4tan−115−tan−11239 is equal to

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Question

4 t a n^{- 1} \frac{1}{5} - t a n^{- 1} \frac{1}{239}

is equal to

Options:

A . π

B . π2

C . π3

D . π4

Answer: Option D
:
D
Since

2 t a n^{- 1} x = t a n^{- 1} \frac{2 x}{1 - x^{2}}

∴ 4 t a n^{- 1} \frac{1}{5} = 2 [2 t a n^{- 1} \frac{1}{5}] = 2 t a n^{- 1} \frac{\frac{2}{5}}{1 - \frac{1}{25}}

= 2 t a n^{- 1} \frac{10}{24} = t a n^{- 1} \frac{\frac{20}{24}}{1 - \frac{100}{576}} = t a n^{- 1} \frac{120}{119}

So,

4 t a n^{- 1} \frac{1}{5} - t a n^{- 1} \frac{1}{239} = t a n^{- 1} \frac{120}{119} - t a n^{- 1} \frac{1}{239}

= t a n^{- 1} \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} . \frac{1}{239}} = t a n^{- 1} \frac{(120 \times 239) - 119}{(119 \times 239) + 120}

\Rightarrow t a n^{- 1} 1 = \frac{π}{4}

.

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