Introduction to...

$${\cos}^{2}{30^\circ}\ {\cos}^{2}{45^\circ}+4\ {\sec}^{2}{60^\circ}+\dfrac{1}{2}\  {\cos}^{2}{90^\circ}-2{\tan}^{2}{60^\circ} =$$

Value of $$\left(3{\cos}^{2}{60^0}+2{\cot}^{2}{30^0}-5{\sin}^{2}{45^0}\right)$$ is:

Value of $$\left(\cos{60^o}\cos{30^o}-\sin{60^o}\sin{30^o}\right)$$ is 

$$\dfrac {2}{3}(\cos^{4} 30^0 - \sin^{4} 45^0) - 3 (\sin^{2}60^0 - \sec^{2} 45^0) + \dfrac {1}{4} \cot^{2} 30^0 = $$

$$\left( \dfrac { 4 }{ 3 } \cot ^{ 2 }{ 30^0 } +3\sin ^{ 2 }{ 60^0 } -2\csc ^{ 2 }{ 60^0 } -\dfrac { 3 }{ 4 } \tan ^{ 2 }{ 30^0 }  \right) =$$

Value of $$\left(\cos{0^0}+\sin{30^0}+\sin{45^0}\right)\left(\sin{90^0}+\cos{60^0}-\cos{45^0}\right)$$ is

Value of $$\left({\sin}^{2}{30}+4{\cot}^{2}{45}-{\sec}^{2}{60}\right)$$ is

Find the value of:

$$(\cos 0^{\circ} + \sin 45^{\circ} + \sin 30^{\circ}) (\sin 90^{\circ} - \cos 45^{\circ} + \cos 60^{\circ}) = $$

In $$\triangle ABC$$, if $$AD\perp BC$$ and $$BD = 10\ cm; \angle B = 60^{\circ}$$ and $$\angle C = 30^{\circ}$$, then $$CD = .........$$