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.