Principle Of Mathematical Induction

Question
CBSEENMA11014286

Prove that:

tan left parenthesis straight x plus straight y right parenthesis space equals space fraction numerator tanx plus tany over denominator 1 minus tanx space tany end fraction comma space space where space straight x comma straight y comma space straight x plus straight y space not equal to space left parenthesis 2 straight n plus 1 right parenthesis straight pi over 2 comma space straight n element of straight Z

Solution

tan left parenthesis straight x plus straight y right parenthesis space equals space fraction numerator sin left parenthesis straight x plus straight y right parenthesis over denominator cos left parenthesis straight x plus straight y right parenthesis end fraction space equals space fraction numerator sinx space cosy plus space cosx space siny over denominator cosx space cosy space minus space sinx space siny end fraction
where,            straight x comma space straight y not equal to space left parenthesis 2 straight n plus 1 right parenthesis straight pi over 2


rightwards double arrow        cos space straight x not equal to 0 space and space cos space straight y not equal to 0 space rightwards double arrow space cosx space cosy space not equal to 0
Dividing numerator and denominator by cosx cosy, we get
           tan left parenthesis straight x plus straight y right parenthesis space equals space fraction numerator begin display style fraction numerator sinx space cosy over denominator cosx space cosy end fraction end style plus begin display style fraction numerator cosx space siny over denominator cosx space cosy end fraction end style over denominator begin display style fraction numerator cosx space cosy over denominator cosx space cosy end fraction end style minus begin display style fraction numerator sinx space siny over denominator cosx space cosy end fraction end style end fraction space equals space fraction numerator tanx plus tany over denominator 1 minus tanx space tany end fraction
Hence,  tan left parenthesis straight x plus straight y right parenthesis space equals space fraction numerator tanx plus tany over denominator 1 minus tanx space tany end fraction

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