Question

Prove that: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]

 

Answer 1

LHS = \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4}cose c^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6}\]
\[ = \tan\left( \frac{11\pi}{3} \right) - 2\sin\left( \frac{4\pi}{6} \right) - \frac{3}{4} \left[ cosec\left( \frac{\pi}{4} \right) \right]^2 + 4 \left[ \cos\left( \frac{17\pi}{6} \right) \right]^2 \]
\[ = \tan\left( \frac{11}{3} \times 180^\circ \right) - 2\sin\left( \frac{4}{6} \times 180^\circ \right) - \frac{3}{4} \left[ cosec\left( \frac{180^\circ}{4} \right) \right]^2 + 4 \left[ \cos\left( \frac{17 \times 180^\circ}{6} \right) \right]^2 \]
\[ = \tan \left( 660^\circ \right) - 2\sin \left( 120^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \cos \left( 510^\circ \right) \right]^2 \]
\[ = \tan \left( 660^\circ \right) - 2\sin \left( 120^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \cos \left( 510^\circ \right) \right]^2 \]
\[ = \tan \left( 90^\circ \times 7 + 30^\circ \right) - 2\sin \left( 90^\circ \times 1 + 30^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \cos\left( 90^\circ \times 5 + 60^\circ \right) \right]^2 \]
\[ = \left[ - \cot \left( 30^\circ \right) \right] - \left[ 2\cos \left( 30^\circ \right) \right] - \frac{3}{4} \left[ cosec \left( 45^\circ \right) \right]^2 + 4 \left[ - \sin\left( 60^\circ \right) \right]^2 \]
\[ = - \cot \left( 30^\circ \right)-2\cos\left( 30^\circ \right) - \frac{3}{4} \left[ cosec\left( 45^\circ \right) \right]^2 + 4 \left[ \sin \left( 60^\circ \right) \right]^2 \]
\[ = - \sqrt{3}-\frac{2\sqrt{3}}{2} - \frac{3}{4} \left[ \sqrt{2} \right]^2 + 4 \left[ \frac{\sqrt{3}}{2} \right]^2 \]
\[ = - \sqrt{3} - \sqrt{3} - \frac{3}{2} + 3\]
\[ = \frac{3 - 4\sqrt{3}}{2}\]
 = RHS
Hence proved .