Question

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

Answer 1

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

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