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If a1, a2,a3, ,an is an A.P. with commo...

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_1a_2))+tan^(-1)(d/(1+a_2a_3))+......+tan^(-1)(d/(1+a_(n-1)a_n))]=((n-1)d)/(1+a_1a_n)`

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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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