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Principle Of Mathematical Induction

Question
CBSEENMA11013395

Show that;

space space tan left parenthesis straight x minus straight y right parenthesis space plus space tan left parenthesis straight y minus straight z right parenthesis plus tan left parenthesis straight z minus straight x right parenthesis space equals space tan left parenthesis straight x minus straight y right parenthesis. space tan left parenthesis straight y minus straight z right parenthesis. space tan left parenthesis straight z minus straight x right parenthesis

Solution

                                     x - y  = (x - y + z - z) = -[(y - z) + (z - x)]
rightwards double arrow                             tan(x - y) = tan[- {(y - z) + (z - x)}]
rightwards double arrow                             tan (x - y) = - tan[(y - z) + (z - x)]              (∵ WiredFaculty)
rightwards double arrow                            tan (x - y) = - space space open square brackets fraction numerator tan left parenthesis straight y space minus space straight z right parenthesis space plus space tan left parenthesis straight z minus straight x right parenthesis over denominator 1 minus tan left parenthesis straight y minus straight z right parenthesis space tan left parenthesis straight z minus straight x right parenthesis end fraction close square brackets
rightwards double arrow         tan(x - y) - tan(x - y) tan(y - z) tan(z - x) = -tan(y - z) - tan(z - x)
Hence,      tan left parenthesis straight x space minus straight y right parenthesis space plus space tan left parenthesis straight y minus straight z right parenthesis plus tan left parenthesis straight z minus straight x right parenthesis space equals space tan left parenthesis straight x minus straight y right parenthesis. space tan left parenthesis straight y minus straight z right parenthesis. space tan left parenthesis straight z minus straight x right parenthesis

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