The value of the given expression $\cos (510°)\cos (330°)+\sin (390°)\cos (120°)\mathrm{is}$

Q. Prove that:
$\cos {510}^{\circ }\cos {330}^{\circ }+\sin {390}^{\circ }\cos {120}^{\circ }=-1$

Q. Prove that $cos{510}^{\circ }cos{330}^{\circ }+sin{390}^{\circ }cos{120}^{\circ }=-1$

Q. Prove that:
(i) tan 720° − cos 270° − sin 150° cos 120° = $\frac{1}{4}$
(ii) sin 780° sin 480° + cos 120° sin 150° = $\frac{1}{2}$
(iii) sin 780° sin 120° + cos 240° sin 390 = $\frac{1}{2}$
(iv) sin 600° cos 390° + cos 480° sin 150° = −1
(v) tan 225° cot 405° + tan 765° cot 675° = 0

Q. If $\frac{\cos {12}^0+\sin {12}^0}{\cos {12}^0-\sin {12}^0}=\tan x^0$, what is the value of $x$?

Q. Find the value of $\frac{sin(-{660}^0)tan\left({1050}^0\right)sec(-{420}^0)}{cos\left({225}^0\right)cosec\left({315}^0\right)cos\left({510}^0\right)}$