`mtan(theta-3 0^0)=ntan(theta+12 0^0)` show that `cos 2theta=(m+n)/(2(m-n))`

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m tan(theta-30^(@))=n tan(theta+120^(@)) show that cos2 theta=(m+n)/(2(m-n))

if m*tan(theta-30)=n*tan(theta+120), show that cos2 theta=(m+n)/(2(m-n))

if m*tan(theta-30)=n*tan(theta+120) , show that cos2theta=(m+n)/(2(m-n)) .

If m tan (theta - 30^(@))= n tan (theta + 120^(@)), show that cos 2 theta = (m +n)/(2 (m -n)).

If m tan (theta - 30^(@)) = n tan (theta + 120^(@)), "show that", 2 cos 2theta = (m+n)/(m-n).

If mtan(theta-30^@)=ntan(theta+120^@) , then cos2theta is equal to

cos2 theta+3cos theta=0

If mtan(theta-30^0)=ntan(theta+120^0) then (m+n)/(m-n) is equal to

If mtan(theta-30^@)=ntan(theta+120^@) , then (m+n)/(m-n)=

If tan(alpha+theta)=ntan(alpha-theta) , show that: (n+1)sin 2theta=(n-1)sin2alpha .