Question

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

• A. $2por1/2p$
• B. $por1/2p$
• C. $2por1/p$
• D. None of these

Answer 1

The correct option is A $2por1/2p$

Given

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

${\sec }^{2}x=p^2+\frac{1}{16p^2}+2\left(p\right)\left(\frac{1}{4p}\right)$

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

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

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

${\tan }^{2}x=p^2+\frac{1}{16p^2}-2\left(p\right)\left(\frac{1}{4p}\right)$

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

$\tan x=\pm (p-\frac{1}{4p})$

$\sec x+\tan x$

$⟹p+\frac{1}{4p}\pm (p-\frac{1}{4p})$

$⟹2p\left(or\right)\frac{1}{2p}$