If$f\left(x\right)$is a function given by

$f\left(x\right)=\left|\begin{array}{l}\sin x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin b\\ \cos x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos b\\ \tan x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan b\end{array}\right|\text{,}$where $0<a<b<\frac{\pi }{2}$ Then the equation $f^{\mathrm{\prime }}\left(x\right)=0$

We know that$\sin x,\cos x$ and $\tan x$ are continuous on$[a, b]$ and differentiable on $(a, b).$

 Therefore,$f\left(x\right)$ is continuous on [a, b] and differentiable on (a, b). 

Also, $f\left(a\right)=f\left(b\right)=0$

Therefore, by Rolle's theorem there exists at least one$c\in (a,b)$ such that $f^{\mathrm{\prime }}\left(c\right)=0$

Hence, $f\text{'} \left(x\right) = 0$ has at least one root in $(a, b).$