Question

If in a triangle $A B C$, then $\frac { 1 } { a + c } + \frac { 1 } { b + c } =$

• A. $\frac { 1 } { a + b + c }$
• B. $\frac { 2 } { a + b + c }$
• C. $\frac { 3 } { a + b + c }$
• D. None of these

Answer 1

Answer: (C)

$\frac { 3 } { a + b + c }$

Answer 2

$\cos C = \frac { \pi } { 3 } \Rightarrow a ^ { 2 } + b ^ { 2 } - c ^ { 2 } = a b$

⇒ $b ^ { 2 } + b c + a ^ { 2 } + a c = a b + a c + b c + c ^ { 2 }$

⇒ $b ( b + c ) + a ( a + c ) = ( a + c ) ( b + c )$

Divide by $( a + c ) ( b + c )$ and add 2 on both sides