Show that f(x) = tan–1(sin x + cos x) is

Solution:

Given, f(x) = tan–1(sin x + cos x) in (0, π/4).

Differentiating both sides w.r.t. x, we got

$f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^{2}} \cdot \frac{d}{d x}(\sin x+\cos x)$

$f^{\prime}(x)=\frac{1 \times(\cos x-\sin x)}{1+(\sin x+\cos x)^{2}}$

$f^{\prime}(x)=\frac{\cos x-\sin x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}$

$\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{1+1+2 \sin x \cos x} \Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{2+2 \sin x \cos x}$

For an increasing function $f^{\prime}(x) \geq 0$

So, $\quad \frac{\cos x-\sin x}{2+2 \sin x \cos x} \geq 0$

$\cos x-\sin x \geq 0 \quad\left[\because \quad(2+\sin 2 x) \geq 0\right.$ in $\left.\left(0, \frac{\pi}{4}\right)\right]$

$\Rightarrow \cos x \geq \sin x$, which is true for $\left(0, \frac{\pi}{4}\right)$

$f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^{2}} \cdot \frac{d}{d x}(\sin x+\cos x)$

$f^{\prime}(x)=\frac{1 \times(\cos x-\sin x)}{1+(\sin x+\cos x)^{2}}$

$f^{\prime}(x)=\frac{\cos x-\sin x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}$

$\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{1+1+2 \sin x \cos x} \quad \Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{2+2 \sin x \cos x}$

For an increasing function $f^{\prime}(x) \geq 0$

So, $\quad \frac{\cos x-\sin x}{2+2 \sin x \cos x} \geq 0$

$\cos x-\sin x \geq 0 \quad\left[\because \quad(2+\sin 2 x) \geq 0\right.$ in $\left.\left(0, \frac{\pi}{4}\right)\right]$

$\Rightarrow \quad \cos x \geq \sin x$, which is true for $\left(0, \frac{\pi}{4}\right)$

Therefore, the given function f(x) is an increasing function in (0, π/4).