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If `tan^(-1)(1/(1+1. 2))+tan^(-1)(1/1+2. 3)++tan^(-1)(1/(1+ndot(n+1)))=tan^(-1)theta,`then findthe value of `thetadot` |
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Answer» General term in the left side of the given equation can be given as, `T_n = tan^-1(1/(1+n(n+1))) = tan^-1((n+1-n)/(1+n(n+1)))` We know, `tan^-1((x-y)/(1+xy)) = tan^-1x - tan^-1y` `:. T_n = tan^-1(n+1)-tan^-1n` So, left side of our given equation becomes, `(tan^-1 2-tan^-1 1)+(tan^-1 3-tan^-1 2)+...(tan^-1 (n+1)-tan^-1 n)` `=(tan^-1 (n+1) -tan^-1 1)` `=tan^-1((n+1-1)/(1+1(n+1)))` `=tan^-1(n/(n+2))` We are given, `tan^-1(n/(n+2)) = tan^-1(theta)` `:. theta = (n/(n+2))` |
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