Use the Double Argument Property #color(red)(tan 2theta=(2tan theta)/(1-tan^2 theta)#
Since #color(red)(cos theta# is positive it means that #color(red)(theta# is either in quadrant I or IV. I will find both and you can pick the answer according to the right quadrant that is given in the problem.
Here it is,
#color(purple)(Quadrant I#.
If #color(red)(cos theta = 5/13# then #color(red)(adjacent = 5 and hypote n use = 13# therefore by using pythagorean property we have #color(red)(opposite = 12#. Therefore, #color(red)(tan theta=(opposite)/(adjacent)=12/5#
Now put it in to the formula for #color(red)(tan 2 theta#. That is,
#color(red)((2tan theta)/(1-tan^2 theta)=color(blue)((2(12/5))/(1-(12/5)^2#
#=color(blue)((24/5)/(1-144/25)#
#=color(blue)((24/5)/((25-144)/25)#
#=color(blue)((24/5)/(-119/25 #
#=color(blue)(24/5 * -25/119#
#=color(blue)(-120/119#
#color( Purple)(Quadrant IV#
If #color(red)(cos theta = 5/13# then #color(red)(adjacent = 5 and hypote n use = 13# therefore the #color(red)(opposite = -12#. Thus, #color(red)(tan theta=-12/5#
#color(red)((2tan theta)/(1-tan^2 theta)=color(blue)((2(-12/5))/(1-(12/5)^2#
#=color(blue)((-24/5)/(1-144/25)#
#=color(blue)((-24/5)/((25-144)/25)#
#=color(blue)((-24/5)/(-119/25 #
#=color(blue)(24/5 * 25/119#
#=color(blue)(120/119#
