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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?

prog_SAHIL
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Dani
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    Might help to simplify the expression (using $\sin^2\theta +\cos^2 \theta =1$ and so on). – lulu Jul 19 '18 at 13:33
  • To address the question of $\theta$ being 45 degrees: the symmetry of replacing $\theta$ by $90-\theta$ in your expression implies that 45 degrees is a critical point. But if you follow down this direction you still need to find out whether the critical point is in fact the global minimum. So while 45 degrees does give you the correct answer, you will not know that for sure unless you do more work. – Willie Wong Jul 19 '18 at 13:59

3 Answers3

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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}

Ng Chung Tak
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$$\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta=3+2(\tan^2\theta+\cot^2\theta)$$

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

lab bhattacharjee
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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.

Willie Wong
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