If $\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$ where a > b > 0, then ${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$ is :

Question

If $\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$

where a > b > 0, then ${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$ is :

${{a - 2b} \over {a + 2b}}$

${{a - b} \over {a + b}}$

${{a + b} \over {a - b}}$

${{2a + b} \over {2a - b}}$

Solution

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