Solve: $tan x tan 2 x + tan 2 x tan 3 x + tan 3 x tan 4 x + . . . . n$ terms =

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Solution

$tan x = tan (2 x - x) = \frac{tan 2 x - tan x}{1 + tan 2 x tan x}$
$∴ T_{1} = tan 2 x tan x = cot x (tan 2 x - tan 1) - 1$
$∴ T_{2} = cot x (tan 3 x - tan 2 x) - 1$ etc.
$∴ S_{n} = cot x [tan (n + 1) x - tan x] - n$
$= cot x tan (n + 1) x - 1 - n$

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tan x tan 2 x + tan 2 x tan 3 x + . . . . . + tan n x tan (n + 1) x

lim n \to \infty \frac{(1 + 2 + 3 + \dots \dots + n terms) . (1^{2} + 2^{2} + \dots \dots n terms)}{n (1^{3} + 2^{3} + . . . + n terms)}

\frac{tan 3 x - tan 2 x}{1 + tan 2 x tan 3 x} = 1

x = n π + \frac{π}{4}, \forall n \in I

tan x

x = n π + \frac{π}{2}, n \in I

\frac{n (3 n + 1)}{2}

3 n^{2} + n - 310 = 0

3 n^{2} + n - 310 = 0

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