If \theta=\tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+(2)(3)}+\tan ^{-1} \frac{1}{1+(3)(4)} +. . .+\tan ^{-1} \frac{1}{1+n(n+1)}, then \tan \theta is equal to

NIMCET 2009 Question Paper with solutions

Question : 1 of 120

Marks: +1, -0

θ=\tan ^{-1} \;{1}/{1+2}+\tan ^{-1} \;{1}/{1+(2)(3)}+\tan ^{-1} \;{1}/{1+(3)(4)}

+. . .+\tan ^{-1} \;{1}/{1+n(n+1)}

, then

\tan θ

is equal to

Solution:

θ=\tan^{-1}1/{1+2}+\tan^{-1}1/{1+(2)(3)}+\tan^{-1}1/{1+(3)(4)}+....+\tan^{-1}/{1+n(n+1)}

=\tan^{-1}{2-1}/{1+1.2}+\tan^{-1}{3-2}/{1+(2)(3)}+\tan^{-1}{4-3}/{1+(3)(4)}+...+\tan^{-1}{(n+1)-n}/{1+n(n+1)}

=(\tan^{-1}2-\tan^{-1}1)+(\tan^{-1}3-\tan^{-1}2)+(\tan^{-1}4-\tan^{-1}3)+.....+(\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}⇒\tanθ=n/{n+2}

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