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if tanA - tanB = x and cotB - CotA = y then cot(A-B)=?
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tan A = 1 / cotA right ?
yes
\[\huge{tanA=\frac{1}{cotA}}\] Ok also the same for tan B
Cot(A-B) = cotA.CotB-1/(cotB - CotA)
cot (A-B) = (cot A cot B + 1)/(cot B - cot A) ok !!!
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put cot B - cot A = y
|dw:1341497660209:dw|
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