Question

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Answer 1

➩ $\sf \dfrac{ {x}^{ - 1} }{ {x}^{ - 1} \: \: + \: \: {y}^{ - 1} } + \dfrac{ {x}^{ - 1} }{ {x}^{ - 1} \: \: - \: \: {y}^{ - 1} } = \dfrac{2y^{2} }{ {y}^{2} - {x}^{2} }$

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➩ $\sf \dfrac{\frac{1}{x}}{\frac{1}{x} + \frac{1}{y}} + \dfrac{\frac{1}{x}}{\frac{1}{x} - \frac{1}{y}} =\dfrac{2y^{2} }{ {y}^{2} - {x}^{2}}$

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➩ $\sf \dfrac{( \frac{1}{x} - \frac{1}{y} ) \frac{1}{x} + ( \frac{1}{x} + \frac{1}{y} ) \frac{1}{x}}{(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y})} = \dfrac{2y^{2} }{ {y}^{2} - {x}^{2}}$

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➩ $\sf \dfrac{\frac{1}{x^2} - \cancel{\frac{1}{yx}} + \frac{1}{x^2} + \cancel{\frac{1}{yx}}}{(\frac{1}{x} + \frac{1}{y}) (\frac{1}{x} - \frac{1}{y})} = \dfrac{2y^{2} }{ {y}^{2} - {x}^{2}}$

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➩ $\sf \dfrac{\frac{1}{x^2} + \frac{1}{x^2}}{\frac{y^2-x^2}{x^2y^2}} = \dfrac{2y^{2} }{ {y}^{2} - {x}^{2}}$

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➩ $\sf \dfrac{\frac{2}{x^2}}{\frac{y^2-x^2}{x^2y^2}} = \dfrac{2y^{2} }{ {y}^{2} - {x}^{2}}$

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➩ $\sf \dfrac{2(x^2y^2)}{(y^2-x^2)x^2} = \dfrac{2y^{2} }{ {y}^{2} - {x}^{2}}$

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➩ $\sf \dfrac{\cancel{2}x^2\cancel{y^2}}{x^2\cancel{(y^2-x^2)}} = \dfrac{\cancel{2}\cancel{y^{2}} }{ \cancel{{y}^{2} - {x}^{2}}}$

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➩ $\sf \dfrac{\cancel x^2}{\cancel x^2} = 1$

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➩ $\sf 1 = 1$

Answer 2

Answer

Step-By-Step-Explanation