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Heya !!
Since, A, B and C are the inner angles of a ∆ABC.
So, A + B + C = 180°
B + C = 180° – A
Dividing by 2 on both sides
(B + C) / 2 = 90° – A/2
Now, tan (B + C) / 2 – L.H.S.
=> tan (90° – A/2)
=> cot A/2 – R.H.S.
Since, A, B and C are the inner angles of a ∆ABC.
So, A + B + C = 180°
B + C = 180° – A
Dividing by 2 on both sides
(B + C) / 2 = 90° – A/2
Now, tan (B + C) / 2 – L.H.S.
=> tan (90° – A/2)
=> cot A/2 – R.H.S.
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