If x= a sec A + b tan A and y= a tan A + b sec A , then prove that x*x-y*y= a*a-b*b.

If x= a sec A + b tan A and y= a tan A + b sec A , then prove that x*x

1.	If x= a sec A + b tan A and y= a tan A + b sec A , then prove that xx-yy= aa-bb.
Answer» L.H.S. =\xa0{tex}{x^2} - {y^2} = {\\left( {a\\sec {\\rm{A}} + b\\tan {\\rm{A}}} \\right)^2} - {\\left( {a\\tan {\\rm{A}} + b\\sec {\\rm{A}}} \\right)^2}{/tex}=\xa0{tex}\\left( {a\\sec {\\rm{A}} + b\\tan {\\rm{A}} + a\\tan {\\rm{A}} + b\\sec {\\rm{A}}} \\right) {\\left( {a\\sec {\\rm{A}} + b\\tan {\\rm{A}} - a\\tan {\\rm{A}} - b\\sec {\\rm{A}}} \\right)}{/tex}=\xa0{tex}\\left[ {a\\left( {\\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right) + b\\left( {\\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right)} \\right]\\left[ {a\\left( {\\sec {\\rm{A}} - \\tan {\\rm{A}}} \\right) + b\\left( { - \\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right)} \\right]{/tex}=\xa0{tex}\\left( {\\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right)\\left( {a + b} \\right)\\left( {\\sec {\\rm{A}} - \\tan {\\rm{A}}} \\right)\\left( {a - b} \\right){/tex}=\xa0{tex}\\left( {{{\\sec }^2}{\\rm{A}} - {{\\tan }^2}{\\rm{A}}} \\right)\\left( {{a^2} - {b^2}} \\right){/tex}=\xa0{tex}{a^2} - {b^2}{/tex}= R.H.S.