Question

maths 1st sum 4th bit

Answer 1

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Answer 2

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$= > \bf \: \tan(60) = \sqrt{3} \\ \\ = > \bf \: \tan(30) = \frac{1}{ \sqrt{3} }$

$\bf \underline{L. H. S} \\ \\ = > \bf \: \tan(a - b) \\ \\ = \tan(60 - 30) = \tan(30) \\ \\ = \bf\frac{1}{ \sqrt{3} } \\ \\ \\ \bf \underline{ R. H. S} \\ \\ = > \bf \: \frac{ \tan(a) - \tan(b) }{1 + \tan(a) \tan(b) } \\ \\ = \frac{ \tan(60) - \tan(30) }{1 + \tan(60) \tan(30) } \\ \\ = \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3 }. \frac{1}{ \sqrt{3} } } = \frac{ \frac{3 - 1}{ \sqrt{3} } }{1 + 1} \\ \\ \bf = \frac{ \frac{2}{ \sqrt{3} } }{2} = \frac{1}{ \sqrt{3} }$

Hence

L. H. S =R. H. S

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