help16

If mtan(A-30)=ntan(A+120), then prove that-

2*cos2A=[m+n]/[m-n]

$\begin{array}{rcll} m\cdot \tan(A-30^{\circ} ) &=& n\cdot \tan(A+120^{\circ}) \\ &=& n\cdot \tan(A+90^{\circ}+ 30^{\circ}) \\ &=& n\cdot \tan[90^{\circ}+( A+30^{\circ}) ] \quad | \quad \tan(90^{\circ}+x) = - \cot(x)\\ &=& -n\cdot \cot( A+30^{\circ}) \\ m\cdot \tan(A-30^{\circ} ) &=& -n\cdot \frac{1}{\tan( A+30^{\circ}) } \\ m\cdot\tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& -n \\ \tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& -\frac{n}{m} \quad | \quad \boxed{~ q = -\frac{n}{m} ~}\\ \tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& q \\ \end{array}$

$\begin{array}{rcll} \tan( A+30^{\circ}) \cdot \tan(A-30^{\circ} ) &=& \left( \frac{ \tan(A) +\tan(30^{\circ} ) } { 1-\tan(A)\tan(30^{\circ}) } \right) \cdot \left( \frac{ \tan(A) -\tan(30^{\circ} )} { 1+\tan(A)\tan(30^{\circ}) } \right) \\ &=& \frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) } \end{array}$

$\boxed{~ \begin{array}{rcll} \frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) } &=& q \end{array} ~}$

$\begin{array}{rcll} \frac{m+n}{m-n} &=& \frac{2m}{m-n}-1 \\ &=& \frac{ 2 } { \frac{m}{m} - \frac{n}{m} } - 1 \\ &=& \frac{ 2 } { 1 - \frac{n}{m} } - 1 \quad | \quad q = -\frac{n}{m}\\ &=& \frac{ 2 } { 1 +q } - 1 \\ &=& \frac{ 2 - (1+q)} { 1 +q } \\ &=& \frac{ 2 - 1-q} { 1 +q } \\ &=& \frac{ 1-q} { 1 +q } \\ \boxed{~ \frac{m+n}{m-n} = \frac{ 1-q} { 1 +q } ~} \end{array}$

$\begin{array}{rcll} \dfrac{m+n}{m-n} &=& \dfrac{ 1-q} { 1 +q } \\\\ &=& \dfrac{ 1-\frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) }} { 1 +\frac{\tan^2(A) -\tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) } } \\\\ &=& \dfrac{ 1-\tan^2(A)\tan^2(30^{\circ}) -\tan^2(A)+ \tan^2(30^{\circ}) } { 1-\tan^2(A)\tan^2(30^{\circ}) +\tan^2(A)- \tan^2(30^{\circ}) } \\\\ \dfrac{m+n}{m-n} &=& \dfrac{ [ 1-\tan^2(A)] [ 1+\tan^2(30^{\circ})] } { [ 1+\tan^2(A)] [ 1-\tan^2(30^{\circ})] } \end{array}$

$\begin{array}{rcll} \sin^2(A)+\cos(A)^2 &=& 1 \quad | \quad :\cos^2(A) \\ \frac{\sin^2(A)}{\cos^2(A)} + \frac{\cos^2(A)}{\cos^2(A)} &=& \frac{1}{\cos^2(A)} \\ \boxed{~ \tan^2(A) + 1 = \frac{1}{\cos^2(A)} ~} \end{array}$

$\begin{array}{rcll} \cos(A)^2 - \sin^2(A) &=& \cos(2A) \quad | \quad :\cos^2(A) \\ \frac{\cos^2(A)}{\cos^2(A)} - \frac{\sin^2(A)}{\cos^2(A)} &=& \frac{\cos(2A)}{\cos^2(A)} \\ \boxed{~ 1-\tan^2(A) = \frac{\cos(2A)}{\cos^2(A)} ~} \end{array}$

$\begin{array}{rcll} \dfrac{m+n}{m-n} &=& \dfrac{ [ 1-\tan^2(A)] [ 1+\tan^2(30^{\circ})] } { [ 1+\tan^2(A)] [ 1-\tan^2(30^{\circ})] } \\\\ \dfrac{m+n}{m-n} &=& \dfrac{ \frac{\cos(2A)}{\cos^2(A)} [ 1+\tan^2(30^{\circ})] } { \frac{1}{\cos^2(A)} [ 1-\tan^2(30^{\circ})] } \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \dfrac{ [ 1+\tan^2(30^{\circ})] } { [ 1-\tan^2(30^{\circ})] } \quad |\quad \tan(30^{\circ}) = \frac{1}{ \sqrt{3} } \quad \tan^2(30^{\circ}) = \frac13 \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \dfrac{ ( 1+\frac13) } { ( 1-\frac13) } \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \dfrac{ \frac43 } { \frac23 } \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \frac43 \cdot \frac32 \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot \frac42 \\\\ \dfrac{m+n}{m-n} &=& \cos(2A)\cdot 2 \\\\ \mathbf{ \dfrac{m+n}{m-n} }& \mathbf{=} & \mathbf{ 2\cdot \cos(2A) } \end{array}$

If mtan(A-30)=ntan(A+120), then prove that-2*cos2A=m+n/m-n

Are there suppose to be more brackets in there?    like this?

If mtan(A-30)=ntan(A+120), then prove that-2*cos2A=[m+n]/[m-n]

If mtan(A-30)=ntan(A+120), then prove that-2*cos2A=[m+n]/[m-n]

$m*tan(A-30)\\ =m*(tanA-tan30)\div (1+tanAtan30)\\ =m*(tanA-\frac{1}{\sqrt3})\div (1+\frac{tanA}{\sqrt3})\\ =m*(\frac{\sqrt3 tanA-1}{\sqrt3})\div (\frac{\sqrt3+tanA}{\sqrt3})\\ =m*(\frac{\sqrt3 tanA-1}{\sqrt3+tanA})\\ $

$ntan(A+120)\\=n*(tanA+tan120)\div (1-tanAtan120)\\ =n*(tanA-tan60)\div (1+tanAtan60)\\ =n*(tanA-\sqrt3)\div (1+\sqrt3 tanA)\\ =-n*(\sqrt3-tanA)\div (\sqrt3 tanA+1)\\ =-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\\ $

$m*\frac{ (\sqrt3 tanA-1)}{(\sqrt3+tanA)}=-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\\ m=-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\div \frac{ (\sqrt3 tanA-1)}{(\sqrt3+tanA)}\\ m=-n*\frac{(\sqrt3-tanA)}{ (\sqrt3 tanA+1)}\times \frac{(\sqrt3+tanA)}{ (\sqrt3 tanA-1)}\\ m=-n*\frac{(3-tan^2A)}{ (3 tan^2A-1)}\\ $

$\frac{m+n}{m-n}\\ =\left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}+n\right] \div \left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}-n\right] \\ =\left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}+\frac{n(3tan^2A-1)}{(3tan^2A-1)}\right] \div \left[\frac{-n(3-tan^2A)}{(3tan^2A-1)}-\frac{n(3tan^2A-1)}{(3tan^2A-1)}\right] \\=\left[\frac{-n(3-tan^2A)+n(3tan^2A-1)}{(3tan^2A-1)}\right] \div \left[\frac{-n(3-tan^2A)-n(3tan^2A-1)}{(3tan^2A-1)}\right] \\=\left[-n(3-tan^2A)+n(3tan^2A-1)\right] \div \left[-n(3-tan^2A)-n(3tan^2A-1)\right] \\=\left[-(3-tan^2A)+(3tan^2A-1)\right] \div \left[-(3-tan^2A)-(3tan^2A-1)\right] \\=\left[-3+tan^2A+3tan^2A-1\right] \div \left[-3+tan^2A-3tan^2A+1\right] \\=\left[-4+4tan^2A\right] \div \left[-2-2tan^2A\right] \\=\left[-2+2tan^2A\right] \div \left[-1-tan^2A\right] \\=\frac{2(1-tan^2A)}{1+tan^2A}$

$=2(1-\frac{sin^2A}{cos^2A})\div (1+\frac{sin^2A}{cos^2A})\\ =2(\frac{cos^2A-sin^2A}{cos^2A})\div (\frac{cos^2A+sin^2A}{cos^2A})\\ =2(cos^2A-sin^2A)\div (cos^2A+sin^2A)\\ =2(cos^2A-sin^2A)\div (1)\\ =2(cos^2A-sin^2A)\\ =2cos2A$

I've lost a negative sign somewhere - oh well, you can find it .....     :))