Minimum  force required on a block of weight $w$ for motion along rough horizontal surface of coefficient of friction $\mu$ is

Let P be the force applied to it an angle θ. 

From the free body diagram

$\begin{array}{l}R+P\sin \theta -mg=0\\ \Rightarrow R=-P\sin \theta +mg\\ \mu R=P\cos \theta \end{array}$

From equation (i)

$\begin{array}{l}\mu (mg-P\sin \theta )=-P\cos \theta \\ \Rightarrow \mu mg=\mu P\sin \theta -P\cos \theta \\ \Rightarrow P=\frac{\mu mg}{\mu \sin \theta +\cos \theta }\end{array}$

Applied force P should be minimum, when $\mu \sin \theta +\cos \theta$ is maximum.

$\begin{array}{l}\frac{d}{d\theta }(\mu \sin \theta +\cos \theta )=0\\ \Rightarrow \mu \cos \theta -\sin \theta =0\\ \Rightarrow \theta ={\tan }^{-1}\mu \end{array}$

$\begin{array}{l}\text{ So, }P=\frac{\mu \mathrm{mg}}{\mu \sin \theta +\cos \theta }\\ =\frac{\frac{\mu \mathrm{mg}}{\cos \theta }}{\frac{\mu \sin \theta }{\cos \theta }+\frac{\cos \theta }{\cos \theta }}\end{array}$

$\begin{array}{l}P=\frac{\mu \mathrm{mgsec}\theta }{1+\mu \tan \theta }\\ =\frac{\mu \mathrm{mgsec}\theta }{1+{\tan }^2\theta }=\frac{\mu \mathrm{mg}}{\sec \theta }\\ =\frac{\mu \mathrm{mg}}{\sqrt{\left(1+{\tan }^2\theta \right)}}=\frac{\mu \mathrm{mg}}{\sqrt{\left(1+{\mu }^2\right)}}\end{array}$