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Sec=17/8 verify that (3-4sin²)/(4cos²-3)=(3-tan²)/(1-3tan²)

Answer» Given\xa0{tex}\\sec A = \\frac{{17}}{8} = \\frac{{AC}}{{AB}}{/tex}Let AC = 17Kand, AB = 8KIn\xa0{tex}\\Delta ABC{/tex}, by Pythagoras theoremBC2 + AB2 = AC2BC2 + (8K)2 = (17K)2BC2 + 64K2 = 289K2BC2 = 289K2 - 64K2BC2 = 225K2{tex}BC = \\sqrt {225{K^2}} = 15K{/tex}{tex}\\therefore \\sin A = \\frac{{BC}}{{AC}} = \\frac{{15K}}{{17K}} = \\frac{{15}}{{17}}{/tex}{tex}\\cos A = \\frac{{AB}}{{AC}} = \\frac{{8K}}{{17K}} = \\frac{8}{{17}}{/tex}{tex}\\tan A = \\frac{{BC}}{{AB}} = \\frac{{15K}}{{8K}} = \\frac{{15}}{8}{/tex}LHS{tex} = \\frac{{3 - 4{{\\sin }^2}A}}{{4{{\\cos }^2}A - 3}}{/tex}{tex} = \\frac{{3 - 4 \\times {{\\left( {\\frac{{15}}{{17}}} \\right)}^2}}}{{4{{\\left( {\\frac{8}{{17}}} \\right)}^2} - 3}}{/tex}{tex} = \\frac{{3 - \\frac{{900}}{{289}}}}{{\\frac{{256}}{{289}} - 3}}{/tex}{tex} = \\frac{{\\frac{{867 - 900}}{{289}}}}{{\\frac{{256 - 867}}{{289}}}}{/tex}{tex} = \\frac{{\\frac{{ - 33}}{{289}}}}{{\\frac{{ - 611}}{{289}}}}{/tex}{tex} = \\frac{{ - 33}}{{289}} \\times \\frac{{289}}{{ - 611}}{/tex}{tex} = \\frac{{33}}{{611}}{/tex}RHS{tex} = \\frac{{3 - {{\\tan }^2}A}}{{1 - 3{{\\tan }^2}A}}{/tex}{tex} = \\frac{{3 - {{\\left( {\\frac{{15}}{8}} \\right)}^2}}}{{1 - 3 \\times {{\\left( {\\frac{{15}}{8}} \\right)}^2}}}{/tex}{tex} = \\frac{{3 - \\frac{{225}}{{64}}}}{{1 - \\frac{{675}}{{64}}}}{/tex}{tex} = \\frac{{\\frac{{192 - 225}}{{64}}}}{{\\frac{{64 - 675}}{{64}}}}{/tex}{tex} = \\frac{{\\frac{{ - 33}}{{64}}}}{{\\frac{{ - 611}}{{64}}}}{/tex}{tex}= \\frac{{ - 33}}{{64}} \\times \\frac{{64}}{{ - 611}}{/tex}{tex} = \\frac{{33}}{{611}}{/tex}Hence verified


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