^2 ( 6 + )- ^2 ( 6 - )= - Vidyadip

MCQ

$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)=$

✓
$\frac{1}{2} \cos 2 \theta$
B
$0$
C
$-\frac{1}{2} \cos 2 \theta$
D
$\frac{1}{2}$

✓

Answer

Correct option: A.

$\frac{1}{2} \cos 2 \theta$

(A)
$\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)$
$=\cos \left(\frac{\pi}{6}+\theta+\frac{\pi}{6}-\theta\right) \cos \left(\frac{\pi}{6}+\theta-\frac{\pi}{6}+\theta\right)$
$\ldots\left[\because \cos ^2 A-\sin ^2 B=\cos (A+B) \cos (A-B)\right]$
$=\cos \frac{2 \pi}{6} \cos 2 \theta=\frac{1}{2} \cos 2 \theta$

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