Solve the following.

$\frac{2 \sin 22^{\circ}}{\cos 68^{\circ}}-\frac{2 \cot 75^{\circ}}{5 \tan 15^{\circ}}-\frac{8 \tan 45^{\circ} \tan 20^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}$

0

Using :-

sinA = cosB  Iff A + B  = 90º

tanA = cotB  Iff A + B  = 90º

& tanA × tanB = 1   Iff A + B  = 90º

Now,

2\(\frac{sin22º}{cos68º}\) - \(\frac{2cot75º}{5tan15º}\) - \(\frac{8tan45º. tan20º.tan40º.tan50º.tan70º}{2}\)

= 2\(\frac{sin22º}{sin22º}\) - \(\frac{2tan15º}{5tan15º}\) - \(\frac{8tan45º. tan20º.tan40º.tan50º.tan70º}{5}\)

= 2 - \(\frac{2}{5}\) - \(\frac{8}{5}\)

= 0