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| 1. |
If x=a sin A +b cos A and y =a cos A+b sin A. Prove thatx^2+y^2=a^2+b^2 |
| Answer» We have, {tex}a\\cos \\theta - b\\sin \\theta = x{/tex}....(i)and {tex}a\\sin \\theta + b\\cos \\theta = y{/tex}...(ii)Squaring Eq. (i) and (ii) and then adding, we get{tex}{x^2} + {y^2} = {(a\\cos \\theta - b\\sin \\theta )^2} + {(a\\sin \\theta + b\\cos \\theta )^2}{/tex}{tex} \\Rightarrow {x^2} + {y^2} = {a^2}{\\cos ^2}\\theta + {b^2}{\\sin ^2}\\theta{/tex}{tex} - 2ab\\cos \\theta \\sin \\theta + {a^2}{\\sin ^2}\\theta + {b^2}{\\cos ^2}\\theta + 2ab\\sin \\theta co\\operatorname{s} \\theta {/tex} [{tex}\\because {/tex} (a\xa0+ b)2 = a2 + b2 + 2ab\xa0and (a\xa0- b)2 = a2 + b2 - 2ab]{tex} \\Rightarrow {x^2} + {y^2} = {a^2}{\\cos ^2}\\theta + {b^2}{\\sin ^2}\\theta + {a^2}{\\sin ^2}\\theta + {b^2}{\\cos ^2}\\theta {/tex}{tex} \\Rightarrow {x^2} + {y^2} = {a^2}({\\cos ^2}\\theta + {\\sin ^2}\\theta ) + {b^2}({\\sin ^2}\\theta + {\\cos ^2}\\theta ){/tex}{tex}[\\because \\sin^2\\theta+\\cos^2\\theta=1]{/tex}{tex}\\Rightarrow x^2+y^2=a^2+b^2{/tex}Hence proved, LHS = RHS | |