if x = a coseco + b coto and
y= a coseco – b coto then eliminate theta​

1 thought on “if x = a coseco + b coto and <br />y= a coseco – b coto then eliminate theta​”

1. [tex]\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{x = acosec\theta + bcot\theta} \\ &\sf{y = acosec\theta – bcot\theta} \end{cases}\end{gathered}\end{gathered}[/tex]

   [tex]\begin{gathered}\begin{gathered}\bf \: To \: eliminate – \begin{cases} &\sf{\theta}\end{cases}\end{gathered}\end{gathered}[/tex]

   [tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex]

   [tex] \boxed{ \sf \: {cosec}^{2}x – {cot}^{2}x = 1} [/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   Given that

   [tex]\rm :\longmapsto\:x = acosec\theta + bcot\theta – – (1)[/tex]

   [tex]\rm :\longmapsto\:y = acosec\theta – bcot\theta – – – (2)[/tex]

   On adding equation (1) and equation (2), we get

   [tex]\rm :\longmapsto\:x + y = 2acosec\theta[/tex]

   [tex]\bf\implies \:cosec\theta = \dfrac{x + y}{2a} – – – (3)[/tex]

   On Subtracting equation (2) from equation (1), we get

   [tex]\rm :\longmapsto\:x – y = 2bcot\theta[/tex]

   [tex]\bf\implies \:cot\theta = \dfrac{x – y}{2b} – – – (4)[/tex]

   Now,

   We know,

   [tex]\rm :\longmapsto\: {cosec}^{2}\theta – {cot}^{2}\theta = 1 [/tex]

   On substituting the values from equation(3) and (4), we get

   [tex]\rm :\longmapsto\: {\bigg(\dfrac{x + y}{2a} \bigg) }^{2} – {\bigg(\dfrac{x – y}{2b} \bigg) }^{2} = 1 [/tex]

   [tex]\rm :\longmapsto\: {\bigg(\dfrac{x + y}{a} \bigg) }^{2} – {\bigg(\dfrac{x – y}{b} \bigg) }^{2} = 4 [/tex]

   Additional Information :-

   [tex] \boxed{ \sf \: {sin}^{2} + {cos}^{2}x = 1} [/tex]

   [tex] \boxed{ \sf \: {sec}^{2}x – {tan}^{2}x = 1} [/tex]

   [tex] \boxed{ \sf \: cosecx = \dfrac{1}{sinx}} [/tex]

   [tex] \boxed{ \sf \: secx = \dfrac{1}{cosx}} [/tex]

   [tex] \boxed{ \sf \: cotx = \dfrac{1}{tanx}} [/tex]

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