If sec x=x

Question:

If $\sec x=x+\frac{1}{4 x}$, then $\sec x+\tan x=$

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

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

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

(d) $-\frac{1}{x}, x$

 

Solution:

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

We have,

$\sec x=x+\frac{1}{4 x}$

$\Rightarrow \sec ^{2} x==x^{2}+\frac{1}{16 x^{2}}+\frac{1}{2}$

$\Rightarrow 1+\tan ^{2} x=1+x^{2}+\frac{1}{16 x^{2}}-\frac{1}{2}$

$\Rightarrow \tan ^{2} x=x^{2}+\frac{1}{16 x^{2}}-\frac{1}{2}$

$\Rightarrow \tan ^{2} x=\left(x-\frac{1}{4 x}\right)^{2}$

$\therefore \tan x=\pm\left(x-\frac{1}{4 x}\right)$

$\Rightarrow \sec x-\tan x=\left(x+\frac{1}{4 x}\right)-\left(x-\frac{1}{4 x}\right)$ or $\left(x+\frac{1}{4 x}\right)-\left[-\left(x-\frac{1}{4 x}\right)\right]$

$=\frac{1}{2 x}$ or $2 x$

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