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

KCETKCET 2006Inverse Trigonometric Functions

Solution:

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