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