If $\sin \theta=\frac{x^2+y^2}{x^2-y^2}$, then

$x \neq y$ and $x \neq 0, y \neq 0$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Analyzing +4 | -1

\sin \theta=\frac{x^2+y^2}{x^2-y^2}

, then

View answer and explanation

Correct option: C

x=0

y=0

Option C (

x=0

y=0

) is correct because the given condition

\sin \theta=\frac{{x^2+y^2}}{{x^2-y^2}}

implies

\left|\frac{{x^2+y^2}}{{x^2-y^2}}\right| \leq 1

, which leads to

x=0

y=0

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