Question

If $a>b>0$, then the value of $Ta{n}^{-1}\left(\frac{a}{b}\right)+Ta{n}^{-1}\left(\frac{a+b}{a-b}\right)$

• A. neither a nor b
• B. a and not b
• C. b and not a
• D. both a and b

Answer 1

Answer: (A)

${\tan }^{-1}\left(\frac{a}{b}\right)+{\tan }^{-1}\left(\frac{a+b}{a-b}\right)$
$={\tan }^{-1}\left(\frac{\frac{a}{b}+\frac{a+b}{a-b}}{1-\frac{a}{b}\left(\frac{a+b}{a-b}\right)}\right)$
$={\tan }^{-1}\left(\frac{a^2-ab+ab+b^2}{ab-b^2-a^2-ab}\right)$
$={\tan }^{-1}\left(-\frac{a^2+b^2}{a^2+b^2}\right)={\tan }^{-1}\left(-1\right)$
$\therefore$ The value is neither depends on a nor b.