If $F = \dfrac{{\mu mg}}{{\cos \theta + \mu \sin \theta }}$ , then $F$ is minimum for
A. $\cos \theta = \mu $
B. $\sin \theta = \mu $
C. $\tan \theta = \mu $
D. $\cot \theta = \mu $

Hint:We can find the minimum value of the given function by using the concept of maxima and minima. We first derivate the given function with respect to $\theta $ and then equate it to zero, to obtain the condition for minimum value of the function. Also, we calculate the minimum value of the function.

Complete step by step answer:
The given function is
$F = \dfrac{{\mu mg}}{{\cos \theta + \mu \sin \theta }}$
Here, $\mu ,m,g$ are constants.
Differentiating the given equation with respect to $\theta $, we get
$\dfrac{{dF}}{{d\theta }} = - \mu mg{\left( {\dfrac{1}{{\cos \theta + \mu \sin \theta }}} \right)^2}\left( { - \sin \theta + \mu \cos \theta } \right)$
Now, For minima, we have to put $\dfrac{{dF}}{{d\theta }} = 0$, then
As $\mu mg{\left( {\dfrac{1}{{\cos \theta + \mu \sin \theta }}} \right)^2} \ne 0$
So,
\[0 = \sin \theta - \mu \cos \theta \]
\[\Rightarrow \mu \cos \theta = \sin \theta \]
We get, \[\tan \theta = \mu \]
The value of function for \[\tan \theta = \mu \] is
$F = \dfrac{{\tan \theta mg}}{{\cos \theta + \tan \theta \sin \theta }}$
$\therefore F = \sin \theta mg$. This is the minimum value of the given function for \[\tan \theta = \mu \] .

Hence, option C is correct.

Note: We should know the rules of differentiation i.e. $\dfrac{d}{{dx}}\left[ {\dfrac{1}{{f(x)}}} \right] = \left[ {\dfrac{{ - 1}}{{{{\left( {f(x)} \right)}^2}}}} \right]f'(x)$. We can also solve this question by putting the various options into the value of $\mu $ , but it will generate false results for every option, so we make use of differentiation.