Question

If tan B = a/x then find the value of
$\frac{x}{ \sqrt{ {a}^{2} + {x}^{2} } }$
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Answer 1

Answer:

tan B = a/x

$\frac{x}{ \sqrt{ {a}^{2} + {x}^{2} } } \\ \\ = \frac{x}{ \sqrt{ {x}^{2} ( \frac{ {a}^{2} }{ {x}^{2} } +1) } } \\ \\ = \frac{x}{x \sqrt{ {( \frac{a}{x}) }^{2} + 1 } } \\ \\ = \frac{1}{ \sqrt{( {tan \: b)}^{2} + 1} } \\ \\ = \frac{1}{ \sqrt{ta {n}^{2}b + 1 } } \\ \\ = \frac{1}{ \sqrt{sec {}^{2} \: b} } \\ \\ = \frac{1}{sec \: b} \\ \\ = cos \: b$

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