Let three Geometric mean exists between ‘4’ and  $\frac{\text{'}1\text{'}}{4}$, then the product of three trigonometric means is  

Let three Geometric mean between ‘4’ and $\frac{\text{'}1\text{'}}{4}$are  $G_1,G_2,G_3$

$4,G_1,G_2,G_3,\frac{1}{4}$

Where  $G_1=ar$

$G_2=a{r}^2$

$G_3=a{r}^3$

Total terms are ‘5’

$a=4,$

$t_n=a{.}^{n-1}$

$t_5=a.r^4$

$\frac{1}{4}=a{r}^4\Rightarrow \frac{1}{4}=4{\left(r\right)}^4$

$r^4=\frac{1}{16}\Rightarrow r^4={\left(\frac{1}{2}\right)}^4$

$\therefore r=\frac{1}{2}$

$\therefore G_1\times G_2\times G_3=ar\times a{r}^2\times a{r}^3$

$=a^3{r}^6={\left(4\right)}^3{\left(\frac{1}{2}\right)}^6=2^6\times \frac{1}{26}=1$