Question

sinA/(1-cosA) +tanA/(secA+1)=secA.cosecA+cotA

Answer 1

The given equation is not an identity, because:

when A=45°, LHS=(1/√2)/(1-1/√2)+1/(1+√2)=1/(√2-1)+1/(1+√2)=1+√2+√2-1=2√2; RHS=2+1=3. Therefore LHS≠RHS.

tanA/(secA+1)=sinA/(1+cosA).

sinA/(1-cosA)+sinA/(1+cosA)=sinA(1+cosA+1-cosA)/(1-cos²A)=

2sinA/sin²A=2/sinA=2cosecA. (When A=45°, 2cosecA=2√2.)

Therefore the identity is sinA/(1-cosA)+tanA/(secA+1)=2cosecA.

So, since the given equation is not an identity:

2cosecA=secAcosecA+cotA. Multiply through by sinAcosA:

2cosA=1+cos²A, cos²A-2cosA+1=(cosA-1)²=0. Therefore cosA=1, A=2πn radians or 360n°, where n is an integer. However, cosA=1 causes division by zero in the original equation, so this solution cannot be validated, implying that there is no solution. I suspect the question has been wrongly stated.