Show that the function

Given:- Function $f(x)=\cot ^{-1}(\sin x+\cos x)$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

$f(x)=\cot ^{-1}(\sin x+\cos x)$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\cot ^{-1}(\sin \mathrm{x}+\cos \mathrm{x})\right\}$

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

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

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

Now, as given

$x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

$\Rightarrow \cos x-\sin x<0 ;$ as here cosine values are smaller than sine values for same angle

$\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0$

$\Rightarrow f^{\prime}(x)<0$

hence, Condition for $f(x)$ to be decreasing

Thus $f(x)$ is decreasing on interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$