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The function f(x) = $T a n^{- 1}$ (sinx + cosx) is an increasing function in

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$(- π / 2, π / 4)$

$(π / 4, π / 2)$

$(0, π / 2)$

$(- π / 2, π / 2)$

answer is A.

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Detailed Solution

$f (x) = \tan^{- 1} (sinx + cosx) f^{I} (x) = 1 / {(sinx + cosx)}^{2} (cosx - sinx)$

$f (x)$ is increasing

$\Rightarrow f^{I} (x) > 0 cosx - sinx > 0 cosx > sinx tanx < 1 x < π / 4$

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