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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

$\begin{array}{ll} \dfrac{1}{1-\cos(x)} - \dfrac{\cos(x)}{1+\cos(x)} &= \dfrac{(1+\cos(x)) - \cos(x)(1-\cos(x))}{(1-\cos(x))(1+\cos(x))} \\ &=\dfrac{1+\cos(x)-\cos(x)+\cos^2(x)}{1-\cos^2(x)} \\ &=\dfrac{1+\cos^2(x)}{\sin^2(x)} \\ &= \dfrac{1+(1-\sin^2(x))}{\sin^2(x)} \\ &=\dfrac{2}{\sin^2(x)} - 1 \\ &= 2\csc^2(x) - 1. \end{array}$