If tan-11-x-tan-11-x-2 then x-

The correct option is A 0 Given that : #160; #160; tan^ 1 (1+x)+tan^ 1(1 x)=π2 #160; #160; #160; #160; #160; #160; #160; ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

If tan-11+x...

Question

If ${tan}^{- 1} (1 + x) + {tan}^{- 1} (1 - x) = \frac{π}{2}$ then $x =$ ?

1

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Solution

The correct option is A $0$
Given that : $t a n^{- 1} (1 + x) + t a n^{- 1} (1 - x) = \frac{π}{2}$

$⟹ t a n^{- 1} (1 + x) = \frac{π}{2} - t a n^{- 1} (1 - x)$

using $t a n^{- 1} x + c o t^{- 1} x = \frac{π}{2} ⟹ c o t^{- 1} x = \frac{π}{2} - t a n^{- 1} x$

$⟹ t a n^{- 1} (1 + x) = c o t^{- 1} (1 - x)$

using $c o t^{- 1} x = t a n^{- 1} \frac{1}{x}$

$⟹ t a n^{- 1} (1 + x) = t a n^{- 1} \frac{1}{(1 - x)}$

$⟹ (1 + x) = \frac{1}{(1 - x)}$

$⟹ 1^{2} - x^{2} = 1$

$⟹ x=0$

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If tan−1(1+x)+tan−1(1−x)=π2 then x=?

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If ${tan}^{- 1} (1 + x) + {tan}^{- 1} (1 - x) = \frac{π}{2}$ then $x =$ ?