Question

If ${\tan }^{2}x=1-{\alpha }^2$, then the possible values of $\alpha$ satisfying the equation $\sec x+{\tan }^{3}x\text{cosec}x=(2-{\alpha }^2{)}^{3/2}$ is

• A. $-\sqrt{2}$
• B. $-1$
• C. $1$
• D. $\sqrt{2}$

Answer 1

Answer: (B), (C)

$1$

$\sec x+{\tan }^{3}x\text{cosec}x=\sec x(1+{\tan }^{3}x\frac{\text{cosec}x}{\sec x})=\sqrt{1+{\tan }^{2}x}(1+{\tan }^{3}x\cot x)=(1+{\tan }^{2}x{)}^{3/2}=(1+1-{\alpha }^2{)}^{3/2}=(2-{\alpha }^2{)}^{3/2}$
So, the given equation is
$\sec x+{\tan }^{3}x\text{cosec}x=(2-{\alpha }^2{)}^{3/2}\Rightarrow (2-{\alpha }^2{)}^{3/2}=(2-{\alpha }^2{)}^{3/2}$
For the square root to be defined,
$2-{\alpha }^2\ge 0\Rightarrow {\alpha }^2\le 2\Rightarrow \alpha \in [-\sqrt{2},\sqrt{2}]\cdots \left(1\right)$
But we know that,
${\tan }^{2}x=1-{\alpha }^2\Rightarrow {\tan }^{2}x\ge 0\Rightarrow 1-{\alpha }^2\ge 0\Rightarrow {\alpha }^2\le 1\Rightarrow \alpha \in [-1,1]\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,
$\therefore \alpha \in [-1,1]$