Question

find
$tan( \frac{\pi}{4} - \tan (inverse )( \frac{1}{8} )$

Answer 1

Hope u like my process
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Formula to be used
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$= > \bf \: \tan(a - b) = \frac{ \tan(a) - \tan(b) }{1 + \tan( a ) \tan(b) }$
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$= > \tan( \frac{\pi}{4} - \tan ^{ - 1} ( \frac{1}{8} ) ) \\ \\ = \frac{ \tan( \frac{\pi}{4} ) - \tan( \tan ^{ - 1} ( \frac{1}{8} ) ) }{1 + \tan( \frac{\pi}{4} ) \tan {}^{ - 1} ( \tan( \frac{1}{8} ) ) } \\ \\ = \frac{1 - \frac{1}{8} }{1 + (1)( \frac{1}{8}) } = \frac{ \frac{7}{8} }{1 + \frac{1}{8} } \\ \\ = \frac{ \frac{7}{8} }{ \frac{9}{8} } = \frac{7}{8} \times \frac{8}{9} \\ \\ = \bf \underline{ \frac{7}{9} }$
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