Prove that: `sin^2pi/6+cos^2pi/3-t a n^2pi/4=-1/2`

Explanation

To prove that \( \sin^2\left(\frac{\pi}{6}\right) + \cos^2\left(\frac{\pi}{3}\right) - \tan^2\left(\frac{\pi}{4}\right) = -\frac{1}{2} \), we will evaluate each trigonometric function step by step.

### Step 1: Evaluate \( \sin^2\left(\frac{\pi}{6}\right) \)

We know that:
\[ \frac{\pi}{6} = 30^\circ \]
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