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Question
If \[\tan A = \frac{m}{m - 1}\text{ and }\tan B = \frac{1}{2m - 1}\], then prove that \[A - B = \frac{\pi}{4}\].
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Solution
We know that
\[\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}\]
\[ = \frac{\frac{m}{m - 1} - \frac{1}{2m - 1}}{1 + \frac{m}{(m - 1)(2m - 1)}}\]
\[ = \frac{2 m^2 - m - m + 1}{2 m^2 - m - 2m + 1 + m}\]
\[ = \frac{2 m^2 - 2m + 1}{2 m^2 - 2m + 1}\]
\[ = 1\]
\[ \Rightarrow A - B = \tan^{- 1} (1) \]
\[ \Rightarrow A - B = \frac{\pi}{4}\]
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