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

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Answer 1 · 2018-06-18T05:33:59

#LHS!= RHS#

We know that

#tan(A+B) = (tanA+tanB)/(1-tanAtanB)#

Thus,

#LHS=(cotA+cotB)/(cotAcotB-1) #

#= (1/tanA + 1/tanB)/(1/tanA * 1/tanB -1)#

#=((tanA+tanB)/(tanAtanB))/((1-tanAtanB)/(tanAtanB))#

#=(tanA+tanB)/(1-tanAtanB)#

#=tan(A+B)#

#RHS=1/(LHS)#

#=1/(tan(A+B))#

#=cot(A+B)#

Except when #tanX = cotX (X = pi/4 +npi),#

#LHS != RHS#