The range of the expression $y=\frac{\left({\cot }^2\theta +5\right)\left({\cot }^2\theta +10\right)}{\left({\cot }^2\theta +1\right)}$

Given $y=\frac{\left({\cot }^2\theta +5\right)\left({\cot }^2\theta +10\right)}{\left({\cot }^2\theta +1\right)}$ Let ${\cot }^2\theta =x$ 

Then $y=\frac{(x+5)(x+10)}{(x+1)}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y=\frac{x^2+15x+50}{x+1}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2+15x+50-xy-y=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2+(15-y)x+(50-y)=0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}D\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(15-y{)}^2-4(50-y)\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}225-30y+y^2-200+4y\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{y}^2-26y+25\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(y-1)(y-25)\ge 0\end{array}$

$\Rightarrow y\le 1$ and $y\ge 25$

Thus $y\ge 25$.