Question

If $\sec \theta =x+\frac{1}{4x}$, find the value of $\sec \theta +\tan \theta$

• A. $2x$
• B. $\frac{1}{2x}$
• C. $4x$
• D. both a and b

Answer 1

The correct option is D both a and b

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

$=\frac{{4x}^2+1}{4x}=\frac{H}{B}$

$P=\sqrt{{({4x}^2+1)}^2-{\left(4x\right)}^2}$

$=\sqrt{({4x}^2+1-4x)({4x}^2+1+4x)}$

$=\sqrt{{(2x+1)}^2{(2x-1)}^2}$

$=\left|(2x+1)(2x-1)\right|$

$\sec \theta +\tan \theta =\frac{{4x}^2+1}{4x}+\frac{(2x+1)(2x-1)}{4x}$ (PF +ve mode open)

$=\frac{{4x}^2+1+{4x}^2-2x+2x-1}{4x}$

$=\frac{{8x}^2}{4x}$

$=2x$

(-ve mode open)

$\sec \theta +\tan \theta =\frac{{4x}^2+1}{4x}-\frac{(2x+1)(2x-1)}{4x}$

$=\frac{{4x}^2+1-{4x}^2+2x-2x+1}{4x}$

$=\frac{2}{4x}=\frac{1}{2x}$