Question

In ΔABC, prove that $tan(\frac{A}{2}) + tan(\frac{B}{2}) + tan(\frac{C}{2}) =\ \frac{bc\ +\ ca\ +\ ab\ -\ s^{2}}{\triangle}$.

Answer 1

Solution :

LHS = tanA/2 + tanB/2 + tanC/2

= [(s-b)(s-c)/∆]

+ [(s-c)(s-a)/∆]

+ [(s-a)(s-b)/∆]

=[(s²-cs-bs+bc)+(s²-as-cs+ac)+(s²-as-bs+ab)]/∆

=(3s²-2as - 2bs - 2cs + bc + ca + ab )/∆

= [ bc + ca + ab + 3s² - 2s( a + b + c )]/∆

= [ bc + ca + ab + 3s² - 2s(2s) ]/∆

= ( bc + ca + ab + 3s² - 4s² ]/∆

= ( bc + ca + ab - s² )/∆

= RHS

•••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

tanA/2 + tanB/2 + tanC/2

=> [(s-b)(s-c)/∆]

+ [(s-c)(s-a)/∆]

+ [(s-a)(s-b)/∆]

=> [(s²-cs-bs+bc)+(s²-as-cs+ac)+(s²-as-bs+ab)]/∆

=> (3s²-2as - 2bs - 2cs + bc + ca + ab )/∆

=> [ bc + ca + ab + 3s² - 2s( a + b + c )]/∆

=> [ bc + ca + ab + 3s² - 2s(2s) ]/∆

=> ( bc + ca + ab + 3s² - 4s² ]/∆

=> ( bc + ca + ab - s² )/∆

I HOPE IT'S HELP YOU DEAR,

THANKS