Generalize this (part-2)

Geometry Level 3

tan1(11+12)+tan1(11+23)++tan1(11+n(n+1))\large\tan^{-1} \left(\dfrac{1}{1 + 1\cdot2}\right) + \tan^{-1} \left(\dfrac{1}{1 + 2\cdot3}\right) + \cdots +\tan^{-1} \left(\dfrac{1}{1 + n(n+1)}\right) If above expression is equal to tan1θ\tan^{-1}\theta,then find θ\theta .

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