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Question: \(\tan^{- 1}\frac{1}{3} + \tan^{- 1}\frac{2}{9} + \tan^{- 1}\frac{4}{33} + ..........\infty\) =...

tan⁡−113+tan⁡−129+tan⁡−1433+..........∞\tan^{- 1}\frac{1}{3} + \tan^{- 1}\frac{2}{9} + \tan^{- 1}\frac{4}{33} + ..........\infty =

A

π/4\pi/4

B

π/2\pi/2

C

π\pi

D

None

Answer

π/4\pi/4

Explanation

Solution

Let θ=cos⁡ec−1(n2+1)(n2+2n+2)\theta = \cos ec^{- 1}\sqrt{\left( n^{2} + 1 \right)\left( n^{2} + 2n + 2 \right)}

⇒ cosec2θ = (n2+1)(n2+2n+2)\left( n^{2} + 1 \right)\left( n^{2} + 2n + 2 \right)

=(n2+1)(n2+1+2n+1)\left( n^{2} + 1 \right)\left( n^{2} + 1 + 2n + 1 \right)

=(n2+1)2+2n(n2+1)+(n2)+1\left( n^{2} + 1 \right)^{2} + 2n\left( n^{2} + 1 \right) + \left( n^{2} \right) + 1 =(n2+n+1)2+1\left( n^{2} + n + 1 \right)^{2} + 1

⇒ cot2θ = (n2+n+1)2\left( n^{2} + n + 1 \right)^{2}

⇒tan⁡θ=1n2+n+1=(n+1)−n1+(n+1)n\tan\theta = \frac{1}{n^{2} + n + 1} = \frac{(n + 1) - n}{1 + (n + 1)n}⇒θ=tan⁡−1[(n+1)−n1+(n+1)n]=tan⁡−1(n+1)−tan⁡−1n\theta = \tan^{- 1}\left\lbrack \frac{(n + 1) - n}{1 + (n + 1)n} \right\rbrack = \tan^{- 1}(n + 1) - \tan^{- 1}nThus, sum to n terms of the given series.

=(tan⁡−12−tan⁡−11)+(tan⁡−13−tan⁡−12)\left( \tan^{- 1}2 - \tan^{- 1}1 \right) + \left( \tan^{- 1}3 - \tan^{- 1}2 \right)

+(tan⁡−14−tan⁡−13+........+[tan⁡−1(n+1)−tan⁡−1n])\left( \tan^{- 1}4 - \tan^{- 1}3 + ........ + \left\lbrack \tan^{- 1}(n + 1) - \tan^{- 1}n \right\rbrack \right)

= tan⁡−1(n+1)−tan⁡−11=tan⁡−1(n+1)−π4\tan^{- 1}(n + 1) - \tan^{- 1}1 = \tan^{- 1}(n + 1) - \frac{\pi}{4}