$$ \frac { \sin heta } { an heta } = \sqrt { 1 - \sin ^ { 2 } heta } = 0 $$

Question

$$ \frac { \sin 	heta } { 	an 	heta } = \sqrt { 1 - \sin ^ { 2 } 	heta } = 0 $$

UpStudy AI · Accepted Answer

To solve the equation

$$
\frac{\sin \theta}{\tan \theta} = \sqrt{1 - \sin^2 \theta} = 0,
$$

we can break it down into two parts.

### Step 1: Solve $$ \frac{\sin \theta}{\tan \theta} = 0 $$

Recall that

$$
\tan \theta = \frac{\sin \theta}{\cos \theta}.
$$

Thus, we can rewrite the left side:

$$
\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta.
$$

So, we have:

$$
\cos \theta = 0.
$$

The cosine function is zero at:

$$
\theta = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}.
$$

### Step 2: Solve $$ \sqrt{1 - \sin^2 \theta} = 0 $$

The expression inside the square root simplifies to:

$$
\sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta}.
$$

This means:

$$
\cos \theta = 0.
$$

This is the same condition we found in Step 1.

### Conclusion

The solutions to the equation

$$
\frac{\sin \theta}{\tan \theta} = \sqrt{1 - \sin^2 \theta} = 0
$$

are:

$$
\theta = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}.
$$
The solutions are $$ \theta = \frac{\pi}{2} + n\pi $$, where $$ n $$ is any integer.

\( \frac { \sin \theta } { \tan \theta } = \sqrt { 1 - \sin ^ { 2 } \theta } = 0 \)