<s>[INST] Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$ 
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$

$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\ 
=\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\
\color{blue}{\text{By using the AM-GM inequlity}} \\
\color{blue}{x+\dfrac{1}{x} \geq 2} \\ 
=2+2+2=6 $
Which is not in options.
But I am not sure if I can use that $ AM-GM$ inequality in this case.
I look for a short and simple way .
I have studied maths upnto $12$th grade .
 [/INST] Let $\sin^2 \theta =u$ and $\cos^2 \theta =v.$ Then $u+v=1.$ The expression exists only when $0<u<1$ and $0<v<1,$ when it is $$(u+v)+\left(\frac {1}{u}+\frac {1}{v}\right)+\left(\frac {u}{v}+\frac {v}{u}\right)=$$ $$=1+\frac {u+v}{uv}+\frac {u^2+v^2}{uv}=$$ $$=1+\frac {1}{uv}+\frac {u^2+v^2}{uv}=$$ $$=1+\frac {1}{uv}+\frac {(u+v)^2-2uv}{uv}=$$ $$=1+\frac {1}{uv}+\frac {1-2uv}{uv}=$$ $$=-1+\frac {2}{uv}=$$ $$=-1+\frac {2}{u(1-u)}.$$ The maximum value of $u(1-u)$ for $0<u<1$ is $\frac {1}{4}.$
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