Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

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

$$\begin{array}{rlrl}& \Rightarrow {\alpha }^2{\cos }^{2}x={\sin }^{2}x+{\alpha }^2-2& \Rightarrow 2={\sin }^{2}x\left(1+{\alpha }^2\right)\Rightarrow {\sin }^{2}x=\frac{2}{1+{\alpha }^2}& \end{array}$$

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