Question

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

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

Answer 1

Answer: (A)

neither a nor b

Answer 2

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