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प्रश्न
If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.
उत्तर
\[\tan^{- 1} \left( \frac{1}{1 + 1 \times 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 \times 3} \right) + . . . + \tan^{- 1} \left[ \frac{1}{1 + n \times \left( n + 1 \right)} \right] = \tan^{- 1} \theta\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2 - 1}{1 + 1 \times 2} \right) + \tan^{- 1} \left( \frac{3 - 2}{1 + 2 \times 3} \right) + . . . + \tan^{- 1} \left[ \frac{\left( n + 1 \right) - n}{1 + n \times \left( n + 1 \right)} \right] = \tan^{- 1} \theta\]
\[ \Rightarrow \tan^{- 1} \left( 2 \right) - \tan^{- 1} \left( 1 \right) + \tan^{- 1} \left( 3 \right) - \tan^{- 1} \left( 2 \right) + . . . + \tan^{- 1} \left( n + 1 \right) - \tan^{- 1} \left( n \right) = \tan^{- 1} \theta\]
\[ \Rightarrow \tan^{- 1} \left( n + 1 \right) - \tan^{- 1} \left( 1 \right) = \tan^{- 1} \theta\]
\[ \Rightarrow \tan^{- 1} \left[ \frac{\left( n + 1 \right) - 1}{1 + \left( n + 1 \right) \times 1} \right] = \tan^{- 1} \theta\]
\[ \Rightarrow \tan^{- 1} \left( \frac{n}{n + 2} \right) = \tan^{- 1} \theta\]
\[ \Rightarrow \theta = \frac{n}{n + 2}\]
Thus, the value of θ is \[\frac{n}{n + 2}\] .
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