InterviewSolution
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InterviewSolution
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If `a_(1), a_(2), a_(3),...., a_(n)` is an A.P. with common difference d, then prove that `tan[tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_( - 1)a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))` |
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Answer» We have `tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_(n -1) a_(n)))` `= tan^(-1) ((a_(2) -a_(1))/(1 + a_(1) a_(2))) + tan^(-1) ((a_(3) -a_(2))/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((a_(n) - a_(n -1))/(1 + a_(n -1) a_(n)))` `= (tan^(-1) a_(2) - tan^(-1) a_(1)) + (tan^(-1) a_(3) - tan^(-1) a_(2)) + ...+ (tan^(-1) a_(n) - tan^(-1) a_(n))` `= tan^(-1) a_(n) - tan^(-1) a_(1) = tan^(-1) ((a_(n) -a_(1))/(1 + a_(n) a_(1))) = tan^(-1) (((n -1))/(1 + a_(1) a_(n)))` `rArr tan[tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ... + tan^(-1) ((d)/(1 + a_(n -1) a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))` |
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