If tan -1 x-1x-2 -tan -1 x-1x-2 -4- then find the value of x-

Given: tan ^ 1 ( x 1x 2 )+tan ^ 1 ( x+1x+2 )=π4 Using tan ^ 1x+tan ^ 1y= ( x+y1 xy ) ⇒tan ^ 1 [ x 1x 2+x+1x+21 x 1x 2×x+1x+2 ]=π4 ⇒x 1x 2+x+1x+21 x 1x 2×x+1x+ ...

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Byju's Answer

Standard XII

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

If tan -1 x-1...

Question

If ${tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}$ , then find the value of $x .$

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Solution

Given: ${tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}$
Using ${tan}^{- 1} x + {tan}^{- 1} y = (\frac{x + y}{1 - x y})$

$\Rightarrow {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎣ \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \times \frac{x + 1}{x + 2}} ⎤ ⎥ ⎥ ⎥ ⎦ = \frac{π}{4}$

$\Rightarrow \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \times \frac{x + 1}{x + 2}} = tan \frac{π}{4}$

$\Rightarrow \frac{\frac{(x - 1) (x + 2) + (x + 1) (x - 2)}{(x - 2) (x + 2)}}{\frac{(x - 2) (x + 2) - (x - 1) (x + 1)}{(x - 2) (x + 2)}} = 1$

$\Rightarrow \frac{(x - 1) (x + 2) + (x + 1) (x - 2)}{(x - 2) (x + 2) - (x - 1) (x + 1)} = 1$

$\Rightarrow \frac{(x - 1) (x + 2) + (x + 1) (x - 2)}{x^{2} - 2^{2} - [x^{2} - 1^{2}]} = 1$

$\Rightarrow \frac{x (x + 2) - (x + 2) + x (x - 2) + (x - 2)}{x^{2} - 4 - x^{2} + 1} = 1$
$\Rightarrow \frac{x^{2} + 2 x - x - 2 + x^{2} - 2 x + x - 2}{- 3} = 1$
$\Rightarrow \frac{2 x^{2} - 4}{- 3} = 1$
$\Rightarrow 2 x^{2} - 4 = - 3$
$\Rightarrow 2 x^{2} = - 3 + 4$
$\Rightarrow 2 x^{2} = 1$
$\Rightarrow x^{2} = \frac{1}{2}$
$∴ x = \pm \frac{1}{\sqrt{2}}$

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Similar questions

Q. If

{tan}^{- 1} \frac{x - 1}{x - 2} + {tan}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}

, then find the value of

x

Prove that: $t a n^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) = \frac{π}{4} - \frac{1}{2} c o s^{- 1} x; - \frac{1}{\sqrt{2}} \leq x \leq 1$ .