Помогите пожалуйста срочно нужно решить

Решите задачу:

$tg \frac{3\pi}{4} + \cos \frac{5\pi}{3} \times \sin \frac{7\pi}{6} - {( \sin \frac{2\pi}{3} ) }^{2} = \\ = tg \frac{4\pi - \pi}{4} + \cos \frac{6\pi - \pi}{3} \times \sin \frac{6\pi + \pi }{6} - \\ - {( \sin \frac{3\pi - \pi}{3} ) }^{2} = tg( \frac{4\pi}{4} - \frac{\pi}{4}) + \\ + \cos( \frac{6\pi}{3} - \frac{\pi}{3}) \times \sin( \frac{6\pi}{6} + \frac{\pi}{6}) - {( \sin (\frac{3\pi}{3} - \frac{\pi}{3}) ) }^{2} = \\ = tg(\pi - \frac{\pi}{4} ) + \cos (2\pi - \frac{\pi}{3} )\times \sin (\pi + \frac{\pi}{6}) - \\ - {( \sin(\pi - \frac{\pi}{3} )) }^{2} = - tg \frac{\pi}{4} + \cos \frac{\pi}{3} \times ( - \sin \frac{\pi}{6}) - {( \sin \frac{\pi}{3} ) }^{2} = \\ = - tg \frac{\pi}{4} - \cos \frac{\pi}{3} \times \sin ( \frac{\pi}{2} - \frac{\pi}{3} )- {( \sin \frac{\pi}{3} ) }^{2} = \\ = - 1 - \cos \frac{\pi}{3} \times \cos \frac{\pi}{3} - {( \sin \frac{\pi}{3} ) }^{2} = - 1 - (\cos \frac{\pi}{3} )^2 - \\ - (\sin \frac{\pi}{3} )^2 = - 1 - (\cos^2 \frac{\pi}{3} + \sin^2 \frac{\pi}{3} )=-1-1=-2$