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Prove that $$(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$$ I tried solving the LHS and RHS seperately but they were not coming out to be equal. Please help me answer this question. And also how does one go about proving such questions? Thanks in advance

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$(\csc\theta-\sec\theta)(\cot\theta-\tan\theta)=\displaystyle\big(\frac{1}{\sin\theta}-\frac{1}{\cos\theta}\big)\big(\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}\big)$

$\displaystyle=\big(\frac{\cos\theta-\sin\theta}{\sin\theta\cos\theta}\big)\big(\frac{\cos^{2}\theta-\sin^2\theta}{\cos\theta\sin\theta}\big)$

$\displaystyle=\frac{(\cos\theta-\sin\theta)^2(\cos\theta+\sin\theta)}{\sin^2\theta\cos^2\theta}$

$\displaystyle=\frac{(\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta)(\cos\theta+\sin\theta)}{\sin^2\theta\cos^2\theta}$

$\displaystyle=\frac{(1-2\sin\theta\cos\theta)(\cos\theta+\sin\theta\big)}{\sin^2\theta\cos^2\theta}$

$\displaystyle=\big(\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}\big)\big(\frac{1-2\sin\theta\cos\theta}{\sin\theta\cos\theta}\big)$

$\displaystyle=\big(\frac{1}{\sin\theta}+\frac{1}{\cos\theta}\big)\big(\frac{1}{\sin\theta\cos\theta}-2\big)$

$\displaystyle=(\csc\theta+\sec\theta)(\csc\theta\sec\theta-2)$

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$$\begin{array}{lll} (\csc\theta - \sec\theta)(\cot\theta - \tan\theta)&=&(\csc\theta - \sec\theta)\bigg(\frac{\csc\theta}{\sec\theta} - \frac{\sec\theta}{\csc\theta}\bigg)\\ &=&(\csc\theta - \sec\theta)\bigg(\frac{\csc^2\theta-\sec^2\theta}{\sec\theta\csc\theta}\bigg)\\ &=&\frac{(\csc\theta-\sec\theta)(\csc\theta-\sec\theta)(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}\\ &=&\frac{(\color{blue}{\csc^2\theta}-2\sec\theta\csc\theta\color{blue}{+\sec^2\theta})(\csc\theta+\sec\theta)}{\sec\theta\csc\theta} \end{array}$$

using the 4th Pythagorean identity $$\sin^2 \theta+\cos^2 \theta = 1$$ $$\frac{\sin^2 \theta+\cos^2}{\cos^2\theta\sin^2\theta} = \frac{1}{\cos^2\theta\sin^2\theta}$$

$$\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$$

we have $$\begin{array}{lll} (\csc\theta - \sec\theta)(\cot\theta - \tan\theta)&=&\frac{(\color{blue}{\sec^2\theta\csc^2\theta}-2\sec\theta\csc\theta)(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}\\ &=&(\csc\theta + \sec\theta)(\sec\theta\csc\theta-2) \end{array}$$

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$$ \left( \mathrm{cosec}\, \theta - \sec \theta\right)\left(\cot \theta - \tan \theta\right) = \frac{\left( \mathrm{cosec}\, \theta + \sec \theta\right)}{\left( \mathrm{cosec}\, \theta + \sec \theta\right)}\left( \mathrm{cosec}\, \theta - \sec \theta\right)\left(\cot \theta - \tan \theta\right) $$ this leads to

$$ \left( \mathrm{cosec}\, \theta + \sec \theta\right)\frac{1-\tan \theta}{1+\tan \theta}\left(1-\tan^2\theta\right)\cot \theta = \left( \mathrm{cosec}\, \theta + \sec \theta\right)\left(1-\tan\theta\right)^2\cot \theta $$ here I have utilized $$\left(1-\tan^2\theta\right) = \left(1-\tan \theta\right)\left(1+\tan \theta\right)$$ next step $$ \left(1-\tan\right)^2\cot \theta = \cot \theta +\tan\theta-2 = \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}-2 = \frac{1}{\sin \theta \cos \theta}-2 $$ thus we obtain $$ \left( \mathrm{cosec}\, \theta + \sec \theta\right)\left(\frac{1}{\sin \theta \cos \theta}-2\right) $$

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$$\frac{(\csc x+\sec x)(\sec x\csc x-2)}{\csc x-\sec x}=\frac{(\cos x+\sin x)(1-2\sin x\cos x)}{(\cos x\sin x)(\cos x-\sin x)}$$

$$=\frac{(\cos x+\sin x)(\cos x-\sin x)^2}{(\cos x\sin x)(\cos x-\sin x)}$$

$$=\frac{\cos^2x-\sin^2x}{\cos x\sin x}=?$$

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  • $\begingroup$ @user166748, How about this? $\endgroup$ – lab bhattacharjee Jul 29 '14 at 9:23

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