MCQ
$\sec^2\text{x}=\frac{4\text{xy}}{(\text{x}+\text{y})^2}$ is true if and only if
- A$\text{x+y}\neq0$
- B$\text{x=y, x}\neq0$
- C$\text{x=y}$
- D$\text{x}\neq0,\text{y}\neq0$
Solution:
We have:
$\sec^2\text{x}=\frac{4\text{xy}}{(\text{x}+\text{y})^2}$
$\Rightarrow\frac{4\text{x}\text{y}}{(\text{x}+\text{y})^2}\geq1 $ $[\therefore\sec^2\text{x}\geq1]$
$\Rightarrow4\text{xy}\geq(\text{x}+\text{y})^2$
$\Rightarrow4\text{xy}\geq\text{x}^2+\text{y}^2+2\text{xy}$
$\Rightarrow2\text{xy}\geq\text{x}^2+\text{y}^2$
$\Rightarrow(\text{x}-\text{y})^2\leq0$
$\Rightarrow(\text{x}-\text{y})\leq0$
$\Rightarrow\text{x}=\text{y}$
For $\text{x}=0,\sec^2\text{x}$ will not be defined,
$\Rightarrow\text{x}\neq 0$
$\therefore\text{x}=\text{y}$
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