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Question Number 194064 by Abdullahrussell last updated on 27/Jun/23

Answered by som(math1967) last updated on 27/Jun/23

$$i)\times asec\theta -ii)\times \frac{cos\theta }{a}$$$$\frac{ay}{b}tan\theta +\frac{by}{a}cot\theta =asec\theta -\frac{cos\theta }{a}+\frac{b^{2}cos\theta }{a}$$$$y\left(\frac{a^2{tan}^2\theta +b^2}{abtan\theta }\right)=\frac{a^2({sec}^2\theta -1)+b^2}{asec\theta }$$$$y\left(\frac{a^2{tan}^2\theta +b^2}{abtan\theta }\right)=\frac{(a^2{tan}^2\theta +b^2)}{asec\theta }$$$$\Rightarrow y=bsin\theta$$$$\therefore \frac{y}{b}=sin\theta$$$$puttingy=bsin\theta in\left(i\right)$$$$\frac{x\cos \theta }{a}+{sin}^2\theta =1$$$$\Rightarrow x=acos\theta$$$$\therefore \frac{x}{a}=cos\theta$$$$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}={cos}^2\theta +{sin}^2\theta =1$$

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