Question

Prove that :

\[ \tan\frac{11\pi}{3} – 2\sin\frac{4\pi}{6} – \frac34\cosec^2\frac{\pi}{4} + 4\cos^2\frac{17\pi}{6} = \frac{3-4\sqrt3}{2} \]

Solution

\[ \tan\frac{11\pi}{3} = \tan\left(2\pi-\frac{\pi}{3}\right) = -\tan\frac{\pi}{3} = -\sqrt3 \]

\[ 2\sin\frac{4\pi}{6} = 2\sin\frac{2\pi}{3} = 2\times\frac{\sqrt3}{2} = \sqrt3 \]

\[ \frac34\cosec^2\frac{\pi}{4} = \frac34\left(\sqrt2\right)^2 = \frac34\times2 = \frac32 \]

\[ 4\cos^2\frac{17\pi}{6} = 4\cos^2\left(2\pi+\frac{5\pi}{6}\right) \]

\[ = 4\cos^2\frac{5\pi}{6} = 4\left(-\frac{\sqrt3}{2}\right)^2 \]

\[ = 4\times\frac34 = 3 \]

Substituting these values,

\[ \begin{aligned} &-\sqrt3-\sqrt3-\frac32+3 \\[4pt] =& -2\sqrt3+\frac32 \\[4pt] =& \frac{3-4\sqrt3}{2} \end{aligned} \]

Hence Proved.