tan−1⁡13+tan−1⁡29+tan−1⁡433+…∞=

tan-1(13)+tan-1(29)+tan-1(433)+........+∞ We know that tan-1(A)-tan-1(B)=tan-1(A-B1+AB) =tan-1(2)-tan-1(1)+tan-1(4)-tan-1(2)+tan-1(8)-tan-1(4)+.....+tan-1(2n-1)-tan-1(2n-2)+....+tan-1(∞) =tan-1(∞)-tan-1(1) =π2-π4 =π4

tan−1⁡13+tan−1⁡29+tan−1⁡433+…∞=

tan-1(13)+tan-1(29)+tan-1(433)+........+∞ We know that tan-1(A)-tan-1(B)=tan-1(A-B1+AB) =tan-1(2)-tan-1(1)+tan-1(4)-tan-1(2)+tan-1(8)-tan-1(4)+.....+tan-1(2n-1)-tan-1(2n-2)+....+tan-1(∞) =tan-1(∞)-tan-1(1) =π2-π4 =π4

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$\tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{2}{9}) + \tan^{- 1} (\frac{4}{33}) + \dots \infty =$

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2 $π$

$\frac{π}{2}$

$\frac{π}{4}$

$π$

answer is B.

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Detailed Solution

$\tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{2}{9}) + \tan^{- 1} (\frac{4}{33}) + . . . . . . . . + \infty W e k n o w t h a t \tan^{- 1} (A) - \tan^{- 1} (B) = \tan^{- 1} (\frac{A - B}{1 + A B}) = \tan^{- 1} (2) - \tan^{- 1} (1) + \tan^{- 1} (4) - \tan^{- 1} (2) + \tan^{- 1} (8) - \tan^{- 1} (4) + . . . . . + \tan^{- 1} (2^{n - 1}) - \tan^{- 1} (2^{n - 2}) + . . . . + \tan^{- 1} (\infty) = \tan^{- 1} (\infty) - \tan^{- 1} (1) = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}$