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| 1. |
Sin u +cos u= m and sec u + cosec u=n then prove that n(m²-1)=2m |
| Answer» Given,\xa0sin{tex}\\theta{/tex}\xa0+ cos{tex}\\theta{/tex}\xa0= m and sec{tex}\\theta{/tex}\xa0+ cosec{tex}\\theta{/tex}\xa0= nConsider,\xa0n(m2 - 1) = (sec{tex}\\theta{/tex}\xa0+ cosec{tex}\\theta{/tex})[(sin{tex}\\theta{/tex}\xa0+ cos{tex}\\theta{/tex})2\xa0-\xa01]{tex}= \\left( \\frac { 1 } { \\cos \\theta } + \\frac { 1 } { \\sin \\theta } \\right){/tex}[(sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) + 2sin{tex}\\theta{/tex}cos{tex}\\theta{/tex}\xa0- 1]{tex}= \\frac { ( \\sin \\theta + \\cos \\theta ) } { \\sin \\theta \\cos \\theta } \\cdot [2 \\sin \\theta \\cos \\theta]{/tex}= 2(sin{tex}\\theta{/tex}\xa0+ cos{tex}\\theta{/tex})= 2m | |