Question

The value of $\sec [{\tan }^{-1}\frac{b+a}{b-a}-{\tan }^{-1}\frac{a}{b}]$ is

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

Answer 1

The correct option is B $\sqrt{2}$

$\sec ({\tan }^{-1}\left(\frac{b+a}{b-a}\right)-{\tan }^{-1}\frac{a}{b})$

$\sec ({\tan }^{-1}\left(\frac{\frac{b+a}{b-a}-\frac{a}{b}}{1+\frac{(b+a)}{(b-a)}\frac{a}{b}}\right)$

$\sec \left({\tan }^{-1}\left(\frac{b(b+a)-a(b-a)}{(b-a)b+(b+a)a}\right)\right)$

$\sec \left({\tan }^{-1}\left(\frac{b^2+ab-ab+a^2}{b^2-ab+a^2+ab}\right)\right)$

$\sec \left({\tan }^{-1}\left(\frac{a^2+b^2}{a^2+b^2}\right)\right)$

$\sec \left({\tan }^{-1}\right(1\left)\right)$

$\sec \left(\frac{\pi }{4}\right)=\sqrt{2}$