Question

Question #e50ba

Answer 1

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`