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1). \(\cot \theta= \frac{{2xy}}{{{x^2} - {y^2}}}\)2). \(\cot \theta= \frac{{2xy}}{{{x^2} + {y^2}}}\)3). \(\cot \theta= \frac{{x - y}}{{{x^2} + {y^2}}}\)4). \(\cot \theta= \frac{{xy\left( {x - y} \right)}}{{{x^2} + {y^2}}}\) |
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Answer» Given, $({\RM{cosec}}\THETA= \frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}})$ As, cot2θ = cosec2θ – 1 $(\begin{array}{l} \Rightarrow co{t^2}\theta= \;{\LEFT( {\frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}} \right)^2} - 1\\ \Rightarrow co{t^2}\theta= \;\frac{{{x^4} + {y^4} + 2{x^2}{y^2}}}{{{x^4} + {y^4} - 2{x^2}{y^2}}} - 1\\ \Rightarrow co{t^2}\theta= \;\frac{{{x^4} + {y^4} + 2{x^2}{y^2} - {x^4} - {y^4} + 2{x^2}{y^2}}}{{{x^4} + {y^4} - 2{x^2}{y^2}}}\\ \Rightarrow co{t^2}\theta= \;\frac{{4{x^2}{y^2}}}{{{{\left( {{x^2} - {y^2}} \right)}^2}}}\\ \Rightarrow \cot \theta= \;\frac{{2xy}}{{{x^2} - {y^2}}}\end{array})$ |
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