Question

If tan x = $x-\frac{1}{4x}$, then sec x − tan x is equal to
(a) $-2x,\frac{1}{2x}$
(b) $-\frac{1}{2x},2x$
(c) 2x
(d) $2x,\frac{1}{2x}$

Answer 1

(a) $-2x,\frac{1}{2x}$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \tan x=x-\frac{1}{4x} \\ \Rightarrow se{c}^{2}x=1+{\tan }^{2}x \\ \Rightarrow se{c}^{2}x=1+{\left(x-\frac{1}{4x}\right)}^2 \\ \Rightarrow se{c}^{2}x=x^2+\frac{1}{16x^2}+\frac{1}{2} \\ \Rightarrow se{c}^{2}x={\left(x+\frac{1}{4x}\right)}^2 \\ \therefore secx=\pm \left(x+\frac{1}{4x}\right) \\ \Rightarrow secx-\tan x=\left(x+\frac{1}{4x}\right)-\left(x-\frac{1}{4x}\right)\mathrm{or}-\left(x+\frac{1}{4x}\right)-\left(x-\frac{1}{4x}\right) \\ =\frac{1}{2x}\mathrm{or}-2x \end{gathered}$$