Principle Of Mathematical Induction

Question
CBSEENMA11014289

Prove that:

space space space tan open parentheses straight pi over 4 minus straight x close parentheses space equals space fraction numerator 1 minus tanx over denominator 1 plus tanx end fraction


Solution

tan left parenthesis straight x minus straight y right parenthesis space equals space fraction numerator tanx minus tany over denominator 1 plus tanx space tany end fraction
Replacing x by space straight pi over 4 and y by x, we get

              space space space tan open parentheses straight pi over 4 minus straight x close parentheses space equals space fraction numerator tan begin display style straight pi over 4 end style minus tanx over denominator 1 plus tan begin display style straight pi over 4 end style space tanx end fraction space equals space fraction numerator 1 minus tanx over denominator 1 plus tanx end fraction                   open parentheses because space tan straight pi over 4 equals 1 close parentheses
Hence, space space space space tan open parentheses straight pi over 4 minus straight x close parentheses space equals space fraction numerator 1 minus tanx over denominator 1 plus tanx end fraction

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