Question

If $\sec \beta =x+\frac{1}{4x}$ then the value of $\sec \beta +\tan \beta$ is

• A. $2x$
• B. $\frac{x}{2}$
• C. $3x$
• D. $\frac{x}{3}$

Answer 1

The correct option is A $2x$

Given that:

$sec\beta =x+\frac{1}{4x}\Rightarrow \frac{4x^2+1}{4x}$

$ta{n}^2\beta =se{c}^2\beta -1$

$ta{n}^2\beta =\frac{(4x^2+1{)}^2}{16x^2}-1$

$ta{n}^2\beta =\frac{(16x^4+1+8x^2-16x^2)}{16x^2}$

$tan\beta =\frac{(4x^2-1)}{4x}$

Hence,

$\Rightarrow sec\beta +tan\beta$

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

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

$\Rightarrow 2x.$