ANSWER :
-tan θ
EXPLANATION :
Note that :
[tex]\begin{gathered} \cot\theta=\frac{\cos\theta}{\sin\theta}\quad,\quad\csc\theta=\frac{1}{\sin\theta} \\ \\ -\sin^2=\cos^2\theta-1 \end{gathered}[/tex]
From the problem, we have :
[tex]\cot\theta-\frac{\csc^2\theta}{\cot\theta}[/tex]
Subsitute the identities :
[tex]\begin{gathered} \frac{\cos\theta}{\sin\theta}-\frac{\frac{1}{\sin^2\theta}}{\frac{\cos\theta}{\sin\theta}}=\frac{\cos\theta}{\sin\theta}-(\frac{1}{\sin^2\theta})(\frac{\sin\theta}{\cos\theta}) \\ =\frac{\cos(\theta)}{\sin(\theta)}-\frac{1}{\sin\theta\cos\theta} \\ =\frac{\cos^2\theta-1}{\sin\theta\cos\theta} \\ =\frac{-\sin^2\theta}{\sin\theta\cos\theta} \\ =\frac{-\sin\theta}{\cos\theta} \\ =-\tan\theta \end{gathered}[/tex]
The answer is -tan θ
