If m = cosecA-sinA and n= secA -tanA prove that (m^2 n)^2/3+(mn^3)^2/3

1.	If m = cosecA-sinA and n= secA -tanA prove that (m^2 n)^2/3+(mn^3)^2/3=1
Answer» We have,cosec{tex}\\theta{/tex}\xa0- sin{tex}\\theta{/tex}\xa0= m and sec{tex}\\theta{/tex}\xa0- cos{tex}\\theta{/tex}\xa0= n{tex}\\Rightarrow \\quad \\frac { 1 } { \\sin \\theta } - \\sin \\theta = m \\text { and } \\frac { 1 } { \\cos \\theta } - \\cos \\theta{/tex}\xa0= n{tex}\\Rightarrow \\quad \\frac { 1 - \\sin ^ { 2 } \\theta } { \\sin \\theta } = m \\text { and } \\frac { 1 - \\cos ^ { 2 } \\theta } { \\cos \\theta }{/tex}\xa0= n{tex}\\Rightarrow \\quad \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta } = m \\text { and } \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta }{/tex}\xa0= n{tex}\\therefore \\quad \\left( m ^ { 2 } n \\right) ^ { 2 / 3 } + \\left( m n ^ { 2 } \\right) ^ { 2 / 3 } = \\left( \\frac { \\cos ^ { 4 } \\theta } { \\sin ^ { 2 } \\theta } \\times \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta } \\right) ^ { 2 / 3 } + \\left( \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta } \\times \\frac { \\sin ^ { 4 } \\theta } { \\cos ^ { 2 } \\theta } \\right) ^ { 2 / 3 }{/tex}= (cos3{tex}\\theta{/tex})2/3\xa0+ (sin3{tex}\\theta{/tex})2/3\xa0= cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}\xa0= 1Hence, (m2n)2/3 + (mn2)2/3 = 1

If m = cosecA-sinA and n= secA -tanA prove that (m^2 n)^2/3+(mn^3)^2/3=1