Is (cotA+cotB)/(cotAcotB-1) equals to (cotAcotB-1)/(cotA+cotB) ?

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Answer 1 · 2018-06-18T05:33:59

$L H S \neq R H S$ $LHS!= RHS$

We know that

$tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ $tan(A+B) = (tanA+tanB)/(1-tanAtanB)$

Thus,

$L H S = \frac{cot A + cot B}{cot A cot B - 1}$ $LHS=(cotA+cotB)/(cotAcotB-1)$

$= \frac{\frac{1}{tan A} + \frac{1}{tan B}}{\frac{1}{tan A} \cdot \frac{1}{tan B} - 1}$ $= (1/tanA + 1/tanB)/(1/tanA * 1/tanB -1)$

$= \frac{\frac{tan A + tan B}{tan A tan B}}{\frac{1 - tan A tan B}{tan A tan B}}$ $=((tanA+tanB)/(tanAtanB))/((1-tanAtanB)/(tanAtanB))$

$= \frac{tan A + tan B}{1 - tan A tan B}$ $=(tanA+tanB)/(1-tanAtanB)$

$= tan (A + B)$ $=tan(A+B)$

$R H S = \frac{1}{L H S}$ $RHS=1/(LHS)$

$= \frac{1}{tan (A + B)}$ $=1/(tan(A+B))$

$= cot (A + B)$ $=cot(A+B)$

Except when $tan X = cot X (X = \frac{π}{4} + n π),$ $tanX = cotX (X = pi/4 +npi),$

$L H S \neq R H S$ $LHS != RHS$