Question

TanA/2+tanB/2+tanC/2=bc+ca+ab-S^2/delta​

Answer 1

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Answer 2

Step-by-step explanation:

Given TanA/2 + tanB/2 + tanC/2 = bc+ca+ab-S^2/delta

So we have  

                Tan A/2 + tan B/2 + tan C/2

• So by using half angle formula we get
•             (s – b) (s – c) / Δ + (s – c) (s – a) / Δ + (s – a) (s – b) / Δ
•        So s^2 – sc – bs + bc + s^2 – cs – sa + ca + s^2 – as – sb + ab / Δ
•        So 3s^2 – 2as – 2bs – 2cs + bc + ca + ab / Δ
•       So bc + ca + ab + 3s^2 – 2s (a + b + c) / Δ
•      So bc + ca + ab + 3s^2 – 4s^2 / Δ (since a + b + c = 2s)
•      So bc + ca + ab – s^2 / Δ (r. h.s)

Hence it is proved. (l. h.s = r. h.s)

Reference link will be

https://brainly.in/question/1768819