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Trigonometric Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions3 Marks
Question
Prove that: $\tan\frac{11\pi}{3}-2\sin\frac{4\pi}{6}-\frac{3}{4}\text{cosec}^2\frac{\pi}{4}+4\cos^2\frac{17\pi}{6}=\frac{3-4\sqrt{3}}{2}$
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Answer

$\text{L.H.S}=\tan\frac{11\pi}{3}-2\sin\frac{4\pi}{6}-\frac{3}{4}\text{cosec}^2\frac{\pi}{4}+\cos^2\frac{17\pi}{6}$ $=\tan\Big(4\pi-\frac{\pi}{3}\Big)-2\sin\frac{2\pi}{3}-\frac{3}{4}\times(\sqrt{2})^2+4\cos^2\Big(3\pi-\frac{\pi}{6}\Big)$ $=-\tan\frac{\pi}{3}-2\sin\Big(\pi-\frac{\pi}{3}\Big)-\frac{3}{4}\times2+4\cos^2\frac{\pi}{6}$ $\Big(\because\tan\Big(4\pi-\frac{\pi}{3}=-\tan\frac{\pi}{3},\cos\Big(3\pi-\frac{\pi}{6}\Big)=-\cos\frac{\pi}{6}\Big)$ $=-\sqrt{3}=2\sin\frac{\pi}{3}-\frac{3}{2}+4\times\Big(\frac{\sqrt{3}}{2}\Big)^2$ $=-\sqrt{3}-2\times\frac{\sqrt{3}}{2}-\frac{3}{2}+4\times\frac{3}{4}$ $=-\sqrt{3}-\sqrt{3}-\frac{3}{2}+3$ $=-2\sqrt{3}\frac{-3+6}{2}$ $=-2\sqrt{3}+\frac{3}{2}$ $=\frac{3-4\sqrt{3}}{2}$ $=\text{R.H.S}$ $\text{Proved}$

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