The function $f (x) = {tan}^{- 1} (sin x + cos x)$ is an increasing function in

(0, \frac{π}{2})

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(- \frac{π}{2}, \frac{π}{2})

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(\frac{π}{4}, \frac{π}{2})

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(- \frac{π}{2}, \frac{π}{4})

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Solution

The correct option is C $(- \frac{π}{2}, \frac{π}{4})$
Differentiating the above function with respect to x, we get
$f^{'} (x) = \frac{c o s x - s i n x}{1 + (s i n x + c o s x)^{2}}$
Now for $f (x)$ to be an increasing function
$f^{'} (x) > 0$
Or
$\frac{c o s x - s i n x}{1 + (s i n x + c o s x)^{2}} > 0$
Or
$c o s x > s i n x$
Or
$c o s (x) - s i n (x) > 0$
Or
$\sqrt{2} s i n (\frac{π}{4} - x) > 0$
This happens in the interval of $x ϵ (\frac{- π}{2}, \frac{π}{4})$

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