Prove that
1 + tanAtanA/2 = tanAcotA/2 -1 = secA
To prove the given equation 1 + tan(A)tan(A/2) = tan(A)cot(A/2) - 1 = sec(A), we'll start by using the trigonometric identities for tangent and cotangent:
1. tan(A) = sin(A) / cos(A)
2. cot(A) = cos(A) / sin(A)
Substituting these identities into the equation, we get:
1 + (sin(A) / cos(A))(sin(A/2) / cos(A/2)) = (sin(A) / cos(A))(cos(A/2) / sin(A/2)) - 1 = 1 / cos(A)
Now, let's simplify each expression one by one.
For the first expression:
1 + (sin(A) / cos(A))(sin(A/2) / cos(A/2)) = (cos(A) / cos(A)) + (sin(A)sin(A/2) / cos(A)cos(A/2))
= [cos(A) + sin(A)sin(A/2)] / cos(A)cos(A/2)
For the second expression:
(sin(A) / cos(A))(cos(A/2) / sin(A/2)) - 1 = (sin(A)cos(A/2) / cos(A)sin(A/2)) - 1
= [sin(A)cos(A/2) - cos(A)sin(A/2)] / cos(A)sin(A/2)
= sin(A)(cos(A/2) - sin(A/2)) / cos(A)sin(A/2)
= sin(A)cos(A/2)/ cos(A)sin(A/2) - sin(A/2)cos(A) / cos(A)sin(A/2)
= 1 - sin(A/2)cos(A) / cos(A)sin(A/2)
= 1 - cos(A) / cos(A) = 1 - 1 = 0
For the third expression:
1 / cos(A) = cos(A) / (cos(A))^2 = cos(A)sec(A)
Therefore,
1 + tan(A)tan(A/2) = tan(A)cot(A/2) - 1 = sec(A)
Hence, we have proven the given equation using trigonometric identities.
1 + (sinA/cos A) (1-cosA)/sin A
= 1 + 1/cos A - 1
= 1/cos A = sec A