Question

In (esa]) =

• A. $\cot C$
• B. $c \cot C$
• C. $\cot \frac { C } { 2 }$
• D. $c \cot \frac { C } { 2 }$

Answer 1

Answer: (D)

$c \cot \frac { C } { 2 }$

Answer 2

$\left\{ \cot \frac { A } { 2 } + \cot \frac { B } { 2 } \right\} \left\{ a \sin ^ { 2 } \frac { B } { 2 } + b \sin ^ { 2 } \frac { A } { 2 } \right\}$

= $\left\{ \frac { \cos \frac { C } { 2 } } { \sin \frac { A } { 2 } \sin \frac { B } { 2 } } \right\} \left\{ a \sin ^ { 2 } \frac { B } { 2 } + b \sin ^ { 2 } \frac { A } { 2 } \right\}$

= $\left\{ \cos \frac { C } { 2 } \right\} \left\{ a \frac { \sin \frac { B } { 2 } } { \sin \frac { A } { 2 } } + b \frac { \sin \frac { A } { 2 } } { \sin \frac { B } { 2 } } \right\}$

$= \sqrt { \frac { s ( s - c ) } { a b } } \left\{ a \frac { \sqrt { \frac { ( s - a ) ( s - c ) } { a c } } } { \sqrt { \frac { ( s - b ) ( s - c ) } { b c } } } + b \frac { \sqrt { \frac { ( s - b ) ( s - c ) } { b c } } } { \sqrt { \frac { ( s - a ) ( s - c ) } { a c } } } \right\}$

$= \sqrt { \frac { s ( s - c ) } { a b } } \left\{ \sqrt { \left( \frac { s - a } { s - b } \right) a b } + \sqrt { \left( \frac { s - b } { s - a } \right) a b } \right\}$

=$\sqrt { s ( s - c ) } \left\{ \frac { s - a + s - b } { \sqrt { ( s - a ) ( s - b ) } } \right\} = \sqrt { s ( s - c ) } \left\{ \frac { 2 s - a - b } { \sqrt { ( s - a ) ( s - b ) } } \right\}$

$= c \sqrt { \frac { s ( s - c ) } { ( s - a ) ( s - b ) } } = c \cot \frac { C } { 2 }$ .

Trick : Such type of unconditional problems can be checked by putting the particular values for $a = 1$ $b = \sqrt { 3 }$ $c = 2$ and $A = 30 ^ { \circ }$ , $B = 60 ^ { \circ } , C = 90 ^ { \circ }$.

Hence expression is equal to 2 which is given by (4).