Principle of Mathematical Induction

Principle of Mathematical Induction

Question

Prove that in triangle ABC,

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Answer

In ΔABC comma space space straight a over sinA space equals space straight b over sinB space equals space straight c over sinC space equals space straight k (say)
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∴      straight L. straight H. straight S. space equals space fraction numerator straight a minus straight b over denominator straight c end fraction space equals space fraction numerator straight k space sinA minus space straight k space sinB over denominator straight k space sinC end fraction space equals space fraction numerator sinA space minus space sinB over denominator sinC end fraction
                  equals fraction numerator 2 cos open parentheses begin display style fraction numerator straight A plus straight B over denominator 2 end fraction end style close parentheses sin open parentheses begin display style fraction numerator straight A minus straight B over denominator 2 end fraction end style close parentheses over denominator 2 sin open parentheses begin display style straight C over 2 end style close parentheses space cos open parentheses begin display style straight C over 2 end style close parentheses end fraction
                       open square brackets because space sinC space minus space sinD space equals space 2 cos space fraction numerator straight C plus straight D over denominator 2 end fraction space sin fraction numerator straight C minus straight D over denominator 2 end fraction comma space sin 2 straight theta space equals space 2 sinθ space cosθ close square brackets
                  equals space fraction numerator cos open parentheses begin display style fraction numerator straight A plus straight B over denominator 2 end fraction end style close parentheses sin open parentheses begin display style fraction numerator straight A minus straight B over denominator 2 end fraction end style close parentheses over denominator sin open parentheses begin display style straight C over 2 end style close parentheses cos open parentheses begin display style straight C over 2 end style close parentheses end fraction space equals space fraction numerator cos open parentheses begin display style straight pi over 2 end style minus begin display style straight C over 2 end style close parentheses sin open parentheses begin display style fraction numerator straight A minus straight B over denominator 2 end fraction end style close parentheses over denominator sin open parentheses begin display style straight C over 2 end style close parentheses cos open parentheses begin display style straight C over 2 end style close parentheses end fraction
                                                                                 open square brackets because space space straight A over 2 plus straight B over 2 plus straight C over 2 space equals space straight pi over 2 close square brackets
                     equals fraction numerator sin open parentheses begin display style straight C over 2 end style close parentheses sin open parentheses begin display style fraction numerator straight A minus straight B over denominator 2 end fraction end style close parentheses over denominator sin open parentheses begin display style straight C over 2 end style close parentheses space cos open parentheses begin display style straight C over 2 end style close parentheses end fraction space equals space fraction numerator sin open parentheses begin display style fraction numerator straight A minus straight B over denominator 2 end fraction end style close parentheses over denominator cos open parentheses begin display style straight C over 2 end style close parentheses end fraction space equals space straight R. straight H. straight S.

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