Question

The general solutions of the equation (\tan^2 \theta + \sec 2\theta = 1) are

• A. (n\pi, n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z})
• B. (n\pi, n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z})
• C. (\frac{n\pi}{4}, \frac{n\pi}{4} \pm \frac{\pi}{3}, n \in \mathbb{Z})
• D. (n\pi, n\pi \pm \frac{\pi}{6}, n \in \mathbb{Z})

Hint:

General solution for $\tan^2 \theta = \tan^2 \alpha$ is $\theta = n\pi \pm \alpha$.

Answer 1

The Correct Option is A

Step 1: Use double angle formula
$\tan^2 \theta + \frac{1+\tan^2 \theta}{1-\tan^2 \theta} = 1$.
$\tan^2 \theta (1-\tan^2 \theta) + (1+\tan^2 \theta) = 1-\tan^2 \theta$.

Step 2: Simplify
$\tan^2 \theta - \tan^4 \theta + 1 + \tan^2 \theta = 1 - \tan^2 \theta$.
$3\tan^2 \theta - \tan^4 \theta = 0$.
$\tan^2 \theta (3 - \tan^2 \theta) = 0$.

Step 3: Solve for $\theta$
Case 1: $\tan^2 \theta = 0 \implies \theta = n\pi$.
Case 2: $\tan^2 \theta = 3 \implies \tan^2 \theta = \tan^2(\pi/3) \implies \theta = n\pi \pm \pi/3$.
Final Answer: (A)