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Question
Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]
Solution
We know that
\[\tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} \left( \frac{x - y}{1 + xy} \right)\]
Now,
\[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right) = \tan^{- 1} \left( \frac{\frac{a}{b} - \frac{a - b}{a + b}}{1 + \frac{a}{b}\frac{a - b}{a + b}} \right)\]
\[ = \tan^{- 1} \left( \frac{\frac{a^2 + ab - ab + b^2}{b\left( a + b \right)}}{\frac{ab + b^2 - ab + a^2}{b\left( a + b \right)}} \right)\]
\[ = \tan^{- 1} \left( 1 \right)\]
\[ = \tan^{- 1} \left( \tan\frac{\pi}{4} \right) \left[ \because \tan\frac{\pi}{4} = 1 \right]\]
\[ = \frac{\pi}{4}\]
∴ \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right) = \frac{\pi}{4}\]
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