Question

If sin0/x = cos0/y then prove that sin 0 – cos 0 = x–y/√x² + y²​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$given : \frac{sin \:θ }{x} = \frac{cos \:θ }{y} \\ \\ to \: prove : sin \:θ - cos \: θ = \frac{x - y}{ \sqrt{x {}^{2} + y {}^{2} } } \\ \\ so \: now \\ \frac{sin \:θ }{cos \:θ } = \frac{x}{y} \: \: \: \: \: \: \: \: \: (1) \\ \\ tan \: θ = \frac{x}{y} \\ \\ we \: know \: that \\ 1 + tan {}^{2}θ = sec {}^{2} θ \\ = 1 + ( \: \frac{x}{y} \: ) {}^{2} \\ \\ = 1 + \frac{x {}^{2} }{y {}^{2} } \\ \\ = \frac{x {}^{2} + y {}^{2} }{y {}^{2} } \\ \\ secθ = \frac{ \sqrt{x {}^{2} + y {}^{2} } }{y} \\ \\ cos \: θ = \frac{y}{ \sqrt{x {}^{2} + y {}^{2} } }$

$applying \: dividendo \: on \: (1) \\ we \: get \\ \\ \frac{sin \:θ - cos \: θ}{cos \: θ} = \frac{x - y}{y} \\ \\ sin \: θ - cos \: θ = \frac{x - y}{y} \times cos \: θ \\ \\ = \frac{x -y }{y} \times \frac{y}{ \sqrt{x {}^{2} + y {}^{2} } } \\ \\ sin \:θ - cos \: θ= \frac{x - y}{ \sqrt{x {}^{2} + y {}^{2} } } \\ \\ thus \: proved$