Sum of All Trig Functions

$f(x)=sinx+cosx+tanx+cotx+secx+cscx$ . $=sinx+cosx+\frac { 1 }{ sinxcosx } +\frac { sinx+cosx }{ sinxcosx }$

We can write $sinx+cosx=\sqrt { 2 } cos(\frac { \pi }{ 4 } -x)$ ; this suggests making the substitution $y=\frac { \pi }{ 4 } -x$ . In this new coordinate,

$sinxcosx=\frac { 1 }{ 2 } sin2x=\frac { 1 }{ 2 } cos2y,$

and writing $c=\sqrt { 2 } cosy$ ,

$f(y)=(1+c)(1+\frac { 2 }{ { c }^{ 2 }-1 } )-1$

$=c+\frac { 2 }{ c-1 }$ .

We must analyse this function of $c$ in the range $\left[ -\sqrt { 2 } ,\sqrt { 2 } \right]$ . Its value at $c=-\sqrt { 2 }$ is $2-3\sqrt {2} < -2.24$ , and at $c=\sqrt {2}$ is $2+3\sqrt {2} >6.24$ . Its derivative is $1-\frac { 2 }{ { (c-1) }^{ 2 } }$ , which vanishes when $c=1\pm \sqrt { 2 }$ . Only $c=1-\sqrt { 2 }$ is in bounds, and the value of $f$ is $1-2\sqrt { 2 }$ . As for the pole at c=1, $f$ decreases as c approaches from below (takes negative values for $c<1$ ) and increases as c approaches from above (takes positive values for $c>1$ ; from this data, we see that $f$ has no sign crossings, so the minimum of $\left| f \right|$ is achieved at a critical point of $f$ . We conclude that the minimum of $\left| f \right|$ is $2\sqrt { 2 }-1$ .

From this, we have $a=2, b=2, c=-1$ and therefore, $abc=\boxed { -4 }$ .