Principle Of Mathematical Induction

Question
CBSEENMA11013401

Prove the following:

fraction numerator tan space left parenthesis straight A plus straight B right parenthesis over denominator cot space left parenthesis straight A minus straight B right parenthesis end fraction space equals space fraction numerator tan squared straight A minus tan squared straight B over denominator 1 minus tan squared straight A space tan squared straight B end fraction


Solution

L.H.S. = fraction numerator tan space left parenthesis straight A plus straight B right parenthesis over denominator cot space left parenthesis straight A minus straight B right parenthesis end fraction equals tan left parenthesis straight A plus straight B right parenthesis space tan left parenthesis straight A minus straight B right parenthesis
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#6 {main}</pre>
          equals fraction numerator tan squared straight A minus tan squared straight B over denominator 1 minus tan squared straight A space tan squared straight B end fraction                  open curly brackets because space left parenthesis straight a plus straight b right parenthesis left parenthesis straight a minus straight b right parenthesis space equals straight a squared minus straight b squared close curly brackets
           = R.H.S.

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