Solve the following.
$\frac{2 \sin 22^{\circ}}{\cos 68^{\circ}}-\frac{2 \cot 75^{\circ}}{5 \tan 15^{\circ}}-\frac{8 \tan 45^{\circ} \tan 20^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}$
Answer & explanation
Correct answer: option 4
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