Question

if sin theta=m/n the show that tan theta =m/root of n^2-m^2

Answer 1

hey mate here is your answer

$\sin( \alpha ) = \frac{m}{n} = \frac{p}{h} \\ p = m \: \: \: \: \: h = n \\ b = \sqrt{ {h}^{2} } - {b}^{2} \\ \: \: \: = \sqrt{ {n}^{2} } - {m}^{2} \\ \\ now \: \\ \tan( \alpha ) = \frac{p}{b} \\ \: \: \: \: \: \: \: \: = \frac{m}{ \sqrt{ {n}^{2} } - {m}^{2} }$


I hope this helps you
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