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Question: In \(\triangle A B C\), \(\left( \cot \frac { A } { 2 } + \cot \frac { B } { 2 } \right) \left( a \s...

In △ABC\triangle A B C, (cot⁡A2+cot⁡B2)(asin⁡2B2+bsin⁡2A2)\left( \cot \frac { A } { 2 } + \cot \frac { B } { 2 } \right) \left( a \sin ^ { 2 } \frac { B } { 2 } + b \sin ^ { 2 } \frac { A } { 2 } \right)equal to

A

cot⁡C\cot C

B

ccot⁡Cc \cot C

C

cot⁡C2\cot \frac { C } { 2 }

D

ccot⁡C2c \cot \frac { C } { 2 }

Answer

ccot⁡C2c \cot \frac { C } { 2 }

Explanation

Solution

{cot⁡A2+cot⁡B2}{asin⁡2B2+bsin⁡2A2}\left\{ \cot \frac { A } { 2 } + \cot \frac { B } { 2 } \right\} \left\{ a \sin ^ { 2 } \frac { B } { 2 } + b \sin ^ { 2 } \frac { A } { 2 } \right\} ={cos⁡C2sin⁡A2sin⁡B2}{asin⁡2B2+bsin⁡2A2}= \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\} ={cos⁡C2}{asin⁡B2sin⁡A2+bsin⁡A2sin⁡B2}= \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\}

=s(s−c)ab{a(s−a)(s−c)ac(s−b)(s−c)bc+b(s−b)(s−c)bc(s−a)(s−c)ac}= \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\} =s(s−c)ab{(s−as−b)ab+(s−bs−a)ab}= \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\} =s(s−c){s−a+s−b(s−a)(s−b)}=s(s−c){2s−a−b(s−a)(s−b)}= \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\} =cs(s−c)(s−a)(s−b)=ccot⁡C2= 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=1a = 1 , b=3b = \sqrt { 3 } , c=2c = 2 and A=30∘A = 30 ^ { \circ } , B=60∘B = 60 ^ { \circ }, C=90∘C = 90 ^ { \circ } , Here expression is equal to 2 which is given by (4).