4 t a n − 1 1 5 − t a n − 1 1 239 is equal to Options: A . &nbsp π B . &nbsp π 2 C . &nbsp π 3 D . &nbsp π 4

Answer: Option D : D Since 2 t a n − 1 x = t a n − 1 2 x 1 − x 2 ∴ 4 t a n − 1 1 5 = 2 [ 2 t a n − 1 1 5 ] = 2 t a n − 1 2 5 1 − 1 25 = 2 t a n − 1 10 24 = t a n − 1 20 24 1 − 100 576 = t a n − 1 120 119 So, 4 t a n − 1 1 5 − t a n − 1 1 239 = t a n − 1 120 119 − t a n − 1 1 239 = t a n − 1 120 119 − 1 239 1 + 120 119 . 1 239 = t a n − 1 ( 120 × 239 ) − 119 ( 119 × 239 ) + 120 ⇒ t a n − 1 1 = π 4 .

4tan−115−tan−11239 is equal to

Discussion Forum : Inverse Trigonometric Functions

Question -

$4 t a n^{- 1} \frac{1}{5} - t a n^{- 1} \frac{1}{239}$ is equal to

Options:

A .

π

B .

\frac{π}{2}

C .

\frac{π}{3}

D .

\frac{π}{4}

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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Discussion Forum : Inverse Trigonometric Functions

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