Answer
$$c = \frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sin x,{\text{ }}g\left( x \right) = \cos x{\text{ }}\underbrace {\left[ {0,\pi /2} \right]}_{\left[ {a,b} \right]} \cr
& {\text{Differentiate both functions}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x} \right] = \cos x \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\cos x} \right] = - \sin x \cr
& {\text{Apply the Extended Mean Value Theorem}} \cr
& \frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( b \right) - f\left( a \right)}}{{g\left( b \right) - g\left( a \right)}} \cr
& {\text{,then}} \cr
& \frac{{\cos c}}{{ - \sin c}} = \frac{{\sin \left( {\pi /2} \right) - \sin \left( 0 \right)}}{{\cos \left( {\pi /2} \right) - \cos \left( 0 \right)}} \cr
& {\text{Simplifying}} \cr
& - \cot c = \frac{1}{{ - 1}} \cr
& \cot c = 1 \cr
& {\text{For the interval }}\left[ {0,\frac{\pi }{2}} \right] \cr
& c = \frac{\pi }{4} \cr} $$
