Confirm that \(\frac{cos A – sin A + 1}{cos A + sin A – 1} = cosec A + cot A\) using the identity \(cosec^2A = 1 + cot^2A\)
Proof:
\(LHS=\frac{cos A – sin A + 1}{cos A + sin A – 1}\)
Dividing both numerator and denominator by \(sin\ A\)
\(=\frac{\frac{cos A}{sin A}-\frac{sin A}{sin A}+\frac{1}{sin A}}{\frac{cos A}{sin A}+\frac{sin A}{sin A}-\frac{1}{sin A}}\)
Multiplying numerator and denominator by \((cot A – 1 + cosec A)\) we get,
\(LHS= \frac{cot^2A + 1 + cosec^2A – 2cot A – 2cosec A + 2cot A cosec A}{cot^2A – (1 + cosec^2A – 2cosec A)}\)
\(=\)
\(=\)
\(=\)
\(= cosec A + cot A\)
\(=RHS\)
Hence \(LHS=RHS\)
Answer variants:
\(\frac{2cosec A(cosec A + cot A) – 2(cosec A + cot A)}{cot^2A – 1 – cosec^2A + 2cosec A}\)
\(\frac{2cosec^2A – 2cot A – 2cosec A + 2cot Acosec A}{cot^2A – 1 – cosec^2A + 2cosec A}\)
\((cosec A + cot A)(2cosec A – 2 ) (2cosec A – 2)\)
\(=\frac{cot A-(1-cosec A)}{cot A+(1-cosec A)}\)
\(=\frac{cot A-1+cosec A}{cot A+1-cosec A}\)