Trigonometry & Inverse Trigonometry Question - None

Question

Example 42: The set of values of $X \in \mathbb{R}$ such that $\tan^2 \theta + \sec \theta = X$ holds for some $\theta$ is

Accepted Answer

Step 1: From $\tan^2 \theta + \sec \theta = X$, use $\tan^2 \theta = \sec^2 \theta - 1$

Step 2: This gives $\sec^2 \theta - 1 + \sec \theta = X$, or $\sec^2 \theta + \sec \theta - 1 - X = 0$

Step 3: For real $\sec \theta$, the discriminant must be non-negative: $1 + 4(1 + X) \geq 0$, so $4X + 5 \geq 0$, giving $X \geq -\frac{5}{4}$

Step 4: The solutions are $\sec \theta = \frac{-1 \pm \sqrt{4X + 5}}{2}$

Step 5: Since $\sec \theta \leq -1$ or $\sec \theta \geq 1$, we need either:

$\frac{-1 - \sqrt{4X + 5}}{2} \leq -1$ which gives $\sqrt{4X + 5} \geq 1$, or

$\frac{-1 + \sqrt{4X + 5}}{2} \geq 1$ which gives $\sqrt{4X + 5} \geq 3$

Step 6: From $\sqrt{4X + 5} \geq 3$, we get $4X + 5 \geq 9$, so $X \geq 1$

∴ The answer is $[1, \infty)$

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