find acute angie A and B satisfying cot(A+B)=1 , cosec(A-B)
To find the acute angles A and B that satisfy the equation cot(A+B) = 1 and cosec(A-B), we need to use the definitions and properties of trigonometric functions.
Let's start with the equation cot(A+B) = 1. Since cotangent is the reciprocal of the tangent function, we can rewrite the equation as tan(A+B) = 1.
Now, to solve for A and B, we can take the inverse tangent (or arctan) of both sides of the equation. Applying the arctan function to both sides gives us A + B = arctan(1).
Since we are looking for acute angles, the arctan function will give us an angle between -π/2 and π/2. In this case, arctan(1) = π/4, which is in the desired range.
So, we have A + B = π/4.
Now, let's consider the second equation, cosec(A-B). Since cosecant is the reciprocal of the sine function, we can rewrite it as sin(A-B).
To find A and B, we can apply the arcsin function to both sides of the equation. This gives us A - B = arcsin(cosec(A-B)).
Since we are looking for acute angles, the arcsine function will give us an angle between -π/2 and π/2. So, A - B will be within this range.
As for the value of cosec(A-B), it is equal to 1/sin(A-B). In order for this value to be defined, sin(A-B) cannot be equal to 0. This means that A-B cannot be equal to 0 or any multiple of π.
Therefore, to find A and B, we need to solve the system of equations:
A + B = π/4
A - B ≠ 0, kπ (where k is an integer)
From the equation A + B = π/4, we can solve for B by subtracting A from both sides:
B = π/4 - A
Substituting this expression into A - B ≠ 0, kπ, we have:
A - (π/4 - A) ≠ 0, kπ
2A - π/4 ≠ 0, kπ
2A ≠ π/4, kπ
A ≠ π/8, kπ/2
From the last equation, A cannot be equal to π/8, kπ/2 for any integer k.
Therefore, there are no acute angles A and B that satisfy both cot(A+B) = 1 and cosec(A-B).