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What is the value of $$sin(\frac{-\pi}{3})+cos(\frac{-\pi}{6})$$ ?
Expression : $$sin(\frac{-\pi}{3})+cos(\frac{-\pi}{6})$$
$$\because sin(-\theta)=-sin\theta$$ and $$cos(-\theta)=cos\theta$$
= $$-sin(\frac{\pi}{3})+cos(\frac{\pi}{6})$$
= $$-\frac{\sqrt3}{2}+\frac{\sqrt3}{2}$$
= $$0$$
=> Ans - (A)
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