Question

If sec $\mathrm{\theta }=\mathrm{x}+\frac{1}{4\mathrm{x}}$, then sec θ + tan θ =
(a) $x,\frac{1}{x}$
(b) $2x,\frac{1}{2x}$
(c) $-2x,\frac{1}{2x}$
(d) $-\frac{1}{x},x$

Answer 1

(b) $2x,\frac{1}{2x}$

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