Prove that
$3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}=1$
Prove that
$3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}=1$
Solution We have
${\text{ L}}{\text{.H}}{\text{.S}}{\text{.}} = 3\sin \frac{\pi }{6}\sec \frac{\pi }{3} - 4\sin \frac{{5\pi }}{6}\cot \frac{\pi }{4}$
$ = 3 \times \frac{1}{2} \times 2 - 4\sin \left( {\pi - \frac{\pi }{6}} \right) \times 1 = 3 - 4\sin \frac{\pi }{6}$
$ = 3 - 4 \times \frac{1}{2} = 1 = R.H.S$
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