Question
Prove that: $3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}=1$

Answer

$\text{L.H.S}=3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}$ $=3\times\frac{1}{2}\times2-4\sin\Big(\pi-\frac{\pi}{6}\Big)\times1$ $=3-4\sin\frac{\pi}{6}$ $=3-4\times\frac{1}{2}$ $= 3 - 2$ $= 1$ $\text{= R.H.S}$ $\text{Proved}$

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