$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right)=$

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right)=$
  1. $-\cos x$
  2. $-\sin x$
  3. $\cos x$
  4. $\sin x$

Solution

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right)$ $=\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x-\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x$ $=\sin x$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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