Prove that:
$
\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = -\frac{1}{2}.
$

Q: Prove that: $ \sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = -\frac{1}{2}. $

Explanation: We know that \[ \sin\frac{\pi}{6} = \frac{1}{2}, \quad \cos\frac{\pi}{3} = \frac{1}{2}, \quad \tan\frac{\pi}{4} = 1. \] Substituting these values: \[ \sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2 \] \[ = \frac{1}{4} + \frac{1}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}. \]

Question

Prove that:  
$
\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4} = -\frac{1}{2}.
$

Accepted Answer

Explanation: 
We know that
$$
\sin\frac{\pi}{6} = \frac{1}{2}, \quad \cos\frac{\pi}{3} = \frac{1}{2}, \quad \tan\frac{\pi}{4} = 1.
$$

Substituting these values:
$$
\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4}
= \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2
$$

$$
= \frac{1}{4} + \frac{1}{4} - 1
= \frac{1}{2} - 1 = -\frac{1}{2}.
$$

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