Question

In ΔABC, prove that $cot(\frac{A}{2}) + cot(\frac{B}{2}) + cot(\frac{C}{2}) =\ \frac{s^{2}}{\triangle}$.

Answer 1

Solution :

LHS = cotA/2 + cotB/2 + cotC/2

= [s(s-a)]/∆ + [s(s-b)]/∆ + [s(s-c)]/∆

= ( s/∆ ) [ s - a + s - b + s - c ]

= ( s/∆ ) [ 3s - ( a + b + c ) ]

= ( s/∆ ) [ 3s - 2s ]

= ( s/∆ ) × s

= s²/∆

= RHS

••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

now, cotA/2 + cotB/2 + cotC/2

=> [s(s-a)]/∆ + [s(s-b)]/∆ + [s(s-c)]/∆

=> ( s/∆ ) [ s - a + s - b + s - c ]

=> ( s/∆ ) [ 3s - ( a + b + c ) ]

=> ( s/∆ ) [ 3s - 2s ]

=> ( s/∆ ) × s

=> s²/∆

I HOPE IT'S HELP YOU DEAR,

THANKS