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Solution:

Given:- Function $f(x)=\cos ^{2} x$

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

(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,

$f(x)=\cos ^{2} x$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^{2} \mathrm{x}\right)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3 \cos \mathrm{x}(-\sin \mathrm{x})$

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

$\Rightarrow f^{\prime}(x)=-\sin 2 x ;$ as $\sin 2 A=2 \sin A \cos A$

Now, as given

$X \in\left(0, \frac{\pi}{2}\right)$

$\Rightarrow 2 x \in(0, \pi)$

$\Rightarrow \operatorname{Sin}(2 x)>0$

$\Rightarrow-\operatorname{Sin}(2 x)<0$

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

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

Thus $f(x)$ is decreasing on interval $\left(0, \frac{\pi}{2}\right)$

Hence proved