Question:
tan A/2-cotA/2+2cotA=0
Answer:
We have to prove that
tanA/2 - cotA/2 + 2cotA=0
LHS:
tanA/2 - cotA/2 + 2cotA
= {sin(A/2)/cos(A/2)} - {cos(A/2)/sin(A/2)} + 2cotA
= {sin2 (A/2) - cos2 (A/2)}/{cos(A/2)*sin(A/2)} + 2cotA
= -{cos2 (A/2) - sin2 (A/2)}/{cos(A/2)*sin(A/2)} + 2cotA
= -{cosA}/{cos(A/2)*sin(A/2)} + 2cotA (since cos2 A - sin2 A = cos2A )
= -{cos(2A/2)}/{cos(A/2)*sin(A/2)} + 2cotA
= -{2*cosA}/{2*cos(A/2)*sin(A/2)} + 2cotA
= -{2cosA}/{sin(2A/2)} + 2cotA
= {-(2cosA)/(sinA)} + 2cotA (since sin2A = 2*sinA*cosA)
= -2cotA + 2cotA
= 0
= RHS
So tan A/2-cotA/2+2cotA=0
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