find acute angles A and B satisfying cot(A+B)=1 , cosec (A-B)=2

To find the acute angles A and B satisfying cot(A + B) = 1 and cosec(A - B) = 2, we can use the trigonometric identities and algebraic manipulations.

Let's start with the first equation:

cot(A + B) = 1

Using the identity cot(x) = 1/tan(x), we can rewrite the equation as:

tan(A + B) = 1

Now, we can find the values of A + B by taking the inverse tangent (tan^(-1)) of both sides:

A + B = tan^(-1)(1)

The inverse tangent of 1 is π/4 (45 degrees). Therefore, we have:

A + B = π/4

Now, let's move on to the second equation:

cosec(A - B) = 2

Using the identity cosec(x) = 1/sin(x), we can rewrite the equation as:

sin(A - B) = 1/2

Now, we can find the values of A - B by taking the inverse sine (sin^(-1)) of both sides:

A - B = sin^(-1)(1/2)

The inverse sine of 1/2 is π/6 (30 degrees). Therefore, we have:

A - B = π/6

Now, we have two equations:

A + B = π/4
A - B = π/6

To solve these equations and find the values of A and B, we can add the two equations together:

(A + B) + (A - B) = π/4 + π/6

Simplifying:

2A = (3π + 2π) / 12
2A = 5π / 12

Dividing both sides by 2:

A = 5π / 24

Next, we can subtract the second equation from the first equation:

(A + B) - (A - B) = π/4 - π/6

Simplifying:

2B = (3π - 2π) / 12
2B = π / 12

Dividing both sides by 2:

B = π / 24

Therefore, the acute angles A and B satisfying cot(A + B) = 1 and cosec(A - B) = 2 are:

A = 5π / 24
B = π / 24

in degrees,

A+B=45
A-B=30

A = 37.5
B = 7.5

Thanks