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

Answer

$\text{L.H.S}=\big(\text{cosec }\text{x}-\sin\text{x}\big)\big(\sec\text{x}-\cos\text{x}\big)\big(\tan\text{x}+\cot\text{x}\big)$
$=\Big(\frac{1}{\sin\text{x}}-\sin\text{x}\Big)\Big(\frac{1}{\cos\text{x}}-\cos\text{x}\Big)\Big(\frac{\sin\text{x}}{\cos\text{x}}+\frac{\cos\text{x}}{\sin\text{x}}\Big)$$\begin{bmatrix}\because\text{cosec}=\frac{1}{\sin\text{x}},\sec\text{x}=\frac{1}{\cos\text{x}},\\\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}} \text{and}\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\end{bmatrix}$
$=\Big(\frac{1-\sin^{2}\text{x}}{\sin\text{x}}\Big)\Big(\frac{1-\cos^{2}\text{x}}{\cos\text{x}}\Big)\Big(\frac{\sin\text{x}}{\cos\text{x}}+\frac{\cos\text{x}}{\sin\text{x}}\Big)$
$=\frac{\cos^{2}\text{x}.\sin^{2}\text{x}.1}{\sin^{2}\text{x}.\cos^{2}\text{x}}$ $\begin{pmatrix}\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\\\Rightarrow1-\sin^{2}\text{x}=\cos^{2}\text{x},\text{and}1-\cos^{2}\text{x}=\sin^{2}\text{x}\end{pmatrix}$
$=1$
$=\text{R.H.S}$
$\text{Proved}$

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