## Product of Tangents

What is the value of the following product?
$\text{tan}\,5^\circ\!\cdot\text{tan}\,15^\circ\!\cdot\text{tan} \,25^\circ\!\cdot\text{tan}\,35^\circ\!\cdot\text{tan}\,45^\circ\!\cdot\text{tan} \,55^\circ\!\cdot\text{tan}\,65^\circ\!\cdot\text{tan}\,75^\circ\!\cdot\text{tan}\,85^\circ$
Source: NCTM Mathematics Teacher, November 2006

SOLUTION
$\text{tan}\,85\cdot\text{tan}\,5=\dfrac{\text{sin}\,85\cdot\text{sin}\,5}{\text{cos}\,85\cdot\text{cos}\,5}$
$= \dfrac{(1/2)\text{cos}\,(85-5)-(1/2)\text{cos}\,(85+5)}{(1/2)\text{cos}\,(85-5)+(1/2)\text{cos}\,(85+5)}$
$=\dfrac{(1/2)\text{cos}\,80-(1/2)\text{cos}\,90}{(1/2)\text{cos}\,80+(1/2)\text{cos}\,90}$
$=\dfrac{(1/2)\text{cos}\,80-0}{(1/2)\text{cos}\,80+0}$
$=1$
Similarly,
$\text{tan}\,75\cdot\text{tan}\,15=\dfrac{(1/2)\text{cos}\,60}{(1/2)\text{cos}\,60}=1$
$\text{tan}\,65\cdot\text{tan}\,25=\dfrac{(1/2)\text{cos}\,40}{(1/2)\text{cos}\,40}=1$
$\text{tan}\,55\cdot\text{tan}\,35=\dfrac{(1/2)\text{cos}\,20}{(1/2)\text{cos}\,20}=1$
$\text{tan}\,45=1$
The product of tangents equals $1$.

Answer: $1$

Alternative solution
$\text{sin}(90-x)=\text{cos}\,x$
$\text{cos}(90-x)=\text{sin}\,x$
$\text{tan}(90-x)\cdot\text{tan}\,x=\dfrac{\text{sin}(90-x)\text{sin}\,x}{\text{cos}(90-x)\text{cos}\,x}$
$=\dfrac{\text{cos}\,x\,\text{sin}\,x}{\text{sin}\,x\,\text{cos}\,x}=1$
Apply the formula to our product
$\text{tan}\,85\cdot\text{tan}\,5=1$
$\text{tan}\,75\cdot\text{tan}\,15=1$
$\text{tan}\,65\cdot\text{tan}\,25=1$
$\text{tan}\,55\cdot\text{tan}\,35=1$
$\text{tan}\,45=1$
The product of tangents equals $1$.