Principle Of Mathematical Induction

Question
CBSEENMA11014297

Prove the following:

cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction plus straight x close parentheses space equals space sinx

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space tan open parentheses fraction numerator 3 straight pi over denominator 2 end fraction plus straight x close parentheses equals negative cotx


Solution

We know that:   cos(x + y) = cosx cosy - sinx siny                                ...(i)
                        sin (x + y) = sinx cosy + cosx siny                              ...(ii)
                        cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses equals 0                                                            ...(iii)
                        sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses equals negative 1                                                          ...(iv)
Replacing x by open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses and y by x in (i), we get
               cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction plus straight x close parentheses space equals space cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses cosx space minus space sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses sinx space equals space 0 left parenthesis cosx right parenthesis space minus space left parenthesis negative 1 right parenthesis space sinx
                                     = sin x                                             [By using (iii) and (iv)]
Hence,    cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction plus straight x close parentheses equals sinx
Replacing x by space space fraction numerator 3 straight pi over denominator 2 end fraction and y by x in (ii), we get
               sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction plus straight x close parentheses space equals space sin open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses space cosx space plus space cos open parentheses fraction numerator 3 straight pi over denominator 2 end fraction close parentheses sinx space equals space left parenthesis negative 1 right parenthesis space cosx space plus space left parenthesis 0 right parenthesis space sinx
                                     = - cosx                                      [By using (iii) and (iv)]
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             tan open parentheses fraction numerator 3 straight pi over denominator 2 end fraction plus straight x close parentheses space equals space fraction numerator sin open parentheses begin display style fraction numerator 3 straight pi over denominator 2 end fraction end style plus straight x close parentheses over denominator cos open parentheses begin display style fraction numerator 3 straight pi over denominator 2 end fraction end style plus straight x close parentheses end fraction space equals space fraction numerator negative cosx over denominator sinx end fraction equals negative cotx
Hence,    tan open parentheses fraction numerator 3 straight pi over denominator 2 end fraction plus straight x close parentheses space equals space minus cotx