If \(\theta\) is eliminated from the equations \(x=a\cos { \left( \theta -\alpha \right) } \) and \(y=b\cos { \left( \theta -\beta \right) }\), which of the following is the resulting equation obtained?

- \(\quad \dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } +\dfrac { 2xy }{ ab } \cos { \left( \alpha -\beta \right) } = \cot ^{ 2 }{ \left( \alpha -\beta \right) } \)
- \(\quad \dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } -\dfrac { 2xy }{ ab } \cos { \left( \alpha -\beta \right) } = \sin ^{ 2 }{ \left( \alpha -\beta \right) } \)
- \(\quad \dfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } +\dfrac { 2xy }{ ab } \cos { \left( \alpha -\beta \right) } = \tan ^{ 2 }{ \left( \alpha -\beta \right) } \)
- \(\quad \dfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } -\dfrac { 2xy }{ ab } \cos { \left( \alpha -\beta \right) } = - \csc ^{ 2 }{ \left( \alpha -\beta \right) } \)

×

Problem Loading...

Note Loading...

Set Loading...