If x = a cos θ and y = b cot θ, show that :
a2x2−b2y2\dfrac{a^2}{x^2} - \dfrac{b^2}{y^2}x2a2−y2b2 = 1
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Substituting value of x and y in L.H.S. of above equation :
⇒a2a2 cos2θ−b2b2 cot2θ⇒1cos2θ−1cot2θ⇒1cos2θ−1cos2θsin2θ⇒1cos2θ−sin2θcos2θ⇒1−sin2θcos2θ⇒cos2θcos2θ⇒1.\Rightarrow \dfrac{a^2}{a^2 \text{ cos}^2 θ} - \dfrac{b^2}{b^2 \text{ cot}^2 θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 θ} - \dfrac{1}{\text{cot}^2 θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 θ} - \dfrac{1}{\dfrac{\text{cos}^2 θ}{\text{sin}^2 θ}} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 θ} - \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{1 - \text{sin}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow 1.⇒a2 cos2θa2−b2 cot2θb2⇒cos2θ1−cot2θ1⇒cos2θ1−sin2θcos2θ1⇒cos2θ1−cos2θsin2θ⇒cos2θ1−sin2θ⇒cos2θcos2θ⇒1.
Since, L.H.S. = R.H.S.
Hence, proved that a2x2−b2y2\dfrac{a^2}{x^2} - \dfrac{b^2}{y^2}x2a2−y2b2 = 1.
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