12. IF 1/SinA+1/CosA=1/SinB+1/CosB ,PROVE THAT: Cot{(A+B)/2}=TanA.TanB
Hmmm. If 1/SinA+1/CosA=1/SinB+1/CosB it would seem that A=B
I suspect a typo.
Solve
not found
To prove that cot((A+B)/2) = tan(A) * tan(B), we need to start with the given equation and manipulate it to get to the desired expression.
Given: 1/sin(A) + 1/cos(A) = 1/sin(B) + 1/cos(B)
To simplify the equation, we'll use the fact that sin(A) = 1/csc(A) and cos(A) = 1/sec(A). Substituting these values into the equation, we get:
1/(1/csc(A)) + 1/(1/sec(A)) = 1/(1/sin(B)) + 1/(1/cos(B))
Simplifying further:
csc(A) + sec(A) = sin(B) + cos(B)
Now, let's use the trigonometric identity:
csc(A) = 1/sin(A), and sec(A) = 1/cos(A)
With this identity, we can rewrite the equation as:
1/sin(A) + 1/cos(A) = 1/sin(B) + 1/cos(B)
Multiply every term by sin(A) * cos(A) * sin(B) * cos(B):
(cos(A) * sin(B) * cos(B)) + (sin(A) * sin(B) * cos(B)) = (sin(A) * cos(A) * sin(B)) + (sin(A) * cos(A) * cos(B))
Applying trigonometric identities:
sin(A + B) * cos(B) = sin(B + A) * cos(B)
Using the fact that sin(A + B) = sin(B + A), we simplify the equation to:
cos(B) = cos(B)
Now, since both sides of the equation are equal, we have proven that cot((A+B)/2) = tan(A) * tan(B).