Question

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Answer 1

Dear student... hope the solution below will help u

Solution:-

mtan(x-π/6)=ntan(x+2π/3)

m/n=tan(x+2π/3)/tan(x-π/6)

applying compendo and dividendo we get

(m+n)/(m-n)={tan(x+2π/3)+tan(x-π/6)}/{tan(x+2π/3)-tan(x-π/6)

converting the tan into sin and cos we get

(m+n)/(m-n)={sin(x+2π/3)/cos(x+2π/3)+sin(x-π/6)/cos(x-π/6)}/{sin(x+2π/3)/(cos(x+2π/3)-sin(x-π/6)/cos(x-π/6)

taking LCM as cos(x-π/6){cos(x+2π/3)}

we get

(m+n)/(m-n)={sin(x+2π/3)cos(x-π/6)+cos(x-π/6)sin(x+2π/3)}/{sin(x+2π/4)sin(x+2π/3)sin(x-π/6)-{sin(x+2π/3){cos(x-π/6)-{sin(x-π/6)cos(x+2π/3)

Now,this fits under the formula

sin(A+B)=SinACosB+CosASinB

sin(A-B)=SinAcosB-Cos(A)Sin(B)

so,

(m+n)/(m-n)={sin(x+2π/3+x-π/6)}/{sin(x+2π/3-x+π/6)

(m+n)/(m-n)={Sin(2x+π/6)}/{sin(5π/6)}

(m+n)/(m-n)=2cos2x

m-n=2sec2x(m+n)

hence proved