Principle Of Mathematical Induction

Question
CBSEENMA11014351

Prove that in any triangle ABC:

(i)   straight r equals left parenthesis straight s minus straight a right parenthesis tan straight A over 2 equals left parenthesis straight s minus straight b right parenthesis tan straight B over 2 space equals space left parenthesis straight s minus straight c right parenthesis tan straight C over 2

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Solution

Proof: (i) Let I be the incentre and D, E, F be the points of contact of the incircle with the sides.
                                   
Join ID, IE, IF such that
ID perpendicular BC,  IE perpendicular CA and IF perpendicular AB
Now,        2s - 2a = AB + CA - BC
                          = (AF + FB) + (CE + EA) - (BD + DC)
                          = (AF + BD) + (CD + AF) - (BD + DC)
                                                        [∵ AE = AF, BF = BD and CD = CE]
                          = 2AF               
rightwards double arrow              AF = s - a                                                                 ...(i)
Now, from right angle triangle AFI,
                  IF over AF space equals space tan straight A over 2
or                           space space straight r equals space AF space tan straight A over 2 equals left parenthesis straight s minus straight a right parenthesis tan straight A over 2                 [By using (i)]               
Similarly, we can show other two results.    
(ii)   Now,      BC space equals space BD space plus space DC space equals space rcot straight B over 2 plus rcot straight C over 2                [From figure]
                    space space space space straight a equals straight r open square brackets cot straight B over 2 plus cot straight C over 2 close square brackets
rightwards double arrow                space space straight a equals straight r open square brackets fraction numerator cos begin display style straight B over 2 end style sin begin display style straight C over 2 end style plus cos begin display style straight C over 2 end style sin begin display style straight B over 2 end style over denominator sin begin display style straight B over 2 end style sin begin display style straight C over 2 end style end fraction close square brackets space equals space straight r open square brackets fraction numerator sin open parentheses begin display style fraction numerator straight B plus straight C over denominator 2 end fraction end style close parentheses over denominator sin begin display style straight B over 2 end style sin begin display style straight C over 2 end style end fraction close square brackets
                         equals straight r open square brackets fraction numerator sin open parentheses begin display style straight pi over 2 end style minus begin display style straight A over 2 end style close parentheses over denominator sin begin display style straight B over 2 end style sin begin display style straight C over 2 end style end fraction close square brackets equals straight r fraction numerator cos begin display style straight A over 2 end style over denominator sin begin display style straight B over 2 end style sin begin display style straight C over 2 end style end fraction
or            straight r equals fraction numerator asin begin display style straight B over 2 end style sin begin display style straight C over 2 end style over denominator cos begin display style straight A over 2 end style end fraction
Similarly, we can show other two results.
                            

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