Verify that \(\frac{\sin A}{\sec A + \tan A - 1} + \frac{\cos A}{\text{cosec}\: A + \cot A - 1} = 1\).
Proof:
LHS \(= \frac{\sin A}{\sec A + \tan A - 1} + \frac{\cos A}{\text{cosec}\: A + \cot A - 1}\)
\(= \frac{\sin A(\text{cosec}\: A + \cot A - 1) + \cos A (\sec A + \tan A - 1)}{(\sec A + \tan A - 1)(\text{cosec}\: A + \cot A - 1)}\)
\(= \frac{\sin A(\text{cosec}\: A + \cot A - 1) + \cos A (\sec A + \tan A - 1)}{(\sec A + \tan A - 1)(\text{cosec}\: A + \cot A - 1)}\)
\(=\)
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\(= 1\)
\(=\) RHS
Hence, proved.
Answer variants:
\(\frac{2 \sin A\cos A}{1 + 2 \sin A \cos A - (\sin^2 A + \cos^2 A)}\)
\(\frac{2 \sin A\cos A}{(1 + \sin A - \cos A)(1 + \cos A - \sin A)}\)
\(\frac{1 + \cos A - \sin A + 1 + \sin A - \cos A}{\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A} - 1\right) \left(\frac{1}{\sin A} + \frac{\cos A}{\sin A} - 1\right)}\)
\(\frac{2}{\left(\frac{1 + \sin A - \cos A}{\cos A}\right)\left(\frac{1 + \cos A - \sin A}{\sin A}\right)}\)
\(\frac{\left(\sin A \cdot \frac{1}{\sin A}\right) + \left(\sin A \cdot \frac{\cos A}{\sin A}\right) - \sin A + \left(\cos A \cdot \frac{1}{\cos A}\right) + \left(\cos A \cdot \frac{\sin A}{\cos A}\right) - \cos A}{(\sec A + \tan A - 1)(\text{cosec}\: A + \cot A - 1)}\)
\(\frac{\sin A \: \text{cosec}\: A + \sin A \cot A - \sin A + \cos A \sec A + \cos A \tan A - \cos A}{(\sec A + \tan A - 1)(\text{cosec}\: A + \cot A - 1)}\)
\(\frac{1 + \cos A - \sin A + 1 + \sin A - \cos A}{(\sec A + \tan A - 1)(\text{cosec}\: A + \cot A - 1)}\)
\(\frac{2 \sin A\cos A}{1 + 2 \sin A \cos A - 1}\)
\(\frac{2 \sin A\cos A}{1 + \cos A - \sin A + \sin A + \sin A \cos A - \sin^2 A - \cos A - \cos^2 A + \cos A \sin A}\)
\(\frac{2 \sin A\cos A}{2 \sin A \cos A}\)