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If tan θ = \(\frac{x}{y},\) then \(\frac{x\,sin\,\theta+y\,cos\,\theta}{x\,sin\,\theta-y\,cos\,\theta}\) is equal to 

(a) \(\frac{x^2+y^2}{x^2-y^2}\)

(b) \(\frac{x^2-y^2}{x^2+y^2}\)

(c) \(\frac{x}{\sqrt{x^2+y^2}}\)

(d) \(\frac{y}{\sqrt{x^2+y^2}}\)

1 Answer

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Best answer

(a) \(\frac{x^2+y^2}{x^2-y^2}\)

\(\frac{x\,sin\,\theta+y\,cos\,\theta}{x\,sin\,\theta-y\,cos\,\theta}\) = \(\frac{x\frac{sin\,\theta}{cos\,\theta}+y\frac{cos\,\theta}{cos\,\theta}}{x\frac{sin\,\theta}{cos\,\theta}-y\frac{cos\,\theta}{cos\,\theta}}\)

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