Question

please give answer of this
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Answer 1

Solution

$= tan {}^{ - 1} ( \frac{a + b \: tanx}{b + a \: tanx} ) \\ = tan {}^{ - 1}( \frac{ \frac{a}{b} + tanx }{1 - \frac{a}{b}tanx } ) \\ now \: using \: the \: identity \: \\ tan {}^{ - 1} x + tan {}^{ - } y = tan {}^{ - }( \frac{x + y}{1 - xy} ) \\ = tan {}^{ - 1} ( \frac{a}{b} ) + tan {}^{ - 1} (tanx) \\ = tan {}^{ - 1} ( \frac{a}{b} ) + x$

Answer 2

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