If tanA= tanB × n , sinA= sinB × m Then, prove that: cos²A = m -1 / no -1

If tanA= tanB × n , sinA= sinB × m Then, prove that: cos²A = m -1 /

1.	If tanA= tanB × n , sinA= sinB × m Then, prove that: cos²A = m -1 / no -1
Answer» {tex}\\sin {\\rm{A}} = \\sin {\\rm{B}} \\times m{/tex} => {tex}m = {{\\sin {\\rm{A}}} \\over {\\sin {\\rm{B}}}}{/tex} => {tex}{1 \\over {\\sin {\\rm{B}}}} = {m \\over {\\sin {\\rm{A}}}}{/tex}=> {tex}\\cos ec{\\rm{B}} = {m \\over {\\sin {\\rm{A}}}}{/tex}{tex}\\tan {\\rm{A}} = \\tan {\\rm{B}} \\times n{/tex} => {tex}n = {{\\tan {\\rm{A}}} \\over {\\tan {\\rm{B}}}}{/tex} => {tex}{1 \\over {\\tan {\\rm{B}}}} = {n \\over {\\tan {\\rm{A}}}}{/tex}=> {tex}\\cot {\\rm{B}} = {n \\over {\\tan {\\rm{A}}}}{/tex}Using identity,\xa0{tex}\\cos e{c^2}{\\rm{B}} - {\\cot ^2}{\\rm{B}} = 1{/tex}=> {tex}{{{m^2}} \\over {{{\\sin }^2}{\\rm{A}}}} - {{{n^2}} \\over {{{\\tan }^2}{\\rm{A}}}} = 1{/tex}=> {tex}{{{m^2}} \\over {{{\\sin }^2}{\\rm{A}}}} - {{{n^2}.{{\\cos }^2}{\\rm{A}}} \\over {{{\\sin }^2}{\\rm{A}}}} = 1{/tex}=> {tex}{m^2} - {n^2}.{\\cos ^2}{\\rm{A}} = {\\sin ^2}{\\rm{A}}{/tex}=> {tex}{m^2} - {n^2}.{\\cos ^2}{\\rm{A}} = 1 - {\\cos ^2}{\\rm{A}}{/tex}=> {tex}{m^2} - 1 = - {\\cos ^2}{\\rm{A}} + {n^2}.{\\cos ^2}{\\rm{A}}{/tex}=> {tex}{\\cos ^2}{\\rm{A}} = {{{m^2} - 1} \\over {{n^2} - 1}}{/tex}