Prove the following identities: cosec x( −1)− (1− )= −

Question

Prove the following identities: $\text{cosec}\text{ x}(\sec\text{x}−1)−\cot\text{x}(1−\cos\text{x})=\tan\text{x}−\sin\text{x}$

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Answer

$\text{L.H.S}=\text{cosec }\text{x}\big(\sec\text{x}-1\big)-\cot\big(1-\cos\text{x}\big)$ $=\frac{1}{\sin\text{x}}\Big(\frac{1}{\cos\text{x}}-1\Big)-\frac{\cos\text{x}}{\sin\text{x}}\big(1-\cos\text{x}\big)$$ \Big[\because\text{cosec }\text{x}=\frac{1}{\sin\text{x}},\sec\text{x}=\frac{1}{\cos\text{x}}\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\Big]$ $=\frac{\big(1-\cos\text{x}\big)}{\sin\text{x}\cos\text{x}}-\frac{\cos\text{x}\big(1-\cos\text{x}\big)}{\sin\text{x}}$ $=\frac{\big(1-\cos\text{x}\big)-\cos^{2}\text{x}\big(1-\cos\text{x}\big)}{\sin\text{x}\cos\text{x}}$ $=\frac{\big(1-\cos\text{x}\big)\big(1-\cos^{2}\text{x}\big)}{\sin\text{x}\cos\text{x}}$ $=\frac{\big(1-\cos\text{x}\big)\sin^{2}\text{x}}{\sin\text{x}\cos\text{x}}$ $=\big(1-\cos\text{x}\big)\frac{\sin\text{x}}{\cos\text{x}}$ $=\frac{\sin\text{x}}{\cos\text{x}}-\sin\text{x}$ $=\tan\text{x}-\sin\text{x}$ $\big(\because1-\cos^{2}\text{x}=\sin^{2}\text{x}\big)$ $=\text{R.H.S} $ $\text{proved}$

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