Question

$$\begin{gathered} \mathrm{Eliminate}\mathrm{\theta },\mathrm{if} \\ x=3\cos ec\theta +4cot\theta ,y=4\cos ec\theta -3cot\theta \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Given}:x=3\mathrm{cosec}\theta +4\cot \theta \mathrm{and}y=4\mathrm{cosec}\theta -3\cot \theta \\ \therefore 4x-3y=4\left(3\mathrm{cosec}\theta +4\cot \theta \right)-3\left(4\mathrm{cosec}\theta -3\cot \theta \right) \\ =12\mathrm{cosec}\theta +16\cot \theta -12\mathrm{cosec}\theta +9\cot \theta \\ =25cot\theta \\ \Rightarrow cot\theta =\frac{4x-3y}{25} \\ \mathrm{Also},3x+4y=3(3\cos ec\theta +4cot\theta )+4(4\cos ec\theta -3cot\theta ) \\ =9\cos ec\theta +12cot\theta +16\cos ec\theta -12cot\theta \\ =25\cos ec\theta \\ \Rightarrow \cos ec\theta =\frac{3x+4y}{25} \\ \mathrm{We}\mathrm{know}\mathrm{that}\cos e{c}^2\theta -co{t}^2\theta =1 \\ \Rightarrow {\left(\frac{3x+4y}{25}\right)}^2-{\left(\frac{4x-3y}{25}\right)}^2=1 \\ \Rightarrow {\left(3x+4y\right)}^2-{\left(4x-3y\right)}^2=625 \end{gathered}$$