What's your question?
  1. We are given that 0° ≤ A, B ≤ 90°, sin A = 1/2, and cos B = 1/2. We need to find the value of A + B. Step 1: Solve for A using sin A = 1/2 The sine function is defined as: sin A = opposite/hypotenuse. From trigonometric values, we know: sin 30° = 1/2. Since 0° ≤ A ≤ 90°, the only possible value forRead more

    We are given that 0° ≤ A, B ≤ 90°, sin A = 1/2, and cos B = 1/2. We need to find the value of A + B.

    Step 1: Solve for A using sin A = 1/2
    The sine function is defined as:
    sin A = opposite/hypotenuse.

    From trigonometric values, we know:
    sin 30° = 1/2.

    Since 0° ≤ A ≤ 90°, the only possible value for A is:
    A = 30°.

    Step 2: Solve for B using cos B = 1/2
    The cosine function is defined as:
    cos B = adjacent/hypotenuse.

    From trigonometric values, we know:
    cos 60° = 1/2.

    Since 0° ≤ B ≤ 90°, the only possible value for B is:
    B = 60°.

    Step 3: Calculate A + B
    Now, add the values of A and B:
    A + B = 30° + 60° = 90°.

    Step 4: Final Answer
    The value of A + B is:
    c) 90°.
    This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/

    See less
    • 25
  2. The correct answer is (d) zero. When a conductor is placed in an electric field, free electrons within it redistribute to cancel the external field inside. This results in electrostatic equilibrium, where the net electric field inside the conductor becomes zero. This phenomenon ensures no electric fRead more

    The correct answer is (d) zero.
    When a conductor is placed in an electric field, free electrons within it redistribute to cancel the external field inside. This results in electrostatic equilibrium, where the net electric field inside the conductor becomes zero. This phenomenon ensures no electric force acts on charges within the conductor.

    For more visit here:
    https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-2/

    See less
    • 15
  3. We are tasked with finding the maximum value of 1/secθ for 0° ≤ θ < 90°. Step 1: Recall the definition of secant The secant function is defined as: secθ = 1/cosθ. Thus, the reciprocal of secant is: 1/secθ = cosθ. Step 2: Analyze the behavior of cosθ in the given range For 0° ≤ θ < 90°: - The cRead more

    We are tasked with finding the maximum value of 1/secθ for 0° ≤ θ < 90°.

    Step 1: Recall the definition of secant
    The secant function is defined as:
    secθ = 1/cosθ.

    Thus, the reciprocal of secant is:
    1/secθ = cosθ.

    Step 2: Analyze the behavior of cosθ in the given range
    For 0° ≤ θ < 90°:
    – The cosine function decreases from cos 0° = 1 to cos 90° = 0 (but does not actually reach 0 since θ < 90°).
    – Therefore, the maximum value of cosθ occurs at θ = 0°.

    At θ = 0°:
    cos 0° = 1.

    Step 3: Conclusion
    The maximum value of 1/secθ is equal to the maximum value of cosθ, which is 1.

    The correct answer is:
    a) 1
    This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/

    See less
    • 25
  4. We are given the equation: sin 2A = 2 sin A. Step 1: Recall the double-angle identity for sine The double-angle identity for sine is: sin 2A = 2 sin A cos A. Substitute this into the given equation: 2 sin A cos A = 2 sin A. Step 2: Simplify the equation Divide both sides of the equation by 2 (assumiRead more

    We are given the equation:
    sin 2A = 2 sin A.

    Step 1: Recall the double-angle identity for sine
    The double-angle identity for sine is:
    sin 2A = 2 sin A cos A.

    Substitute this into the given equation:
    2 sin A cos A = 2 sin A.

    Step 2: Simplify the equation
    Divide both sides of the equation by 2 (assuming sin A ≠ 0):
    sin A cos A = sin A.

    Rearrange the terms:
    sin A cos A – sin A = 0.

    Factor out sin A:
    sin A (cos A – 1) = 0.

    Step 3: Solve for A
    This equation is satisfied if either:
    1. sin A = 0, or
    2. cos A – 1 = 0.

    Case 1: sin A = 0
    The sine function is zero when A = 0°, 180°, etc. Among the given options, A = 0° satisfies this condition.

    Case 2: cos A – 1 = 0
    Solve for cos A:
    cos A = 1.

    The cosine function equals 1 when A = 0°, 360°, etc. Again, among the given options, A = 0° satisfies this condition.

    Step 4: Verify the solution
    Substitute A = 0° into the original equation:
    sin 2(0°) = 2 sin(0°).

    The left-hand side:
    sin 2(0°) = sin 0° = 0.

    The right-hand side:
    2 sin(0°) = 2(0) = 0.

    Since both sides are equal, A = 0° satisfies the equation.

    Step 5: Final Answer
    The value of A is 0°.

    The correct answer is:
    a) 0°
    This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/

    See less
    • 20
  5. We are given: 16 cot x = 12. Step 1: Solve for cot x Rearrange the equation to solve for cot x: cot x = 12/16 = 3/4. Step 2: Express tan x in terms of cot x Using the identity cot x = 1/tan x, we can write: tan x = 1/cot x = 1/(3/4) = 4/3. Step 3: Express sin x and cos x in terms of tan x Using theRead more

    We are given:
    16 cot x = 12.

    Step 1: Solve for cot x
    Rearrange the equation to solve for cot x:
    cot x = 12/16 = 3/4.

    Step 2: Express tan x in terms of cot x
    Using the identity cot x = 1/tan x, we can write:
    tan x = 1/cot x = 1/(3/4) = 4/3.

    Step 3: Express sin x and cos x in terms of tan x
    Using the identity tan x = sin x / cos x, we can write:
    sin x = 4k and cos x = 3k,
    where k is a positive constant such that sin²x + cos²x = 1 (Pythagorean identity).

    Substitute sin x = 4k and cos x = 3k into the identity:
    (4k)² + (3k)² = 1
    16k² + 9k² = 1
    25k² = 1
    k² = 1/25
    k = √(1/25)
    k = 1/5.

    Thus:
    sin x = 4k = 4/5,
    cos x = 3k = 3/5.

    Step 4: Simplify the given expression
    We are tasked with finding the value of:
    (sin x – cos x) / (sin x + cos x).

    Substitute sin x = 4/5 and cos x = 3/5 into the expression:

    Numerator:
    sin x – cos x = (4/5) – (3/5)
    = (4 – 3)/5
    = 1/5.

    Denominator:
    sin x + cos x = (4/5) + (3/5)
    = (4 + 3)/5
    = 7/5.

    Thus, the entire expression becomes:
    (sin x – cos x) / (sin x + cos x) = (1/5) / (7/5).

    Simplify:
    (1/5) / (7/5) = 1/7.

    Step 5: Final Answer
    The value of (sin x – cos x) / (sin x + cos x) is 1/7.

    The correct answer is:
    a) 1/7
    This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/

    See less
    • 24