Write the value

Question:

Write the value of $\tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)$.

Solution:

We know that $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$.

Now,

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

$=\tan ^{-1}\left(\frac{\frac{a^{2}+a b-a b+b^{2}}{b(a+b)}}{\frac{a b+b^{2}-a b+a^{2}}{b(a+b)}}\right)$

$=\tan ^{-1}(1)$

$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$                    $\left[\because \tan \frac{\pi}{4}=1\right]$

$=\frac{\pi}{4}$

$\therefore \tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)=\frac{\pi}{4}$

 

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