What is the least value of $\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta$? What is the least value of this expression?
$$\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta$$
Will putting $\theta=45^{\circ}$ give right answer?
 A: Let $s=\sin \theta$ and $c=\cos \theta$, then
\begin{align}
  E &= \tan^2 \theta+\cot^2 \theta+\sin^2 \theta+
  \cos^2 \theta+\sec^2 \theta+\csc^2 \theta \\
  &= \frac{s^2}{c^2}+\frac{c^2}{s^2}+1+\frac{1}{c^2}+\frac{1}{s^2} \\
  &= \frac{s^4+c^4+s^2c^2+s^2+c^2}{s^2c^2} \\
  &= \frac{(s^2+c^2)^2-s^2c^2+1}{s^2c^2} \\
  &= \frac{2-s^2c^2}{s^2c^2} \\
  &= \frac{2}{\sin^2 \theta \cos^2 \theta}-1 \\
  &= \frac{8}{\sin^2 2\theta}-1 \\
  & \ge 7
\end{align}
A: $$\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta=3+2(\tan^2\theta+\cot^2\theta)$$
Now $\tan^2\theta+\cot^2\theta=(\cot\theta-\tan\theta)^2+2\ge?$
the equality occurs if $\cot\theta-\tan\theta=0\iff\tan^2\theta=1\iff\cos2\theta=0\iff2\theta=(2m+1)\dfrac\pi2$ where $m$ is any integer
A: One can also answer using the "guess" that the value is attained at $\theta= \pi/4$. 
Guessing $\theta$
The expression you care about is


*

*periodic with period $\pi/2$; and

*symmetric when you replace $\theta$ by $\pi/2 - \theta$. 


So you just need to look at the range $\theta \in (0,\pi/2)$; the endpoints are ruled out since your function is infinite there. The symmetry implies that $\theta = \pi/4$ is a critical point. 
$\pi/4$ is the only critical point
Observe that $\sin^2\theta + \cos^2\theta = 1$. Observe further that $\sec, \csc, \tan,\cot$ are all


*

*non-negative on $(0,\pi/2)$

*convex on $(0,\pi/2)$


these imply that $\tan^2$ etc. are all convex functions on $(0,\pi/2)$. Hence their sum is convex. A convex function can only have at most one critical point, and the critical point, if exists, is the global minimum. 
