Question

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

• (A) 2
• (B) √2
• (C) 4
• (D) 1

Answer 1

The Correct Option is B

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