M = sec theta,show that m^2-1/m^2+1 =1/cosec theta

1.	M = sec theta,show that m^2-1/m^2+1 =1/cosec theta
Answer» We\xa0have,(m2\xa0- 1) = (sec{tex}\\theta{/tex}\xa0+ tan{tex}\\theta{/tex})2\xa0- 1= sec2{tex}\\theta{/tex}\xa0+ tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0- 1= (sec2{tex}\\theta{/tex} - 1\xa0) + tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}= 2tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0[{tex}\\because{/tex}\xa0sec2{tex}\\theta{/tex}\xa0- 1 = tan2{tex}\\theta{/tex}]= 2tan{tex}\\theta{/tex}(tan{tex}\\theta{/tex}\xa0+ sec{tex}\\theta{/tex}) ..(i)(m2 +1) = (sec{tex}\\theta{/tex}\xa0+ tan{tex}\\theta{/tex})2 + 1= sec2{tex}\\theta{/tex}\xa0+ tan2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0+ 1= (1 + tan2{tex}\\theta{/tex}) + sec2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}= 2sec2{tex}\\theta{/tex}\xa0+ 2sec{tex}\\theta{/tex}tan{tex}\\theta{/tex}\xa0[{tex}\\because{/tex}\xa01 + tan2{tex}\\theta{/tex}\xa0= sec2{tex}\\theta{/tex}]= 2sec{tex}\\theta{/tex}(sec{tex}\\theta{/tex}\xa0+ tan{tex}\\theta{/tex}) ..(ii)From (i) and (ii), we get{tex}\\frac { \\left( m ^ { 2 } - 1 \\right) } { \\left( m ^ { 2 } + 1 \\right) } = \\frac { \\tan \\theta } { \\sec \\theta } = \\left( \\frac { \\sin \\theta } { \\cos \\theta } \\times \\cos \\theta \\right) = \\sin \\theta{/tex}Hence,\xa0{tex}\\frac { \\left( m ^ { 2 } - 1 \\right) } { \\left( m ^ { 2 } + 1 \\right) } = \\sin \\theta{/tex}

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