A geometry problem by Rakhi Bhattacharyya

If we put $z = \sqrt{2}\cos(x - \tfrac14\pi) = \cos x + \sin x$ , then $z^2-1 = 2\sin x \cos x$ and hence $\sin x + \cos x + \sec x + \mathrm{cosec}\,x + \tan x + \cot x \; = \; z + \frac{\sin x + \cos x}{\sin x \cos x} + \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \; = \; z + \frac{2(z+1)}{z^2-1} \; = \; \frac{z^2-z+2}{z-1}$ and so we simply need to minimize the modulus of $g(z) \; = \; \frac{z^2-z+2}{z-1}$ over the interval $[-\sqrt{2},\sqrt{2}]$ . Now $g(z)$ has no real zeros, and since $g'(z) \; = \; \frac{(z-1)^2-2}{(z-1)^2}$ we see that the only turning point of $g(z)$ in the interval $[-\sqrt{2},\sqrt{2}]$ occurs when $z = 1 - \sqrt{2}$ .

It is easy to see that $g(z)$ is:

• increasing and negative for $-\sqrt{2} \le z \le 1-\sqrt{2}$ ,
• decreasing and negative for $1-\sqrt{2} \le z < 1$ ,
• decreasing and positive for $1 < z \le \sqrt{2}$ .

Since $g(\sqrt{2}) = 2+3\sqrt{2}$ and $g(1-\sqrt{2}) = 1 - 2\sqrt{2}$ , it is clear that the minimum value of $|g(z)|$ , over the interval $[-\sqrt{2},\sqrt{2}]$ is $2\sqrt{2}-1 = \boxed{1.828427125}$ .