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| 1. |
If SECa =X+1/4x, prove that SECa+ TANa =2x+1/2x |
| Answer» Given:\xa0{tex}\\sec {\\rm{A}} = x + {1 \\over {4x}}{/tex}Then,\xa0{tex}\\tan {\\rm{A}} = \\sqrt {{{\\sec }^2}{\\rm{A}} - 1} {/tex}\xa0=> {tex}\\tan {\\rm{A}} = \\sqrt {{x^2} + {1 \\over 2} + {1 \\over {16{x^2}}} - 1} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{{16{x^4} + 8{x^2} + 1 - 16{x^2}} \\over {16{x^2}}}} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{{16{x^4} - 8{x^2} + 1} \\over {16{x^2}}}} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{x^2} - {1 \\over 2} + {1 \\over {16{x^2}}}} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{{\\left( {x - {1 \\over {4x}}} \\right)}^2}} {/tex}=> {tex}\\tan {\\rm{A}} = \\pm \\left( {x - {1 \\over {4x}}} \\right){/tex}Therefore,\xa0{tex}\\sec {\\rm{A}} + \\tan {\\rm{A}} = x + {1 \\over {4x}} + x - {1 \\over {4x}} = 2x{/tex}Or\xa0{tex}\\sec {\\rm{A}} + \\tan {\\rm{A}} = x + {1 \\over {4x}} - x + {1 \\over {4x}} = {1 \\over {2x}}{/tex} | |