How to simplify $\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $ How to simplify the following expression :
$$\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
 A: Write $1$ in the numerator as : $$\sec^2(\theta) - \tan^2(\theta)$$
$$\frac{(\sec\theta -\tan\theta)^2+\sec^2\theta - \tan^2\theta}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta -\tan\theta)^2+(\sec\theta - \tan\theta)(\sec\theta + \tan\theta)}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta - \tan\theta)(\sec\theta - \tan\theta + \sec\theta + \tan\theta)}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
$$\frac{(\sec\theta - \tan\theta)(2 \sec\theta)}{\csc\theta(\sec\theta - \tan\theta)}$$
$$2 \tan\theta$$
Hence the simplified result is: $$2 \tan\theta$$
Hope the answer is clear !
A: The numerator becomes 
$(\sec\theta -\tan\theta)^2+1=\sec^2\theta+\tan^2\theta-2\sec\theta\tan\theta+1=2\sec\theta(\sec\theta -\tan\theta)$
So, $$\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta}$$
$$=\frac{2\sec\theta(\sec\theta -\tan\theta)}{\csc\theta(\sec\theta -\tan\theta)}=2\frac{\sec\theta}{\csc\theta}(\text{ assuming } \sec\theta -\tan\theta\ne0)$$
$$=2\frac{\sin\theta}{\cos\theta}=2\tan\theta$$
A: Here is a detailed solution.Ready, set, go!
$$\require{cancel}\begin{align}\frac{\left(\sec\theta-\tan\theta\right)^2+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\theta+\color{blue}{\tan^2\theta+1}}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\theta+\color{blue}{\sec^2\theta}}{\sec\theta\csc\theta-\tan\theta\csc\theta}\tag{1}\label{ko-eq1}\\&=\frac{2\sec^2\theta-2\sec\theta\tan\theta}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{2\sec\theta\cancel{\left(\sec\theta-\tan\theta\right)}}{\csc\theta\cancel{\left(\sec\theta-\tan\theta\right)}}\\&=\frac{2\sec\theta}{\csc\theta}\\&=\frac{\frac{2}{\cos\theta}}{\frac{1}{\sin\theta}}\\&=\frac{2}{\cos\theta}\sin\theta\\&=\frac{2\sin\theta}{\cos\theta}\\&=2\tan\theta\end{align}$$
In equation $\eqref{ko-eq1}$, $\tan^2\theta+1=\sec^2\theta$.I hope this helps.
