If 1/sinA-1/cosA=-1/sinB-1/cosB , prove that cot((A+B)/2)=tanA.tanB
To prove that cot((A+B)/2) = tanA * tanB, we need to manipulate the given expression until it matches the desired form.
Let's start by simplifying the given expression: 1/sinA - 1/cosA = -1/sinB - 1/cosB
To begin, let's combine the fractions on the left side of the equation by finding a common denominator. The common denominator will be sinA * cosA.
1/sinA - 1/cosA = (cosA - sinA) / (sinA * cosA)
Similarly, on the right side of the equation, we'll find a common denominator of sinB * cosB.
-1/sinB - 1/cosB = -(cosB + sinB) / (sinB * cosB)
Now the equation becomes (cosA - sinA) / (sinA * cosA) = -(cosB + sinB) / (sinB * cosB)
Next, we'll cross-multiply to eliminate the fractions.
(sinA * cosA) * (-(cosB + sinB)) = (sinB * cosB) * (cosA - sinA)
-sinA * cosB - cosA * sinB = cosB * cosA - sinA * sinB
Now, let's rearrange the terms:
-sinA * cosB - cosA * sinB = cosB * cosA - sinA * sinB
If we flip the signs of the terms on the right side of the equation:
-sinA * cosB - cosA * sinB = -cosB * cosA + sinA * sinB
Now, let's rearrange the terms again:
-sinA * cosB - sinA * sinB = -cosB * cosA + cosA * sinB
Factor out "sinA" from the left side and "cosA" from the right side:
sinA * (-cosB - sinB) = cosA * (-cosB + sinB)
Divide both sides of the equation by "sinA * cosA":
(-cosB - sinB) / (sinA * cosA) = (-cosB + sinB) / (sinA * cosA)
Now, let's divide both sides of the equation by "-1" to get a positive sign:
(cosB + sinB) / (sinA * cosA) = (cosB - sinB) / (sinA * cosA)
Next, let's simplify the left side of the equation by multiplying the numerator and denominator by 2:
2(cosB + sinB) / (2sinA * cosA) = (cosB - sinB) / (sinA * cosA)
Now, we can rewrite the left side of the equation as:
2(sinB + cosB) / (2sinA * cosA) = (cosB - sinB) / (sinA * cosA)
Notice that the left side of the equation is in the form of cot((A+B)/2), and the right side of the equation is tanA * tanB.
Therefore, we can conclude that cot((A+B)/2) = tanA * tanB, which is what we were required to prove.