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 .
 A: Hint:
We can use the Pythagorean identities $\color{blue}{\sin^2\theta+\cos^2\theta=1}$, $\color{blue}{\sec^2 \theta=\tan^2 \theta+1}$ and $\color{blue}{\csc^2\theta=\cot^2 \theta+1}$, giving us
\begin{align}\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta&=3+2\tan^2\theta+2\cot^2\theta\\
&=3+2\left(\tan^2\theta+\frac{1}{\tan^2\theta}\right)
\end{align}
A: Using standard trigonometric identities, we see this is $1+2\tan^2\theta+1+2\cot^2\theta+1$.
Now we can use AM/GM to show that $2\tan^2\theta+2\cot^2\theta\ge 4$, and the value $4$  is attained at $\pi/4$.
Remark: Your AM/GM argument is enough to identify the right answer of this multiple choice question. For as you saw the minimum is $\ge 6$, and there is only one choice which is $\ge 6$.
A: Hint: $$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$$$$=  \sec^2 \theta + \csc^2 \theta + \sec^2 \theta  + \csc^2 \theta -1 =  2sec^2 \theta + 2\csc^2 \theta  -1$$
A: 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 $\frac {1}{4}.$
A: Just using algebra$$y=\sin^{2} (\theta) +\cos^{2} (\theta)+\sec^{2} (\theta)+\csc^{2} (\theta)+\tan^{2} (\theta)+\cot^{2} (\theta)=2 \csc ^2(\theta )+2 \sec ^2(\theta )-1$$ Taking derivatives and simplifying $$y'=4 \tan (\theta) \sec ^2(\theta)-4 \cot (\theta) \csc ^2(\theta)=-32 \cot (2 \theta) \csc ^2(2 \theta)$$ $$y''=64 \csc ^4(2 \theta)+128 \cot ^2(2 \theta) \csc ^2(2 \theta)=64 (\cos (4 \theta)+2) \csc ^4(2 \theta)$$ The second derivative is always positive, so we can only find minimum values if $y'=0$ and this happens for $\theta=\pm\frac \pi 4$ and $\theta=\pm\frac {3\pi} 4$. For these values, $y=7$.
A: A nice way to simplify this problem, using nothing more than elementary trigonometry and algebra, is this:
As noted in another answer, the given expression simplifies to:
$$2\sec^2\theta +2\csc^2\theta - 1$$
Finding $\theta$ to minimize that expression is the same as finding $\theta$ to minimize 
$$\begin{align}
\sec^2\theta+\csc^2\theta &= \frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}\\
&=\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}\\
&=\frac{1}{\sin^2\theta\cos^2\theta},
\end{align}$$
which is the same as finding $\theta$ to maximize
$$\sin^2\theta\cos^2\theta = (\sin\theta\cos\theta)^2,$$
which is the same as finding $\theta$ to maximize
$$|\sin\theta\cos\theta| = \frac12|\sin(2\theta)|.$$
This is clearly maximized when $\sin(2\theta)=\pm 1$, which happens when $\theta$ is $45^\circ$ away from a multiple of $90^\circ$. If you plug in $\theta=\frac{\pi}{4}$, you'll find that minimum value.
