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Section 9.1 #11: Simplify
$$-\tan(-x)\cot(-x).$$
Solution: Recall the "odd property" $\sin(-x)=-\sin(x)$ and the "even property" $\cos(-x)=\cos(x)$. Now simplify
$$-\tan(-x)\cot(-x) = -\dfrac{\sin(-x)}{\cos(-x)} \dfrac{\cos(-x)}{\sin(-x)}= - \dfrac{-\sin(x)}{\cos(x)} \dfrac{\cos(x)}{-\sin(x)}= -1.$$
Section 9.1 #12: Simplify
$$\dfrac{-\sin(-x)\cos(x)\sec(x)\csc(x)\tan(x)}{\cot(x)}.$$
Solution: Calculate
$$\begin{array}{ll}
\dfrac{-\sin(-x)\cos(x)\sec(x)\csc(x)\tan(x)}{\cot(x)} &= \dfrac{-(-\sin(x))\cos(x)\frac{1}{\cos(x)}\frac{1}{\sin(x)}\frac{\sin(x)}{\cos(x)}}{\frac{\cos(x)}{\sin(x)}} \\
&=\dfrac{\frac{\sin(x)}{\cos(x)}}{\frac{\cos(x)}{\sin(x)}} \\
&= \dfrac{\cos^2(x)}{\sin^2(x)}.
\end{array}$$
Section 9.1 #15: Simplify
$$\dfrac{1-\cos^2(x)}{\tan^2(x)}+2\sin^2(x).$$
Solution: Recall the Pythagorean identity $\cos^2(x)+\sin^2(x)=1$. Rearranging that identity yields $\sin^2(x)=1-\cos^2(x)$. So, Compute
$$\begin{array}{ll}
\dfrac{1-\cos^2(x)}{\tan^2(x)}+2\sin^2(x) &= \dfrac{\sin^2(x)}{\frac{\sin^2(x)}{\cos^2(x)}} + 2\sin^2(x) \\
&= \left( \sin^2(x) \right) \left( \dfrac{\cos^2(x)}{\sin^2(x)} \right) + 2\sin^2(x) \\
&=\cos^2(x) + 2\sin^2(x) \\
&= \left( \cos^2(x) + \sin^2(x) \right) + \sin^2(x) \\
&= 1 + \sin^2(x).
\end{array}$$
Section 9.1 #20: Write the first expression in terms of the second expression:
$$\dfrac{1}{1-\cos(x)} - \dfrac{\cos(x)}{1+\cos(x)}; \csc(x).$$
Solution: Calculate
$$\begin{array}{ll}
\dfrac{1}{1-\cos(x)} - \dfrac{\cos(x)}{1+\cos(x)} &= \dfrac{(1+\cos(x)) - \cos(x)(1-\cos(x))}{(1-\cos(x))(1+\cos(x))} \\
&=\dfrac{1+\cos(x)-\cos(x)+\cos^2(x)}{1-\cos^2(x)} \\
&=\dfrac{1+\cos^2(x)}{\sin^2(x)} \\
&= \dfrac{1+(1-\sin^2(x))}{\sin^2(x)} \\
&=\dfrac{2}{\sin^2(x)} - 1 \\
&= 2\csc^2(x) - 1.
\end{array}$$
