If tan -1x-1-tan -1x-1-tan -18-31- then x is equal to

The correct option is C 1/4We have, tan^ 1[x+1+x 1/1(x+1)(x 1) ]=tan^ 18/31 ⇒2x/1 (x^2 1)=8/31⇒ 62x=16 8x^2 ⇒ 8x^2+62x 16=0⇒ 4x^2+31x 8=0 ⇒ 4x^2+32x x 8=0⇒ 4x( ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If tan -1x+1+...

Question

If $t a n^{- 1} (x + 1) + t a n^{- 1} (x - 1) = t a n^{- 1} (\frac{8}{31})$ , then x is equal to

1/2

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-1/2

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1/4

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Solution

The correct option is C 1/4
We have, $t a n^{- 1} [\frac{x + 1 + x - 1}{1 (x + 1) (x - 1)}] = t a n^{- 1} \frac{8}{31}$
$\Rightarrow \frac{2 x}{1 - (x^{2} - 1)} = \frac{8}{31} \Rightarrow 62 x = 16 - 8 x^{2}$
$\Rightarrow 8 x^{2} + 62 x - 16 = 0 \Rightarrow 4 x^{2} + 31 x - 8 = 0$
$\Rightarrow 4 x^{2} + 32 x - x - 8 = 0 \Rightarrow 4 x (x - 8) - 1 (x + 8) = 0$
$\Rightarrow (x + 8) (4 x - 1) = 0 \Rightarrow x = - 8. \frac{1}{4}$
$∴ x = \frac{1}{4} . N e g l e c t i n g x = - 8. a s . x^{2} - 1 < 1$

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