Question

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

Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{x = acosec\theta + bcot\theta} \\ &\sf{y = acosec\theta - bcot\theta} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: eliminate - \begin{cases} &\sf{\theta}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \sf \: {cosec}^{2}x - {cot}^{2}x = 1}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x = acosec\theta + bcot\theta - - (1)$

$\rm :\longmapsto\:y = acosec\theta - bcot\theta - - - (2)$

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

$\rm :\longmapsto\:x + y = 2acosec\theta$

$\bf\implies \:cosec\theta = \dfrac{x + y}{2a} - - - (3)$

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

$\rm :\longmapsto\:x - y = 2bcot\theta$

$\bf\implies \:cot\theta = \dfrac{x - y}{2b} - - - (4)$

Now,

We know,

$\rm :\longmapsto\: {cosec}^{2}\theta - {cot}^{2}\theta = 1$

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

$\rm :\longmapsto\: {\bigg(\dfrac{x + y}{2a} \bigg) }^{2} - {\bigg(\dfrac{x - y}{2b} \bigg) }^{2} = 1$

$\rm :\longmapsto\: {\bigg(\dfrac{x + y}{a} \bigg) }^{2} - {\bigg(\dfrac{x - y}{b} \bigg) }^{2} = 4$

Additional Information :-

$\boxed{ \sf \: {sin}^{2} + {cos}^{2}x = 1}$

$\boxed{ \sf \: {sec}^{2}x - {tan}^{2}x = 1}$

$\boxed{ \sf \: cosecx = \dfrac{1}{sinx}}$

$\boxed{ \sf \: secx = \dfrac{1}{cosx}}$

$\boxed{ \sf \: cotx = \dfrac{1}{tanx}}$