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.
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.
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.
We are given that the angles A, B, and C of a triangle △ABC form an increasing arithmetic progression (AP). We need to find the value of sin B. Step 1: Properties of angles in a triangle The sum of the angles in any triangle is: A + B + C = 180°. Since A, B, and C form an increasing arithmetic progrRead more
We are given that the angles A, B, and C of a triangle △ABC form an increasing arithmetic progression (AP). We need to find the value of sin B.
Step 1: Properties of angles in a triangle
The sum of the angles in any triangle is:
A + B + C = 180°.
Since A, B, and C form an increasing arithmetic progression, let the common difference of the AP be d. Then we can write:
A = B – d, B = B, C = B + d.
Substitute these into the angle sum property:
(B – d) + B + (B + d) = 180°.
Simplify:
3B = 180°.
Solve for B:
B = 60°.
Step 2: Find sin B
Now that we know B = 60°, we use the standard trigonometric value:
sin 60° = √3/2.
Step 3: Final Answer
The value of sin B is:
√3/2.
The correct answer is:
b) √3/2
This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.
To determine which of the given trigonometric expressions is not defined, we analyze each option based on the definitions and properties of trigonometric functions. Step 1: Analyze each option a) cos 0° The cosine function is defined for all real numbers. The value of cos 0° is: cos 0° = 1. Thus, coRead more
To determine which of the given trigonometric expressions is not defined, we analyze each option based on the definitions and properties of trigonometric functions.
Step 1: Analyze each option
a) cos 0°
The cosine function is defined for all real numbers. The value of cos 0° is:
cos 0° = 1.
Thus, cos 0° is defined.
b) tan 45°
The tangent function is defined as tan θ = sin θ / cos θ. For θ = 45°:
tan 45° = sin 45° / cos 45° = (1/√2) / (1/√2) = 1.
Thus, tan 45° is defined.
c) sec 90°
The secant function is defined as sec θ = 1 / cos θ. For θ = 90°:
cos 90° = 0.
Since division by zero is undefined, sec 90° = 1 / cos 90° = 1 / 0 is not defined.
Thus, sec 90° is not defined.
d) sin 90°
The sine function is defined for all real numbers. The value of sin 90° is:
sin 90° = 1.
Thus, sin 90° is defined.
Step 2: Final Answer
The expression that is not defined is:
c) sec 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 0° ≤ θ < 90°, the maximum value of 1/secθ is
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:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/
sin 2A = 2 sinA is true when A =
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:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/
If 16 cot x = 12, then sin x – cos x/ sin x + cos x equals
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:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/
If angles A, B, C of a Triangle ABC from an increasing Ao, then sin B =
We are given that the angles A, B, and C of a triangle △ABC form an increasing arithmetic progression (AP). We need to find the value of sin B. Step 1: Properties of angles in a triangle The sum of the angles in any triangle is: A + B + C = 180°. Since A, B, and C form an increasing arithmetic progrRead more
We are given that the angles A, B, and C of a triangle △ABC form an increasing arithmetic progression (AP). We need to find the value of sin B.
Step 1: Properties of angles in a triangle
The sum of the angles in any triangle is:
A + B + C = 180°.
Since A, B, and C form an increasing arithmetic progression, let the common difference of the AP be d. Then we can write:
A = B – d, B = B, C = B + d.
Substitute these into the angle sum property:
(B – d) + B + (B + d) = 180°.
Simplify:
3B = 180°.
Solve for B:
B = 60°.
Step 2: Find sin B
Now that we know B = 60°, we use the standard trigonometric value:
sin 60° = √3/2.
Step 3: Final Answer
The value of sin B is:
√3/2.
The correct answer is:
b) √3/2
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:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/
Which of the following is not defined?
To determine which of the given trigonometric expressions is not defined, we analyze each option based on the definitions and properties of trigonometric functions. Step 1: Analyze each option a) cos 0° The cosine function is defined for all real numbers. The value of cos 0° is: cos 0° = 1. Thus, coRead more
To determine which of the given trigonometric expressions is not defined, we analyze each option based on the definitions and properties of trigonometric functions.
Step 1: Analyze each option
a) cos 0°
The cosine function is defined for all real numbers. The value of cos 0° is:
cos 0° = 1.
Thus, cos 0° is defined.
b) tan 45°
The tangent function is defined as tan θ = sin θ / cos θ. For θ = 45°:
tan 45° = sin 45° / cos 45° = (1/√2) / (1/√2) = 1.
Thus, tan 45° is defined.
c) sec 90°
The secant function is defined as sec θ = 1 / cos θ. For θ = 90°:
cos 90° = 0.
Since division by zero is undefined, sec 90° = 1 / cos 90° = 1 / 0 is not defined.
Thus, sec 90° is not defined.
d) sin 90°
The sine function is defined for all real numbers. The value of sin 90° is:
sin 90° = 1.
Thus, sin 90° is defined.
Step 2: Final Answer
The expression that is not defined is:
c) sec 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:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/