Question

The value of $\sec \left[{\tan }^{-1}\left(\frac{b+a}{b-a}\right)-{\tan }^{-1}\left(\frac{a}{b}\right)\right]$ is;

• A. 2
• B. $\sqrt{2}$
• C. 4
• D. 1
• E. $\frac{a}{b}$

Answer 1

Answer: (B)

$\sec \left[{\tan }^{-1}\left(\frac{b+a}{b-a}\right)-{\tan }^{-1}\left(\frac{a}{b}\right)\right]$
$=\sec \left[{\tan }^{-1}\left\{\frac{\frac{b+a}{b-a}-\frac{a}{b}}{\left(\frac{b+a}{b-a}\right)\left(\frac{a}{b}\right)}\right\}\right]$
$=\sec [{\tan }^{-1}\{1\}]$
$=\sec \frac{\pi }{4}=\sqrt{2}$