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 .

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}

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 $\ge 6$.

Hint: $$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$$$$= \sec^2 \theta + \csc^2 \theta + \sec^2 \theta + \csc^2 \theta -1 = 2sec^2 \theta + 2\csc^2 \theta -1$$

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 $\frac {1}{4}.$