Question

prove that c/a-b=tan a/2+tan b/2 /tan a/2-tan b/2​

Answer 1

Given: The expression: c/a-b = tan A/2 + tan B/2 / tan A/2 - tan B/2​

To find: Prove the above expression.

Solution:

• We know the formula:

               a / sin A = b / sin B = c / sin C

• From this we have:

               a/c = sin A / sin C and b/c = sin B / sin C   .............(i)

• Now consider LHS, we have: c / (a - b)

• It can be written as:

               1 / (a/c - b/c)

• Putting (i) in above equation, we get:

               1 / ((sin A / sin C) - (sin B / sin C))

               sin C / sin A - sin B

• Now consider RHS, we have: (tan A/2 + tan B/2) / (tan A/2 - tan B/2)

• Converting tan to sin and cos, we get:

               (sin A/2/cos A/2 + sin B/2/cos B/2) / (sin A/2/cos A/2 - sin B/2/cos B/2)

               (sin A/2 . cos B/2 + sin B/2 . cos A/2) / (sin A/2 . cos B/2 - sin B/2 . cos A/2)

• It can be written as:

               2 sin(A + B/2) cos (A + B/2) / 2 sin(A - B/2) cos (A + B/2)

               sin (A + B) / sin A - sin B

               sin (180 - A - B) / sin A - sin B

              sin C / sin A - sin B

               LHS.......hence proved.

Answer:

          So we proved that  c/a-b = tan A/2 + tan B/2 / tan A/2 - tan B/2​