$Find the angle where tan( heta) = -1 in the unit circle$

To find the angle where $\tan(\theta) = -1$ in the unit circle, we need to look for the values of $\theta$ where the tangent function is negative and equals -1.

We know that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. For $\tan(\theta) = -1$, this means $\sin(\theta) = -\cos(\theta)$.

This occurs in the second and fourth quadrants.

In the second quadrant: $\theta = \pi – \frac{\pi}{4} = \frac{3\pi}{4}$

In the fourth quadrant: $\theta = 2\pi – \frac{\pi}{4} = \frac{7\pi}{4}$

Hence, the angles are $\theta = \frac{3\pi}{4}$ and $\theta = \frac{7\pi}{4}$.

Let’s solve for $ heta$ where $ an( heta) = -1$.

We know that $ an( heta) = frac{sin( heta)}{cos( heta)} = -1$ implies $sin( heta) = -cos( heta)$.

This occurs in the second and fourth quadrants of the unit circle.

In the second quadrant, the reference angle is $frac{pi}{4}$. So, $ heta = pi – frac{pi}{4} = frac{3pi}{4}$.

In the fourth quadrant, the reference angle is again $frac{pi}{4}$. Thus, $ heta = 2pi – frac{pi}{4} = frac{7pi}{4}$.

Therefore, the angles $ heta$ that satisfy $ an( heta) = -1$ are $frac{3pi}{4}$ and $frac{7pi}{4}$.

To find $ heta$ where $ an( heta) = -1$, set $sin( heta) = -cos( heta)$.

In the second quadrant: $ heta = frac{3pi}{4}$

In the fourth quadrant: $ heta = frac{7pi}{4}$

The angles are $frac{3pi}{4}$ and $frac{7pi}{4}$.

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