Question

Eliminate t between a cosec t + b cot t= x , a cot t + b cosec t = y​​

Answer 1

Answer:

${a}^{2} - {b}^{2} = {x}^{2} - {y}^{2}$

Step-by-step explanation:

$a \csc(t) + b \cot(t) = x \\ \\ a \cot(t) + b \csc(t) = y \\ \\ a \csc(t) \cot(t) + b { \cot(t) }^{2} = x \cot(t) \\ \\ a \csc(t) \cot(t) + b { \csc(t) }^{2} = y \csc(t) \\ \\ \\ \\ solving \: these \: equations \\ \\ \\ b( { \csc(t) }^{2} - { \cot(t) }^{2} ) = ( y\csc(t) - x \cot(t) ) \\ \\b = ( y\csc(t) - x \cot(t) ) \\ \\ \\ \\ \\ a \csc(t) + b \cot(t) = x \\ \\ a \cot(t) + b \csc(t) = y \\ \\ \\ squaring \: and \: subtracting \\ \\ {(a \csc(t)) }^{2} + {(b \cot(t)) }^{2} - {(a \cot(t)) }^{2} + {(b \csc(t)) }^{2} = {x}^{2} - {y}^{2} \\ \\ {a}^{2} - {b}^{2} = {x}^{2} - {y}^{2}$