To solve the problem given, let's go through the properties of the inverse tangent function: Step 1: Define tan⁻¹x = y The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x Step 2: Range of tan⁻¹x The principalRead more
To solve the problem given, let’s go through the properties of the inverse tangent function:
Step 1: Define tan⁻¹x = y
The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x
Step 2: Range of tan⁻¹x
The principal range of the inverse tangent function tan⁻¹x (or arctan) is:
-π/2 ≤ y ≤ π/2
This signifies that y is in the interval (-π/2, π/2).
We are given to determine sin(tan⁻¹x), where |x| < 1 . Step 1: Let us take θ = tan⁻¹x . This gives us the following equations: tan(θ) = x Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below: -Opposite side= x Adjacent siRead more
We are given to determine sin(tan⁻¹x), where |x| < 1 .
Step 1: Let us take θ = tan⁻¹x .
This gives us the following equations:
tan(θ) = x
Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below:
-Opposite side= x
Adjacent side = 1
Step2: Applying Pythagorean theorem
To find the hypotenuse, use the Pythagorean theorem:
Hypotenuse = √(1² + x²) = √(1 + x²)
Step 3: Calculate sin(θ)
We know that:
sin(θ) = Opposite / Hypotenuse = x / √(1 + x²)
Final Answer:
Thus, the value of sin(tan⁻¹x) is:
x / √(1 + x²)
We are given to find the interval where sin⁻¹x > cos⁻¹x. Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π]. For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy: sin⁻¹x > cos⁻¹x Step 2:Read more
We are given to find the interval where sin⁻¹x > cos⁻¹x.
Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π].
For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy:
sin⁻¹x > cos⁻¹x
Step 2: Use the identity sin⁻¹x + cos⁻¹x = π/2 From the identity:
sin⁻¹x + cos⁻¹x = π/2 We can subtract cos⁻¹x from both sides to get:
sin⁻¹x = π/2 – cos⁻¹x Thus, for sin⁻¹x > cos⁻¹x, we need:
π/2 – cos⁻¹x > cos⁻¹x That is:
π/2 > 2cos⁻¹x which gives:
cos⁻¹x 1/√2
Hence, the required condition sin⁻¹x > cos⁻¹x holds when x belongs to:
(1/√2, 1)
We are given that y = cot⁻¹x and x < 0. We need to find the range of y. Step 1: Recall the range of cot⁻¹x The range of the inverse cotangent function cot⁻¹x is (0, π) for all real x. Step 2: Analyze the condition x < 0 When x < 0, the value of y = cot⁻¹x lies in the interval (π/2, π), becaRead more
We are given that y = cot⁻¹x and x < 0. We need to find the range of y.
Step 1: Recall the range of cot⁻¹x
The range of the inverse cotangent function cot⁻¹x is (0, π) for all real x.
Step 2: Analyze the condition x < 0
When x < 0, the value of y = cot⁻¹x lies in the interval (π/2, π), because the cotangent function is negative in this interval.
Final Answer:
Thus, the range of y is π/2 < y ≤ π.
The position vector of a center of mass in an n-particle system can be determined by a weighted average of the positions of the particles, and their masses play roles as weights. This average is influenced by the total mass in the system. According to Newton's second law, the motion of the center ofRead more
The position vector of a center of mass in an n-particle system can be determined by a weighted average of the positions of the particles, and their masses play roles as weights. This average is influenced by the total mass in the system. According to Newton’s second law, the motion of the center of mass is given by the fact that the total external force acting on the system is equal to the mass of the system multiplied by the acceleration of the center of mass. The acceleration of the center of mass is also determined by the net external forces acting on each individual particle so that it can behave like a single point mass.
If tan⁻¹ x = y, then:
To solve the problem given, let's go through the properties of the inverse tangent function: Step 1: Define tan⁻¹x = y The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x Step 2: Range of tan⁻¹x The principalRead more
To solve the problem given, let’s go through the properties of the inverse tangent function:
Step 1: Define tan⁻¹x = y
The function tan⁻¹x = y is a tangent function inverse. The definition simply states that for any real number x, angle y is such that:tan(y) = x
Step 2: Range of tan⁻¹x
The principal range of the inverse tangent function tan⁻¹x (or arctan) is:
-π/2 ≤ y ≤ π/2
This signifies that y is in the interval (-π/2, π/2).
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-2
sin (tan⁻¹x), where |x| < 1, is equal to
We are given to determine sin(tan⁻¹x), where |x| < 1 . Step 1: Let us take θ = tan⁻¹x . This gives us the following equations: tan(θ) = x Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below: -Opposite side= x Adjacent siRead more
We are given to determine sin(tan⁻¹x), where |x| < 1 .
Step 1: Let us take θ = tan⁻¹x .
This gives us the following equations:
tan(θ) = x
Since tangent of any angle is the ratio of the opposite side to the adjacent side, we can depict it in a right triangle as below:
-Opposite side= x
Adjacent side = 1
Step2: Applying Pythagorean theorem
To find the hypotenuse, use the Pythagorean theorem:
Hypotenuse = √(1² + x²) = √(1 + x²)
Step 3: Calculate sin(θ)
We know that:
sin(θ) = Opposite / Hypotenuse = x / √(1 + x²)
Final Answer:
Thus, the value of sin(tan⁻¹x) is:
x / √(1 + x²)
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-2
If sin⁻¹ x > cos⁻¹ x, then x, should lie in the interval
We are given to find the interval where sin⁻¹x > cos⁻¹x. Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π]. For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy: sin⁻¹x > cos⁻¹x Step 2:Read more
We are given to find the interval where sin⁻¹x > cos⁻¹x.
Step 1: Recall the properties of inverse trigonometric functions The range of sin⁻¹x is [-π/2, π/2] and the range of cos⁻¹x is [0, π].
For the condition sin⁻¹x > cos⁻¹x to be true, the values of x must satisfy:
sin⁻¹x > cos⁻¹x
Step 2: Use the identity sin⁻¹x + cos⁻¹x = π/2 From the identity:
sin⁻¹x + cos⁻¹x = π/2 We can subtract cos⁻¹x from both sides to get:
sin⁻¹x = π/2 – cos⁻¹x Thus, for sin⁻¹x > cos⁻¹x, we need:
π/2 – cos⁻¹x > cos⁻¹x That is:
π/2 > 2cos⁻¹x which gives:
cos⁻¹x 1/√2
Hence, the required condition sin⁻¹x > cos⁻¹x holds when x belongs to:
(1/√2, 1)
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/chapter-2/
If y = cot⁻¹ x, x < 0, then
We are given that y = cot⁻¹x and x < 0. We need to find the range of y. Step 1: Recall the range of cot⁻¹x The range of the inverse cotangent function cot⁻¹x is (0, π) for all real x. Step 2: Analyze the condition x < 0 When x < 0, the value of y = cot⁻¹x lies in the interval (π/2, π), becaRead more
We are given that y = cot⁻¹x and x < 0. We need to find the range of y.
Step 1: Recall the range of cot⁻¹x
The range of the inverse cotangent function cot⁻¹x is (0, π) for all real x.
Step 2: Analyze the condition x < 0
When x < 0, the value of y = cot⁻¹x lies in the interval (π/2, π), because the cotangent function is negative in this interval.
Final Answer:
Thus, the range of y is π/2 < y ≤ π.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/chapter-2/
Write an expression for the position vector of the centre of mass of n-particle system. Also write the equations of motion which govern the motion of the centre of mass.
The position vector of a center of mass in an n-particle system can be determined by a weighted average of the positions of the particles, and their masses play roles as weights. This average is influenced by the total mass in the system. According to Newton's second law, the motion of the center ofRead more
The position vector of a center of mass in an n-particle system can be determined by a weighted average of the positions of the particles, and their masses play roles as weights. This average is influenced by the total mass in the system. According to Newton’s second law, the motion of the center of mass is given by the fact that the total external force acting on the system is equal to the mass of the system multiplied by the acceleration of the center of mass. The acceleration of the center of mass is also determined by the net external forces acting on each individual particle so that it can behave like a single point mass.
See less