Question

Evaluate:

$\frac{2\sin 68°}{\cos 22°}-\frac{2\tan \left(90-15°\right)}{5\cot 15°}-\frac{3\tan 45°\tan 20°\tan 40°\tan 50°\tan 70°}{5\left({\sin }^{2}70°+{\sin }^{2}20°\right)}$

Answer 1

$$\begin{gathered} \frac{2\sin 68°}{\cos 22°}-\frac{2\tan \left(90°-15°\right)}{5\cot 15°}-\frac{3\tan 45°.\tan 20°.\tan 40°.\tan 50°.\tan 70°}{5\left({\sin }^{2}70°+{\sin }^{2}20°\right)} \\ =\frac{2\sin \left(90°-22°\right)}{\cos 22°}-\frac{2}{5}\times \frac{\cot 15°}{\cot 15°}-\frac{3}{5}\times \frac{\tan 45°.\tan 20°.\tan 40°.\tan \left(90°-40°\right).\tan \left(90°-20°\right)}{\left[{\sin }^2\left(90°-20°\right)+{\sin }^{2}20°\right]} \\ =\frac{2\cos 22°}{\cos 22°}-\frac{2}{5}\times 1-\frac{3}{5}\times \frac{1\times \tan 20°.\tan 40°.\cot 40°.\cot 20°}{\left({\cos }^{2}20°+{\sin }^{2}20°\right)} \\ =2-\frac{2}{5}-\frac{3}{5}\times \frac{\tan 20°.\tan 40°.{\displaystyle \frac{1}{\tan 40°}}.{\displaystyle \frac{1}{\tan 20°}}}{1} \\ =2-\frac{2}{5}-\frac{3}{5} \\ =\frac{10-2-3}{5} \\ =\frac{5}{5} \\ =1 \end{gathered}$$