Tg’2 -1=0

To solve the equation \( \tan^2 \theta - 1 = 0 \), we can start by rearranging the equation:

\[ \tan^2 \theta = 1 \]

Taking the square root of both sides, we get:

\[ \tan \theta = \pm 1 \]

The solutions to \( \tan \theta = 1 \) occur when:

\[ \theta = \frac{\pi}{4} + k\pi \]

The solutions to \( \tan \theta = -1 \) occur when:

\[ \theta = \frac{3\pi}{4} + k\pi \]

where \( k \) is any integer (i.e., \( k \in \mathbb{Z} \)).

Thus, the general solutions to the equation \( \tan^2 \theta - 1 = 0 \) are:

\[ \theta = \frac{\pi}{4} + k\pi \quad \text{and} \quad \theta = \frac{3\pi}{4} + k\pi \]

for any integer \( k \).