<s>[INST] Trignometry-Prove that $(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$ 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
 [/INST] $$\begin{array}{lll}
(\csc\theta - \sec\theta)(\cot\theta - \tan\theta)&=&(\csc\theta - \sec\theta)\bigg(\frac{\csc\theta}{\sec\theta} - \frac{\sec\theta}{\csc\theta}\bigg)\\
&=&(\csc\theta - \sec\theta)\bigg(\frac{\csc^2\theta-\sec^2\theta}{\sec\theta\csc\theta}\bigg)\\
&=&\frac{(\csc\theta-\sec\theta)(\csc\theta-\sec\theta)(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}\\
&=&\frac{(\color{blue}{\csc^2\theta}-2\sec\theta\csc\theta\color{blue}{+\sec^2\theta})(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}
\end{array}$$
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
$$\begin{array}{lll}
(\csc\theta - \sec\theta)(\cot\theta - \tan\theta)&=&\frac{(\color{blue}{\sec^2\theta\csc^2\theta}-2\sec\theta\csc\theta)(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}\\
&=&(\csc\theta + \sec\theta)(\sec\theta\csc\theta-2)
\end{array}$$
</s>