If sec A = x + …

Given, sec {tex}\theta{/tex} = x + {tex}\frac{1}{4 x}{/tex}
We know that, tan {tex}\theta{/tex} = {tex}\pm \sqrt{\sec ^{2} \theta-1}{/tex}
{tex}=\pm \sqrt{\left(x+\frac{1}{4 x}\right)^{2}-1}{/tex}
{tex}=\pm \sqrt{x^2 + \frac{1}{(4x)^2}+2.x. \frac{1}{(4x)}-1}{/tex}
{tex}=\pm \sqrt{x^2 + \frac{1}{(4x)^2}+ \frac{1}{(2)}-1}{/tex}
{tex}=\pm \sqrt{x^2 + \frac{1}{(4x)^2}- \frac{1}{(2)}}{/tex}
{tex}=\pm \sqrt {\left(x-\frac{1}{4 x}\right)^2}{/tex}
{tex}=\pm\left(x-\frac{1}{4 x}\right){/tex}
Now, sec {tex}\theta{/tex} + tan {tex}\theta{/tex} = {tex}\left(x+\frac{1}{4 x}\right) \pm\left(x-\frac{1}{4 x}\right){/tex}
i) sec {tex}\theta{/tex} + tan {tex}\theta{/tex} = {tex}\left(x+\frac{1}{4 x}\right) +\left(x-\frac{1}{4 x}\right){/tex}
{tex}=2x{/tex}
ii) sec {tex}\theta{/tex} + tan {tex}\theta{/tex} = {tex}\left(x+\frac{1}{4 x}\right) -\left(x-\frac{1}{4 x}\right){/tex}
{tex}=\frac{1}{4x}-(-\frac{1}{4x}) \\=\frac{1}{4x}+\frac{1}{4x}\\=\frac{2}{4x}{/tex}
{tex}=\frac{1}{2 x}{/tex}