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प्रश्न
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
विकल्प
- \[\frac{1}{2} \cos 2 x\]
0
- \[- \frac{1}{2} \cos 2 x\]
- \[\frac{1}{2}\]
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उत्तर
\[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\]
\[ = \cos\left( \frac{\pi}{6} + x + \frac{\pi}{6} - x \right)\cos\left( \frac{\pi}{6} + x - \frac{\pi}{6} + x \right) \left[\text{ Using }\cos(A + B) \cos(A - B) = \cos^2 A - \sin^2 B \right]\]
\[ = \cos\frac{2\pi}{6}\cos2x\]
\[ = \frac{1}{2}\cos2x \left[ \text{ As }\cos\frac{\pi}{3} = \frac{1}{2} \right]\]
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