Principle of Mathematical Induction

In any triangle ABC, prove that (i) (ii) (iii) - Wired Faculty

Question

In any triangle ABC, prove that

(i) $tan open parentheses fraction numerator straight A minus straight B over denominator 2 end fraction close parentheses space equals space fraction numerator straight a minus straight b over denominator straight a plus straight b end fraction cot straight C over 2$

(ii) $tan open parentheses fraction numerator straight B minus straight C over denominator 2 end fraction close parentheses space equals space fraction numerator straight b minus straight c over denominator straight b plus straight c end fraction cot straight A over 2$

(iii) $tan open parentheses fraction numerator straight C minus straight A over denominator 2 end fraction close parentheses space equals space fraction numerator straight c minus straight a over denominator straight c plus straight a end fraction cot straight B over 2$

Solution

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$space space equals space cot open parentheses fraction numerator straight A plus straight B over denominator 2 end fraction close parentheses tan open parentheses fraction numerator straight A minus straight B over denominator 2 end fraction close parentheses cot straight C over 2 space equals space cot open parentheses fraction numerator straight pi minus straight C over denominator 2 end fraction close parentheses space tan open parentheses fraction numerator straight A minus straight B over denominator 2 end fraction close parentheses space cot straight C over 2$
[∵ $straight A plus straight B plus straight C equals straight pi$ ]
 $equals cot open parentheses straight pi over 2 minus straight C over 2 close parentheses tan open parentheses fraction numerator straight A minus straight B over denominator 2 end fraction close parentheses cot straight C over 2 equals tan straight C over 2 tan fraction numerator straight A minus straight B over denominator 2 end fraction cot straight C over 2$
 $space space equals space tan open parentheses fraction numerator straight A minus straight B over denominator 2 end fraction close parentheses equals straight L. straight H. straight S.$
Similarly, we can prove
(ii) $tan open parentheses fraction numerator straight B minus straight C over denominator 2 end fraction close parentheses space equals space fraction numerator straight b minus straight c over denominator straight b plus straight c end fraction cot straight A over 2$
and (iii) $tan open parentheses fraction numerator straight C minus straight A over denominator 2 end fraction close parentheses space equals space fraction numerator straight c minus straight a over denominator straight c plus straight a end fraction cot straight B over 2$

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Principle Of Mathematical Induction

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