Question

If secθ=(m+n)/√mn, then find sinθ

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ sin \: θ \: (in \: terms \: of \: m) \\ \\ \: secθ \: = \frac{m + n}{ \sqrt{mn} } \\ \\ therefore \: then\: \\ \\ cos \: θ = \frac{ \sqrt{mn} }{m + n} \\ \\ so \: we \: know \: that \\ sin {}^{2} \: θ = 1 - cos {}^{2} \: θ \\ \\ = 1 - ( \frac{ \sqrt{mn} }{m + n} \: ) {}^{2} \\ \\ = 1 - \frac{mn}{m {}^{2} + n {}^{2} + 2mn} \\ \\ = \frac{m {}^{2} + n {}^{2} + 2mn - mn}{m {}^{2} + n {}^{2} + 2mn} \\ \\ = \frac{m {}^{2} + n {}^{2} + mn}{m {}^{2} + n {}^{2} + 2mn } \\ \\ = \frac{m {}^{2} + n {}^{2} + mn}{(m + n) {}^{2} } \\ \\ taking \: square \: root \: on \: both \: sides \\ we \: get \\ \\ sin \: θ = ± \: \frac{ \sqrt{m {}^{2} + n {}^{2} + mn} }{m + n}$