Trig problem 3

Let the other equation of ${ x }^{ 2 }-(tan\theta -\frac { 1 }{ tan\theta } )x+1=0$ as $\alpha$ .

Using the relationship of roots and coefficients, we have $\alpha (2+\sqrt { 3 } )$ =1

$\alpha$ = $\frac { 1 }{ 2+\sqrt { 3 } }$ = $2-\sqrt { 3 }$

Then $\alpha +(2+\sqrt { 3 } )$ = $(2-\sqrt { 3 } )+(2+\sqrt { 3 } )=4=tan\theta +\frac { 1 }{ tan\theta }$

So that $tan\theta +\frac { 1 }{ tan\theta } =\frac { sin\theta }{ cos\theta } +\frac { cos\theta }{ sin\theta } =\frac { 1 }{ sin\theta cos\theta } =4$

$sin\theta cos\theta$ = $\boxed { 0.25 }$