cosec^2 = 4xy (x +y)^2 is true if and only if - Vidyadip

MCQ

$cosec^2\theta $ = $\frac{4xy}{(x +y)^2}$ is true if and only if

A
$x + y$ $\neq$ $0$
✓
$x = y$, $x$ $\neq$ $0$
C
$x = y$
D
$x$ $\neq$ $0$, $y$ $\neq$ $0$

✓

Answer

Correct option: B.

$x = y$, $x$ $\neq$ $0$

b
$\frac{4xy}{(x +y)^2}$ $\leq$ $1$ $\Rightarrow$ $(x - y)^2$ < $0$ $x$= $y$ but $x$ $\neq$ $0$ ($cosec^2\theta$ $\neq$ $0$)

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