Question 1077833
Think of {{{ 5*pi/3 }}} as {{{ 6*pi/3 - pi/3 }}}
which is {{{ 2*pi - pi/3 }}}
This makes a negative 60 degree angle with
the 0 degree vector, so
{{{ sec( 5*pi/3 ) = 1/cos( -pi/3 ) }}}
{{{ 1/cos( -pi/3 ) = 2 }}}
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{{{ tan( 5*pi/4 ) = tan( 4*pi/4 + pi/4 ) }}}
This is a tangent in the 3rd quadrant which
is (-)/(-) and is positive
{{{ tan( 5*pi/4 ) = 1 }}}
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{{{ cot( 2*pi/3 ) = cot( 3*pi/3 - pi/3 ) }}}
This is a cotangent in the 2nd quadrant
which is negative
{{{ cot( 2*pi/3 ) = -1/sqrt(3) }}}
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{{{ sin( -pi/3 ) }}} is a sine function  in the 4th quadrant 
which is negative.
{{{ sin( -pi/3 ) = -sqrt(3)/2 }}}
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Putting it all together:
{{{ 2*1 - ( -1/sqrt(3) )*( -sqrt(3)/2 ) }}}
{{{ 2 - 1/2 = 3/2 }}}
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Check the math. I just used trig functions of 
{{{ pi/3 }}} in different quadrants
and also a function of {{{ pi/4 }}} in the 3rd quadrant