IF

(1/SinA)+(1/cosA)= (1/sinB)+(1/cosB)
Prove that cot((A+B)/2)=tanA tanB
Please helppppppppppp

To prove that cot((A+B)/2) = tanA * tanB, we will start by manipulating the given equation and expressing the left-hand side in terms of tanA and tanB.

Given equation: (1/sinA) + (1/cosA) = (1/sinB) + (1/cosB)

We will work on the left-hand side of the equation:

1/sinA + 1/cosA

To manipulate this expression, we will multiply the numerator and denominator of the first term (1/sinA) by cosA, and the numerator and denominator of the second term (1/cosA) by sinA:

(cosA * 1) / (sinA * cosA) + (sinA * 1) / (cosA * sinA)

Simplifying:

cosA / (sinA * cosA) + sinA / (cosA * sinA)

Now, we can simplify further by canceling out the common factors in the denominators:

cosA / (sinA * cosA) + sinA / (cosA * sinA)
= (cosA + sinA) / (sinA * cosA)

We will perform the same manipulation on the right-hand side of the equation:

1/sinB + 1/cosB

Multiply the numerator and denominator of the first term (1/sinB) by cosB, and the numerator and denominator of the second term (1/cosB) by sinB:
(cosB * 1) / (sinB * cosB) + (sinB * 1) / (cosB * sinB)

Simplifying:

cosB / (sinB * cosB) + sinB / (cosB * sinB)

Canceling out the common factors in the denominators:

cosB / (sinB * cosB) + sinB / (cosB * sinB)
= (cosB + sinB) / (sinB * cosB)

Now, we have both the left-hand side and the right-hand side of the given equation written in the same form:

((cosA + sinA) / (sinA * cosA)) = ((cosB + sinB) / (sinB * cosB))

Since the left-hand side and the right-hand side of the equation are equal, we can set them equal to a variable, let's say 'k':

(cosA + sinA) / (sinA * cosA) = (cosB + sinB) / (sinB * cosB) = k

Now, we can solve for cosA and sinA in terms of k:

cosA = k * sinA * cosA
sinA = k * sinA * cosA

Dividing both sides by cosA:

tanA = k * sinA

Similarly, for B:

tanB = k * sinB

Substituting these values into the original equation:

k * sinA * k * sinB = k^2 * sinA * sinB = tanA * tanB

Now, we can substitute the value of k into the equation:

k^2 * sinA * sinB = tanA * tanB
k^2 = cot((A+B)/2)

Hence, we have proved that cot((A+B)/2) = tanA * tanB.