First slide

Trigonometric Identities

Question

$\sin^{2} θ = \frac{(x + y)^{2}}{4 x y}$ , where $x, y \in R$ , gives real $θ$ if and only if

Moderate

A

$x + y = 0$
B

$x = y$
C

$| x | = | y | \neq 0$
D

none of these

Solution

We know that
$\begin{array}{l} 0 \leq \sin^{2} θ \leq 1 \\ \Rightarrow 0 \leq \frac{(x + y)^{2}}{4 x y} \leq 1 \\ \Rightarrow 0 \leq (x + y)^{2} \leq 4 x y [∵ 4 x y > 0] \\ \Rightarrow (x - y)^{2} \leq 0 \Rightarrow x = y \end{array}$
Thus, we have $x = y$ and $x y > 0$
$∴ | x | = | y | \neq 0$

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