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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
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Verified Answer
The correct answer is:
neither $a$ nor $b$
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 $b$.
$\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1}\left(\frac{a+b}{a-b}\right)$
$=\tan ^{-1}\left[\frac{\frac{a}{b}+\frac{a+b}{a-b}}{1-\left(\frac{a}{b}\right)\left(\frac{a+b}{a-b}\right)}\right]$
$=\tan ^{-1}\left[\frac{\frac{a^{2}-a b+a b+b^{2}}{b(a-b)}}{\frac{b(a-b)-a(a+b)}{b(a-b)}}\right]$
$=\tan ^{-1}\left[\frac{a^{2}+b^{2}}{a b-b^{2}-a^{2}-a b}\right]$
$=\tan ^{-1}\left[\frac{a^{2}+b^{2}}{-\left(a^{2}+b^{2}\right)}\right]=\tan ^{-1}(-1)$
It does not depends neither $a$ nor $b$.
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