Problem: Let {{\theta}_1}, {{\theta}_2}, … , {{\theta}_{10}} be any values in the closed interval {[0,\pi]}. Show that
{F} = \( {(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2)………(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})} \) {\displaystyle{\le({\frac{9}{4}})^{10}}}.
What is the maximum value attainable by {F} and at what values of {{\theta}_1}, {{\theta}_2}, … , {{\theta}_{10}}, is the maximum value attained?

Solution:
{F} = {\displaystyle{(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2).........(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})}}
Now we will show that for any {\theta \in} {[0,\pi]} {\displaystyle{(1 + {\sin}^2 \theta)(1 + {cos}^2 \theta) < {\frac{9}{4}}}}
{\Leftrightarrow} {\displaystyle{2 + {{\sin}^2{\theta}}{{\cos}^2{\theta}}}} {\displaystyle{< {\frac{9}{4}}}}
{\Leftrightarrow} {\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}} {\displaystyle{< {\frac{1}{4}}}}
Now {\displaystyle{{{\sin}^2{\theta}} + {{\cos}^2{\theta}} = 1}}
{\Rightarrow} {\displaystyle{({{\sin}^2{\theta}} + {{\cos}^2{\theta}})^2 = 1}}
{\Rightarrow} {\displaystyle{{\sin}^4{\theta} + {\cos}^4{\theta} + 2 {{\sin}^2{\theta}{\cos}^2{\theta}} = 1}}
{\Rightarrow} {\displaystyle{4 {{\sin}^2{\theta}}{{\cos}^2{\theta}}}} {\displaystyle{< 1}} [ as {{a^2 + b^2} > {2ab}} ]
{\Rightarrow} {\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}} {\displaystyle{< {\frac{1}{4}}}}
So for any {\theta \in} {[0,\pi]} {\displaystyle{(1 + {\sin}^2 \theta)(1 + {\cos}^2 \theta) {\le} {\frac{9}{4}}}}
So {F {\le} ({\frac{9}{4}})^{10}} (proved)
Maximum value attained by {F} is {({\frac{9}{4}})^{10}} and will be attained for {{\theta_i} = {\frac{\pi}{2}} \pm {\frac{\pi}{4}}} for {i = 1, 2, ...., 10}