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

1 Answer
Jun 18, 2018

LHSRHS

Explanation:

We know that

tan(A+B)=tanA+tanB1tanAtanB

Thus,

LHS=cotA+cotBcotAcotB1

=1tanA+1tanB1tanA1tanB1

=tanA+tanBtanAtanB1tanAtanBtanAtanB

=tanA+tanB1tanAtanB

=tan(A+B)

RHS=1LHS

=1tan(A+B)

=cot(A+B)

Except when tanX=cotX(X=π4+nπ),

LHSRHS