There is a property of the #tan# function that states:
if #tan(x/2) = t# then
#sin(x) = (2t)/(1+t^2)#
From here you write the equation
#(2t)/(1+t^2) = 3/5#
#rarr 5*2t = 3(1+t^2)#
#rarr 10t = 3t^2+3#
#rarr 3t^2-10t+3 = 0#
Now you find the roots of this equation:
#Delta = (-10)^2 - 4*3*3 = 100-36 = 64#
#t_(-) = (10-sqrt(64))/6 = (10-8)/6 = 2/6 = 1/3#
#t_(+) = (10+sqrt(64))/6 = (10+8)/6 = 18/6 = 3#
Finaly you have to find which of the above answers is the right one. Here is how you do it:
Knowing that #90°< x <180°# then #45°< x/2 <90°#
Knowing that on this domain, #cos(x)# is a decreasing function and #sin(x)# is an increasing function, and that #sin(45°) = cos(45°)#
then #sin(x/2) > cos(x/2)#
Knowing that #tan(x) = sin(x)/cos(x)# then in our case #tan(x/2) > 1#
Therefore, the correct answer is #tan(x/2) = 3#
