Given #pi < x<(3pi)/2 and tanx=3/4#
#pi < x<(3pi)/2#
#=>pi/2 < x/2<(3pi)/4->x/2in " 2nd quadrant"#
This means
#sin(x/2)->+ve#
#cos(x/2)->-ve#
#tan(x/2)->-ve#
Now #tanx=3/4#
#=>(2tan(x/2))/(1-tan^2(x/2))=3/4#
#=>8tan(x/2)=3-3tan^2(x/2)#
#=>3tan^2(x/2)+8tan(x/2)-3=0#
#=>3tan^2(x/2)+9tan(x/2)-tan(x/2)-3=0#
#=>3tan(x/2)(tan(x/2)+3)-1(tan(x/2)+3)=0#
#=>(3tan(x/2)-1)(tan(x/2)+3)=0#
This means
#tan(x/2)=1/3->"not acceptable as "tan(x/2)->-ve#
So #tan(x/2)->-3#
Now
#cos(x/2)=1/sec(x/2)=-1/sqrt(1+tan^2(x/2)#
#=-1/sqrt(1+(-3)^2)=-1/sqrt10#
Again
#sin(x/2)=tan(x/2)xxcos(x/2)#
#=-3xx(-1/sqrt10)=3/sqrt10#