First slide

Trigonometric Identities

Question

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

Moderate

A

$x + y \neq 0$
B

$x = y, x \neq 0$
C

$x = y$
D

$x \neq 0, y \neq 0$

Solution

Given, $\sec^{2} θ = \frac{4 xy}{(x + y)^{2}}$
$\begin{array}{l} Now \sec^{2} θ \geq 1 \Rightarrow \frac{4 xy}{(x + y)^{2}} \geq 1 \\ or (x + y)^{2} \leq 4 xy \\ or (x + y)^{2} - 4 xy \leq 0 \\ o r (x - y)^{2} \leq 0 \end{array}$
But for real values of x and y,
$\begin{array}{ll} (x - y)^{2} \geq 0 or (x - y)^{2} = 0 \\ ∴ x = y \end{array}$
Also $x + y \neq 0 \Rightarrow x \neq 0 . y \neq 0$

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