If SECa =X+1/4x, prove that SECa+ …

Given: {tex}\sec {\rm{A}} = x + {1 \over {4x}}{/tex}

Then, {tex}\tan {\rm{A}} = \sqrt {{{\sec }^2}{\rm{A}} - 1} {/tex} 

=>          {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, {tex}\sec {\rm{A}} + \tan {\rm{A}} = x + {1 \over {4x}} + x - {1 \over {4x}} = 2x{/tex}

Or {tex}\sec {\rm{A}} + \tan {\rm{A}} = x + {1 \over {4x}} - x + {1 \over {4x}} = {1 \over {2x}}{/tex}