Question

If $\sec x+\tan x=p,$ then the value of $\frac{\sec x-\tan x}{2\sec x}$ is

• A. $\frac{p}{1+p^2}$
• B. $\frac{1}{1+p^2}$
• C. $\frac{p}{1-p^2}$
• D. $\frac{1}{1-p^2}$

Answer 1

The correct option is B $\frac{1}{1+p^2}$

$\sec x+\tan x=p\cdots \left(i\right)$
We know that,
${\sec }^{2}x-{\tan }^{2}x=1\Rightarrow (\sec x+\tan x)(\sec x-\tan x)=1\Rightarrow (\sec x-\tan x)=\frac{1}{p}\cdots \left(ii\right)$
Adding equations $\left(i\right)$ and $\left(ii\right)$, we get
$2\sec x=p+\frac{1}{p}=\frac{p^2+1}{p}$

Therefore,
$\frac{\sec x-\tan x}{2\sec x}=\frac{\frac{1}{p}}{\frac{p^2+1}{p}}=\frac{1}{1+p^2}$