If ^ -1 1 1+2 + ^ -1 1 1+(2)(3) + ^ -1 1 1+(3)(4) + . .+ ^ -1 1 1+n(n+1) = ^ -1 , then θ =
Mathematics · JEE Main · NTA Exams — Sets, Relations and Functions
If \(\tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+(2)(3)}+\tan ^{-1} \frac{1}{1+(3)(4)}\)
\(+\ldots . .+\tan ^{-1} \frac{1}{1+n(n+1)}=\tan ^{-1} \theta,\) then θ =
\(+\ldots . .+\tan ^{-1} \frac{1}{1+n(n+1)}=\tan ^{-1} \theta,\) then θ =
- \(\frac{n}{n+1}\)
- \(\frac{n+1}{n+2}\)
- \(\frac{n}{n+2}\)
- \(\frac{n-1}{n+2}\)
Answer
(C) n n+2
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