Question

$\frac{{\alpha }^2}{1-{\tan }^{2}x}=\frac{{\sin }^{2}x+{\alpha }^2-2}{\cos 2x}$ has a solution if

• A. $\alpha \le -1$
• B. $\alpha \ge 1$
• C. $\alpha =1/2$
• D. $\alpha$ is any real number

Answer 1

Answer: (A), (B)

The correct options are
A $\alpha \le -1$
B $\alpha \ge 1$

$\frac{{\alpha }^2}{1-ta{n}^{2}x}=\frac{{\sin }^{2}x+{\alpha }^2-2}{\cos 2x}$
$\frac{{\alpha }^2}{1-{\tan }^{2}x}=\frac{({\sin }^{2}x+{\alpha }^2-2)(1+{\tan }^{2}x)}{1-{\tan }^{2}x}$

$\frac{{\alpha }^2}{1-{\tan }^{2}x}=\frac{({\sin }^{2}x+{\alpha }^2-2)\left({\sec }^{2}x\right)}{1-{\tan }^{2}x}$
$\Rightarrow {\alpha }^2{\cos }^{2}x={\sin }^{2}x+{\alpha }^2-2$

$\Rightarrow 2={\sin }^{2}x(1+{\alpha }^2)$

$\Rightarrow {\sin }^{2}x=\frac{2}{1+{\alpha }^2}$
Since, $0\le {\sin }^{2}x\le 1$

$\Rightarrow 0\le \frac{2}{1+{\alpha }^2}\le 1$

$\Rightarrow {\alpha }^2\ge 1$

$\Rightarrow \alpha \le -1$ or $\alpha \ge 1$