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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
\[\frac{1}{2} \cos 2x\]
0
\[- \frac{1}{2} \cos 2x\]
\[\frac{1}{2}\]
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उत्तर
\[\frac{1}{2} \cos 2x\]
\[\text{ We have, } \]
\[ \cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\]
\[ = \cos^2 \left( \frac{\pi}{6} + x \right) - \cos^2 \left[ \frac{\pi}{2} - \left( \frac{\pi}{6} - x \right) \right]\]
\[ = \cos^2 \left( \frac{\pi}{6} + x \right) - \cos^2 \left( \frac{\pi}{3} + x \right)\]
\[ = \left[ \cos\left( \frac{\pi}{6} + x \right) + \cos\left( \frac{\pi}{3} + x \right) \right]\left[ \cos\left( \frac{\pi}{6} + x \right) - \cos\left( \frac{\pi}{3} + x \right) \right]\]
\[ = 2\cos\left( \frac{\frac{\pi}{6} + x + \frac{\pi}{3} + x}{2} \right) \cos\left( \frac{\frac{\pi}{6} + x - \frac{\pi}{3} - x}{2} \right) 2\sin\left( \frac{\frac{\pi}{6} + x + \frac{\pi}{3} + x}{2} \right) \sin\left( \frac{\frac{\pi}{3} + x - \frac{\pi}{6} - x}{2} \right)\]
\[ = 4\cos\left( \frac{\pi}{4} + x \right)\cos\left( - \frac{\pi}{12} \right) \sin\left( \frac{\pi}{4} + x \right) \sin\left( \frac{\pi}{12} \right)\]
\[ = 4\cos\left( \frac{\pi}{4} + x \right)\cos\left( \frac{\pi}{12} \right) \sin\left( \frac{\pi}{4} + x \right) \sin\left( \frac{\pi}{12} \right)\]
\[ = \left[ 2\sin\left( \frac{\pi}{4} + x \right)\cos\left( \frac{\pi}{4} + x \right) \right]\left[ 2 \sin\left( \frac{\pi}{12} \right)\cos\left( \frac{\pi}{12} \right) \right]\]
\[ = \sin\left( \frac{\pi}{2} + 2x \right)\sin\frac{\pi}{6}\]
\[ = \cos2x \times \frac{1}{2}\]
\[ = \frac{1}{2}\cos2x\]
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