Use the identity \(\sin^2\alpha+\cos^2\alpha=1\).

\(\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{21}{25}=\dfrac{4}{25}\).

So \(\sin\alpha=\pm\dfrac{2}{5}\).

Since \(\dfrac{\pi}{2}<\alpha<\pi\), the angle is in quadrant II, where \(\sin\alpha>0\).

Therefore, \(\sin\alpha=\dfrac{2}{5}\).

\(\tan 10^\circ\cdot\cot 10^\circ=1\).

\(\sin 30^\circ=\dfrac{1}{2}\).

So \(1+4\cdot\dfrac{1}{2}=1+2=3\).

\(\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{15}{16}=\dfrac{1}{16}\).

So \(\cos\alpha=\pm\dfrac{1}{4}\).

Since \(-\dfrac{\pi}{2}<\alpha<0\), the angle is in quadrant IV, where \(\cos\alpha>0\).

Therefore, \(\cos\alpha=\dfrac{1}{4}\).

\(\tan 60^\circ=\sqrt{3}\), so \(\sqrt{3}\cdot\sqrt{3}=3\).

Also, \(\sin^2 20^\circ+\cos^2 20^\circ=1\).

Therefore, \(3-1=2\).

\(\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\).

So \(\sin\alpha=\pm\dfrac{4}{5}\).

Since \(\pi<\alpha<\dfrac{3\pi}{2}\), the angle is in quadrant III, where \(\sin\alpha<0\).

Therefore, \(\sin\alpha=-\dfrac{4}{5}\).

\(\tan 80^\circ=\dfrac{\sin 80^\circ}{\cos 80^\circ}\), so

\(\tan 80^\circ\cdot\dfrac{\cos 80^\circ}{\sin 80^\circ}=1\).

\(\cos 120^\circ=-\dfrac{1}{2}\).

Therefore, \(1+6\cdot\left(-\dfrac{1}{2}\right)=1-3=-2\).

\(\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{51}{100}=\dfrac{49}{100}\).

So \(\cos\alpha=\pm\dfrac{7}{10}\).

In quadrant II, \(\cos\alpha<0\).

Therefore, \(\cos\alpha=-\dfrac{7}{10}\).

\(\sin^2 110^\circ+\cos^2 110^\circ=1\).

\(\sin 45^\circ=\dfrac{\sqrt{2}}{2}\).

Then \(5\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=5\).

Therefore, \(1+5=6\).

\(-0.6=-\dfrac{3}{5}\).

\(\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\).

So \(\sin\alpha=\pm\dfrac{4}{5}\).

In quadrant II, \(\sin\alpha>0\), so \(\sin\alpha=\dfrac{4}{5}\).

\(\cot 80^\circ\cdot\tan 80^\circ=1\).

\(\cos 60^\circ=\dfrac{1}{2}\).

Therefore, \(1+12\cdot\dfrac{1}{2}=1+6=7\).

\(\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{7}{16}=\dfrac{9}{16}\).

So \(\cos\alpha=\pm\dfrac{3}{4}\).

In quadrant IV, \(\cos\alpha>0\), so \(\cos\alpha=\dfrac{3}{4}\).

\(\cot 30^\circ=\sqrt{3}\), so \(2\sqrt{3}\cdot\sqrt{3}=6\).

\(\sin^2 140^\circ+\cos^2 140^\circ=1\).

Therefore, \(6-1=5\).

\(\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{24}{25}=\dfrac{1}{25}\).

So \(\sin\alpha=\pm\dfrac{1}{5}\).

In quadrant III, \(\sin\alpha<0\), so \(\sin\alpha=-\dfrac{1}{5}\).

\(\cot 40^\circ=\dfrac{\cos 40^\circ}{\sin 40^\circ}\), so

\(\cot 40^\circ\cdot\dfrac{\sin 40^\circ}{\cos 40^\circ}=1\).

\(\cos 120^\circ=-\dfrac{1}{2}\).

Therefore, \(1-2\cdot\left(-\dfrac{1}{2}\right)=1+1=2\).

\(\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{91}{100}=\dfrac{9}{100}\).

So \(\cos\alpha=\pm\dfrac{3}{10}\).

In quadrant II, \(\cos\alpha<0\), so \(\cos\alpha=-\dfrac{3}{10}\).

\(\sin^2 200^\circ+\cos^2 200^\circ=1\).

\(\cos 45^\circ=\dfrac{\sqrt{2}}{2}\).

Then \(3\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=3\).

Therefore, \(1+3=4\).