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Answer 1 · 2016-07-25T05:38:33

Given #tanx=9/40 and x->"In 3rd quadrant"#
So #x=pi+alpha."where alpha acute angle"#

#:.tanx=9/40=>tan(pi+alpha)=9/40#
#=>tanalpha=9/40#

#color(red)(2x=2pi+2alpha->"in 1st or in 2nd quadrant.")#

#color(blue)("So "sin2x" will be positive.")#

Now
#sin2x=sin(2pi+2alpha)=sin2alpha=(2tanalpha)/(1+tan^2alpha)#

#=(2*9/40)/(1+(9/40)^2)#

#=(2*9/40*40^2)/(40^2+9^2)#

#=720/1681#

Now #x/2=1/2(pi+alpha)=(pi/2+alpha/2)#
#:.tan(x/2)=tan(pi/2+alpha/2)=-cot(alpha/2)#

Again #tanalpha=9/40#
#=>(2tan(alpha/2))/(1-tan^2(alpha/2))=9/40#

If #tan(alpha/2)=a#
then the above relation becomes
#(2a)/(1-a^2)=9/40#

#=>9a^2+80a-9=0#

#=>a=(-80+sqrt(80^2-4*9*(-9)))/(2*9)#

#color(red)([a>0 " since " alpha/2 " acute"])#

#=(-80+82)/18=2/18=1/9#
#:.tan(alpha/2)=1/9#
#=>cot(alpha/2)=9#

Hence #tan(x/2)=-cot(alpha/2)=-9#