$\cos \theta+\sec \theta=2$ হলে $\cos ^{\mathrm{n}} \theta+\sec ^{\mathrm{n}} \theta$ এর মান কোনটি?

$\cos \theta+\sec \theta=2 \Rightarrow \cos \theta+\frac{1}{\cos \theta}=2$ $\begin{array}{l}\Rightarrow \cos ^{2} \theta-2 \cos \theta+1=0 \\\Rightarrow(\cos \theta-1)^{2}=0 \Rightarrow \cos \theta=1 \Rightarrow \sec \theta=1\end{array}$ $\therefore \cos ^{\mathrm{n}} \theta+\sec ^{\mathrm{n}} \theta=1^{\mathrm{n}}+1^{\mathrm{n}}=1+1=2$