Simplifying trionometric expressions or identities often involves a little trial and error, so it's hard to come up with a strategy that works every time. A lot of times you have to try multiple strategies and see which one helps.

Often, if you have any form of \(\displaystyle \tan\theta,\) \(\displaystyle \cot\theta,\) \(\displaystyle \csc\theta,\) or \(\displaystyle \sec\theta,\) in an expression, it helps to rewrite it in terms of sine and cosine. In this problem, we can use the identities \(\displaystyle \cot\theta=\frac{\cos\theta }{\sin\theta }\) and \(\displaystyle \csc\theta=\frac{1}{\sin\theta }\).

\(\displaystyle (1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot(\frac{1}{\sin\theta})^2\)

\(\displaystyle =(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}\).

This doesn't seem to help a whole lot. However, we should recognize that \(\displaystyle (1-\cos^2\theta )=\sin^2\theta\) because of the Pythagorean identity \(\displaystyle \sin^2\theta+\cos^2\theta=1\).

\(\displaystyle (1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=(\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}\)

We can cancel the \(\displaystyle \sin^2\theta\) terms in the numerator and denominator.

\(\displaystyle (\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=\frac{\cos\theta }{\sin\theta }=\cot\theta\).

When working with basic trigonometric identities, it's easiest to remember the mnemonic: \(\displaystyle SOHCAHTOA\).

\(\displaystyle \sin=\frac{opposite}{hypotenuse}\)

\(\displaystyle \cos=\frac{adjacent}{hypotenuse}\), 

\(\displaystyle \tan=\frac{opposite}{adjacent}\)  

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

\(\displaystyle \cot(x) = \cos(x)/\sin(x)\).  Thus: \(\displaystyle [\cos(x)/\sin(x)]\times\sin(x)=\cos(x)\)

\(\displaystyle \cot(x)\times\tan(x) = 1\) 

and

 \(\displaystyle 1/(\cos(x)) = \sec(x)\).

Remember that \(\displaystyle \sin^2x+\cos^2 x=1\). We can rearrange this to simplify our given equation:

\(\displaystyle \frac{\cos^2x-1}{\sin x}\)

\(\displaystyle =\frac{-\sin^2x}{\sin x}=-\sin x\)

Whenever you see a trigonometric function squared, start looking for a Pythagorean identity.

The two identities used in this problem are \(\displaystyle \sin^2 x + \cos ^2 x=1\) and \(\displaystyle \tan^2 x +1 =\sec^2 x\).

Substitute and solve.

\(\displaystyle \sin^2x+\cos^2x+\tan^2x=?\)

\(\displaystyle (\sin^2x+\cos^2x)+\tan^2x=?\)

\(\displaystyle (1)+\tan^2x=\sec^2 x\)

To reduce \(\displaystyle \frac{\sin^2x-9}{\sin x-3}\), factor the numerator: \(\displaystyle \frac{(\sin x-3)(\sin x+3)}{\sin x-3}\)

Notice that we can cancel out a \(\displaystyle \sin x -3\).

This leaves us with \(\displaystyle \sin x +3\).

To simplify \(\displaystyle \sin x\cot x\), break them into their SOHCAHTOA parts:

\(\displaystyle \frac{\text{opposite}}{\text{hypotenuse}}*\frac{\text{adjacent}}{\text{opposite}}\).

Notice that the opposite's cancel out, leaving \(\displaystyle \frac{\text{adjacent}}{\text{hypotenuse}}=\cos x\).