JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

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Analyzing +4 | -1

All the pairs (x, y) that satisfy the inequality

{2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.\frac{1}{{{4^{{{\sin }^2}y}}}} \le 1

also satisfy the equation

View answer and explanation

Correct option: B

\sin x = \left| {\sin y} \right|

Option B (

\sin x = \left| {\sin y} \right|

) is correct because the inequality holds only when

\sin x = 1

and

\sin^2 y = 1

, which implies

\sin x = |\sin y|

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