We have to prove (cosecθ – secθ)(cotθ – tanθ) = (cosecθ + secθ)(secθ cosecθ – 2)
\(\Rightarrow \frac{\mathrm{cosec} \theta - \sec\theta}{\mathrm{cosec}\theta +\sec\theta}\) \(=\frac{\sec\theta\mathrm{cosec}\theta-2}{\cot\theta-\tan\theta}\)
L.H.S \(=\frac{\mathrm{cosec}\theta-\sec\theta}{\mathrm{cosec}\theta+\sec\theta}\)
\(=\cfrac{\frac{1}{\sin\theta}-\frac{1}{\cos\theta}}{\frac{1}{\sin\theta}+\frac{1}{\cos\theta}}\) \(=\cfrac{\frac{\cos\theta - \sin\theta}{\sin\theta\cos\theta}}{\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}}\)
\(=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\)
R.H.S \(=\frac{\sec\theta\mathrm{cosec}\theta-2}{\cot\theta-\tan\theta}\)
\(=\cfrac{\frac{1}{\cos\theta\sin\theta}-2}{\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}}\) \(=\frac{1-2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}\)
\(=\frac{\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta}{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta}\) \((\because \sin^2\theta+\cos^2\theta = 1\) & \(a^2-b^2=(a+b)(a-b))\)
\(=\frac{(\cos\theta-\sin\theta)^2}{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}\) \((\because a^2+b^2-2ab = (a-b)^2)\)
\(=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\) = L.H.S
Hence proved
