Q99 / 180 · COMEDK 2016 · Mathematics

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

  • Correct answer
    A
    neither $a$ nor $b$
  • B
    $a$ and not $b$
  • C
    $b$ and not $a$
  • D
    both $a$ and $b$

Solution

We have, $\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1}\left(\frac{a+b}{a-b}\right)$ $=\tan ^{-1}\left[\frac{\frac{a}{b}+\frac{a+b}{a-b}}{1-\left(\frac{a}{b}\right)\left(\frac{a+b}{a-b}\right)}\right]$ $=\tan ^{-1}\left[\frac{\frac{a^{2}-a b+a b+b^{2}}{b(a-b)}}{\frac{b(a-b)-a(a+b)}{b(a-b)}}\right]$ $=\tan ^{-1}\left[\frac{a^{2}+b^{2}}{a b-b^{2}-a^{2}-a b}\right]$ $=\tan ^{-1}\left[\frac{a^{2}+b^{2}}{-\left(a^{2}+b^{2}\right)}\right]=\tan ^{-1}(-1)$ It does not depends neither $a$ nor $b$.

Hence, the correct answer is option: 'A'.

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