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If 0lta1lta2lt....ltan , then prove that...

If `0lta_1lta_2lt....lta_n ,` then prove that `tan^(-1)((a_1x-y) /(x+a_1y))+tan^(-1)((a_2-a_1) /(1+a_2a_1))+tan^(-1)((a_3-a_2)/(1+a_3a_2))+.......+tan^(-1)((a_n-a_(n-1)) /(1+a_n a_(n-1)))+tan^(-1)(1/(a_n))=tan^(-1)(x/y)dot`

Text Solution

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Here,
`tan^(_1) ((a_(1) x- y)/(a + a_(1) y)) = tan^(-1) ((a_(1) - (y)/(x))/(1 + a_(1) (y)/(x))) = tan^(-1) a_(1) - tan^(-1) (y)/(x)`
`tan^(-1) ((a_(2) -a_(1))/(1 + a_(2) a_(1))) = tan^(-1) a_(2) - tan^(-1) a_(1)`
`{:(tan^(-1) ((a_(3) - a_(2))/(1 + a_(3) a_(2))) = tan^(-1) a_(3) - tan^(-1) a_(2)),(vdots),(tan^(-1) ((a_(n) - a_(n -1))/(1 + a_(n) a_(n -1))) = tan^(-1) a_(n) - tan^(-1) a_(n -1)):}`
`tan^(-1) ((1)/(a_(n))) = cot^(-1) a_(n)`
Adding, we get
`L.H.S. = tan^(-1) a_(n) - tan^(-1) (y)/(x) = (pi)/(2) - tan^(-1) (y)/(x)`
`= cot^(-1) (y)/(x) = tan^(-1) (x)/(y) = R.H.S`
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