If tan(π9),x,tan(7π18) are in arithmetic progression and tan(π9),y,tan(5π18) are also in arithmetic progression, then |x−2y| is equal to
A
1
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B
0
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C
3
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D
4
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Solution
The correct option is B0 x−2y=tan20∘+tan70∘2−(tan20∘+tan50∘) 12(tan70∘−tan20∘−2tan50∘) =12[(tan70∘−tan50∘)−(tan20∘+tan50∘)] =12[sin20∘cos70∘cos50∘−sin70∘cos20∘cos50∘] =12[1cos50∘−1cos50∘]=0