Prove that, function $f (x) = {tan}^{- 1} (sin x + cos x)$ , is increasing function in interval $(0, \frac{π}{4})$ .

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Solution

$\begin{matrix} f (x) = {tan}^{- 1} (S i n x + C o s x) d i f f e r e n t i a t e W . r . t o x f (x) = \frac{1}{1 + {(S i n x + C o s x)}^{2}} \times (C o s x - S i n x) C o m p a r e w i t h z e r o f (x) \geq 0 \frac{C o s x - S i n x}{(1 + {S i n}^{2} x + {cos}^{2} x + 2 S i n x - C o s x)} = 0 \Rightarrow \frac{cos x - sin x}{2 (1 + sin x . cos x)} = 0 \Rightarrow cos x - sin x = 0 \Rightarrow tan x = 1 x = \frac{π}{4} 1 + S i n x \times C o s x > 0 S i n x . C o s x > - 1 H e n c e f (x) i s i n c r e a sin g (0, \frac{π}{4}) \end{matrix}$

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