Question

If $\sec \theta =2p$ and $\tan \theta =\frac{2}{p}$, find the value of $2(p-\frac{1}{p^2})$.

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{1}{2}$

we know that,

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

So, $2(p^2-\frac{1}{p^2})=\frac{1}{2}$

Therefore, Answer is $\frac{1}{2}$