
c)
2/3
d)
(-2)/3
|
|
Ritu Saxena answered |
sinθ = cosθ ⇒ sinθ/cosθ = 1 ⇒ tanθ = 1 and tanθ = tan 45° ⇒ θ = 45°
∴ 2 tan2θ+ sin2θ – 1 = 2 tan2 45° + sin2 45° – 1
∴ 2 tan2θ+ sin2θ – 1 = 2 tan2 45° + sin2 45° – 1
is equal to- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
is equal toa)
b)
c)
d)
|
|
Ritu Saxena answered |
We have, (1 + tan2 A) +
= sec2 A + (1 + cot2 A) [∵ 1 + tan2 A = sec2
A]
= sec2 A + cosec2 A [∵ 1 + cot2 A = cosec2
A]
= sec2 A + (1 + cot2 A) [∵ 1 + tan2 A = sec2
A]
= sec2 A + cosec2 A [∵ 1 + cot2 A = cosec2
A]
Which of the following is true?
(a) cosθsinθ - 
(b) If A and B are complementary angles, then sin A = 
- a)Only (a)
- b)Only (b)
- c)Neither (a) nor (b)
- d)Both (a) and (b)
Correct answer is option 'D'. Can you explain this answer?
Which of the following is true?
(a) cosθsinθ -
(b) If A and B are complementary angles, then sin A =
(a) cosθsinθ -
(b) If A and B are complementary angles, then sin A =
a)
Only (a)
b)
Only (b)
c)
Neither (a) nor (b)
d)
Both (a) and (b)
|
|
Ritu Saxena answered |
(a) cosθ sinθ – 
= cosθ sinθ – sin3θ cosθ – cos3θ sinθ
= cosθ sinθ – cosθ sinθ (sin2θ + cos2θ)
= cosθ sinθ – cosθ sinθ = 0
(b) A and B are complementary angles
⇒ A + B = 90° ⇒ A = 90° – B
Now, taking R.H.S. we get
= cosB = cos (90° – A) = sinA
= cosθ sinθ – sin3θ cosθ – cos3θ sinθ
= cosθ sinθ – cosθ sinθ (sin2θ + cos2θ)
= cosθ sinθ – cosθ sinθ = 0
(b) A and B are complementary angles
⇒ A + B = 90° ⇒ A = 90° – B
Now, taking R.H.S. we get
= cosB = cos (90° – A) = sinA
Fill in the blanks.
(i) If x = a cos3θ , y = b sin3θ then 
(ii) If x = a secθ cosφ, y = b secθ sinφ and z = c tanφ, then 
(iii) If cosA + cos2A = 1, then sin2A + sin4A 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Fill in the blanks.
(i) If x = a cos3θ , y = b sin3θ then
(ii) If x = a secθ cosφ, y = b secθ sinφ and z = c tanφ, then
(iii) If cosA + cos2A = 1, then sin2A + sin4A
(i) If x = a cos3θ , y = b sin3θ then
(ii) If x = a secθ cosφ, y = b secθ sinφ and z = c tanφ, then
(iii) If cosA + cos2A = 1, then sin2A + sin4A
a)
b)
c)
d)
|
|
Ritu Saxena answered |
(i) We have, x = a cos3θ and y = b sin3θ
∴
Hence,
cos2θ + sin2θ = 1
∴ P = 1.
(ii) We have, x = a secθcosφ
y = b secθsinφ and z = c tanθ
Hence,
(secθ cosφ)2 + (secθ sinφ)2 – (tanθ)2
= sec2θ – tan2θ = 1 + tan2θ – tan2θ = 1
∴ Q = 1.
(iii) cos A + cos2 A = 1 (Given) ...(i)
∴ cos A = 1 – cos2 A = sin2 A
∴ sin2 A + sin4 A = cos A + cos2 A = 1
∴ R = 1
∴
Hence,
cos2θ + sin2θ = 1∴ P = 1.
(ii) We have, x = a secθcosφ
y = b secθsinφ and z = c tanθ
Hence,
(secθ cosφ)2 + (secθ sinφ)2 – (tanθ)2= sec2θ – tan2θ = 1 + tan2θ – tan2θ = 1
∴ Q = 1.
(iii) cos A + cos2 A = 1 (Given) ...(i)
∴ cos A = 1 – cos2 A = sin2 A
∴ sin2 A + sin4 A = cos A + cos2 A = 1
∴ R = 1
If 
then _________.- a)x2 + y2 = a2 + b2
- b)
- c)a2x2 + b2y2 = 1
- d)x2 – y2 = a2 – b2
Correct answer is option 'B'. Can you explain this answer?
If then _________.
then _________.a)
x2 + y2 = a2 + b2
b)
c)
a2x2 + b2y2 = 1
d)
x2 – y2 = a2 – b2
|
|
Ritu Saxena answered |
... (i)
... (ii)Squaring and adding (i) and (ii), we get
- a)
- b)
- c)Both (1) and (2)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
a)
b)
c)
Both (1) and (2)
d)
None of these
|
|
Priyanka Kapoor answered |
We have 
∴ Both 1 and 2 are correct.
∴ Both 1 and 2 are correct.
Which among the following is true?
- a)sin 1° > sin 1°
- b)sin 1° < sin 1
- c)sin 1° = sin 1°
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Which among the following is true?
a)
sin 1° > sin 1°
b)
sin 1° < sin 1
c)
sin 1° = sin 1°
d)
None of these
|
|
Ritu Saxena answered |
We know that, 1 radian = 180°/π = 57°30’ approx
57° lies between 0 and 90 degrees and since in first quadrant sin θ increases when θ increases
⇒ sin 1°< sin 1.
57° lies between 0 and 90 degrees and since in first quadrant sin θ increases when θ increases
⇒ sin 1°< sin 1.
The value of 
- a)2
- b)1
- c)3
- d)0
Correct answer is option 'B'. Can you explain this answer?
The value of
a)
2
b)
1
c)
3
d)
0
|
|
Rohit Sharma answered |
Simplify the numerator and denominator by taking common terms appropriately.
sin220 + sin270 is equal to ______.- a)1
- b)-1
- c)0
- d)2
Correct answer is option 'A'. Can you explain this answer?
sin220 + sin270 is equal to ______.
a)
1
b)
-1
c)
0
d)
2
|
|
Priyanka Kapoor answered |
Given
sin220 + sin270 = sin220 + [cos(90 - 20)]2
= sin220 + cos220 = 1
sin220 + sin270 = sin220 + [cos(90 - 20)]2
= sin220 + cos220 = 1
Chapter doubts & questions for Trigonometry - International Mathematics Olympiad (IMO) for Class 10 2025 is part of Class 10 exam preparation. The chapters have been prepared according to the Class 10 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Class 10 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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International Mathematics Olympiad (IMO) for Class 10
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