mtan(a-30)=ntan(a+30)
=>m/n=tan(a+30)/tan(a-30)
=>(m+n)/(m-n)=(tan(a+30)+tan(a-30))/(t…
=>(m+n)/(m-n)=(8tana)/(2√3+2√3tan^2a)
Note that
tan(a+30)=(tana+tan30)/(1-tanatan30)
=(√3tana+1)/(√3-tana)
similarly, tan(a-30)=(√3tana-1)/(√3+tana)
so,tan(a+30)+tan(a-30)=(8tana)/(3-tan^2a…
and tan(a+30)-tan(a-30)=(2√3+2√3tan^2a)/(3-t…
so,(m+n)/(m-n)=(2/√3)(2tana)/(+tan^2a)
=>(√3/2)(m+n)/(m-n)=sin2a
=>cos^2(2a)=1-3(m+n)^2/4(m-n)^2
=>cos^2(2a)=[(m+n)/2(m-n)]^2
=>cos2a=(m+n)/2(m-n)
=>m/n=tan(a+30)/tan(a-30)
=>(m+n)/(m-n)=(tan(a+30)+tan(a-30))/(t…
=>(m+n)/(m-n)=(8tana)/(2√3+2√3tan^2a)
Note that
tan(a+30)=(tana+tan30)/(1-tanatan30)
=(√3tana+1)/(√3-tana)
similarly, tan(a-30)=(√3tana-1)/(√3+tana)
so,tan(a+30)+tan(a-30)=(8tana)/(3-tan^2a…
and tan(a+30)-tan(a-30)=(2√3+2√3tan^2a)/(3-t…
so,(m+n)/(m-n)=(2/√3)(2tana)/(+tan^2a)
=>(√3/2)(m+n)/(m-n)=sin2a
=>cos^2(2a)=1-3(m+n)^2/4(m-n)^2
=>cos^2(2a)=[(m+n)/2(m-n)]^2
=>cos2a=(m+n)/2(m-n)