If \frac{1+ \sin x}{1- \sin x}=\frac{(1+ \sin y)^3}{(1- \sin y)^3} for some real values x and y, then \frac{ \sin x}{ \sin y}=

Correct Answer is : 3+sin2y1+3sin2y

\frac{3+ \sin ^2 y}{1+3 \sin ^2 y}

\frac{3+\cos ^2 y}{1+3 \cos ^2 y}

\frac{3+ \sin ^2 y}{1-3 \sin ^2 y}

\frac{3+ \sin ^2 y}{1-3 \cos ^2 y}

\frac{1+3 \sin ^2 y}{1-3 \cos ^2 y}

KEAM 2023 Math Solved Paper

Question : 58 of 120

Marks: +1, -0

If ‌

1+sin‌x

1−sin‌x

=‌

(1+sin‌y)3

(1−sin‌y)3

for some real values x and y, then ‌

sin‌x

sin‌y

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