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If 2cos theta -sin theta =x and cos theta - 3sin theta =y,prove that 2xsquare +y square -2xy =5 |
| Answer» Given, 2 cos{tex}\\theta {/tex} - sin{tex}\\theta {/tex}\xa0{tex}= x\xa0{/tex} andcos{tex}\\theta {/tex}\xa0- 3 sin{tex}\\theta {/tex}\xa0{tex}= y{/tex}Put the values of x and y in {tex}2x^2\xa0+ y^2\xa0− 2xy{/tex}\xa0(LHS) , we get\xa0= 2(2 cos{tex}\\theta {/tex}\xa0− sin{tex}\\theta {/tex})2\xa0+ (cos{tex}\\theta {/tex}\xa0− 3 sin{tex}\\theta{/tex})2\xa0− 2(2 cos{tex}\\theta {/tex}\xa0− sin{tex}\\theta {/tex})(cos{tex}\\theta {/tex}\xa0− 3 sin{tex}\\theta{/tex})= 2(4cos2{tex}\\theta {/tex}− 4cos{tex}\\theta {/tex}\xa0sin{tex}\\theta {/tex}\xa0+ sin2{tex}\\theta {/tex}) + (cos2\xa0{tex}\\theta {/tex}− 6cos{tex}\\theta {/tex}\xa0sin{tex}\\theta {/tex}\xa0+ 9sin2{tex}\\theta{/tex}) − 2(2cos2\xa0\xa0{tex}\\theta{/tex}− 7cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0+ 3sin2{tex}\\theta{/tex})= 8cos2{tex}\\theta{/tex}\xa0− 8cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0+ 2sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}\xa0− 6cos{tex}\\theta{/tex} sin{tex}\\theta{/tex}\xa0+ 9sin2{tex}\\theta{/tex}\xa0− 2(2cos2\xa0\xa0{tex}\\theta{/tex}− 7cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0+ 3sin2{tex}\\theta{/tex})= 8cos2{tex}\\theta{/tex}\xa0− 8cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0+ 2sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}\xa0− 6cos{tex}\\theta{/tex} sin{tex}\\theta{/tex}\xa0+ 9sin2{tex}\\theta{/tex}\xa0− 4cos2{tex}\\theta{/tex}\xa0+ 14cos{tex}\\theta{/tex}\xa0sin{tex}\\theta{/tex}\xa0− 6sin2{tex}\\theta{/tex}= 5cos2{tex}\\theta{/tex}\xa0+ 5sin2{tex}\\theta{/tex}= 5(cos2{tex}\\theta{/tex}\xa0+ sin2{tex}\\theta{/tex}){tex}=5{/tex}= RHS | |