Question

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

• A. 2
• B. $\sqrt{2}$
• C. 4
• D. 1

Answer 1

Answer: (B)

$ta{n}^{-1}\frac{b+a}{b-a}-ta{n}^{-1}\frac{a}{b}=ta{n}^{-1}\frac{\frac{b+a}{b-a}-\frac{a}{b}}{1+\frac{b+a}{b-a}\cdot \frac{a}{b}}$
$=ta{n}^{-1}\frac{b^2+ab-ab+a^2}{b^2-ab+ab+a^2}$
$=ta{n}^{-1}\frac{a^2+b^2}{a^2+b^2}$
$=ta{n}^{-1}=\frac{\pi }{4}$
$\therefore$ required value $=sec\left(\frac{\pi }{4}\right)=\sqrt{2}$