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Principle Of Mathematical Induction

Question
CBSEENMA11013851

Show that:

space space open parentheses fraction numerator cosA plus cosB over denominator sinA minus sinB end fraction close parentheses to the power of straight n plus open parentheses fraction numerator sinA plus sinB over denominator cosA minus cosB end fraction close parentheses to the power of straight n space equals space open curly brackets table attributes columnalign left end attributes row cell 2 cot to the power of straight n open parentheses fraction numerator straight A minus straight B over denominator 2 end fraction close parentheses space space when space straight n space is space even end cell row cell space space space space space 0 space space space space space space space space space space space space space space space space when space straight n space is space odd end cell end table close

                                                              open square brackets Hint colon space straight L. straight H. straight S. equals cot to the power of straight n open parentheses fraction numerator straight A minus straight B over denominator 2 end fraction close parentheses open parentheses left parenthesis 1 right parenthesis to the power of straight n plus left parenthesis negative 1 right parenthesis to the power of straight n close parentheses close square brackets

Solution

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