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**Section 3.4 #20:** Prove the identity
$$\sec(\theta)\sin(\theta)=\tan(\theta).$$
*Solution:* We will start from the left-hand side and end up on the right-hand side. Write the lefth and side in terms of sines and cosines and then proceed:
$$\begin{array}{ll}
\sec(\theta)\sin(\theta) &= \dfrac{1}{\cos(\theta)} \sin(\theta) \\
&= \dfrac{\sin(\theta)}{\cos(\theta)} \\
&= \tan(\theta),
\end{array}$$
because the identity $\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)}$ is a well-known "basic trigonometric identity", as was to be shown.

**Section 3.4 #37:** Prove the identity
$$\sec(u)-\tan(u) = \dfrac{\cos(u)}{1+\sin(u)}.$$
*Solution:* It is not clear which side is "more difficult" -- on the right-hand side we have a denominator with a sum and on the left side we do not have this, so it is reasonable to think of the right-hand side as "more difficult". We will start with the right side and try to derive the left side.

In order to remove the sum in the denominator, we will use the "multiply by the conjugate trick". After that, we will use the basic triongometric identity $\sin^2(u)+\cos^2(u)=1$ rearranged to $1-\sin^2(u)=\cos^2(u)$. To finalize, we will use the basic trigonometric identites $\dfrac{1}{\cos(u)}=\sec(u)$ and $\tan(u)=\dfrac{\sin(u)}{\cos(u)}$ to see
$$\begin{array}{ll}
\dfrac{\cos(u)}{1+\sin(u)} &= \dfrac{\cos(u)}{1+\sin(u)} \left( \dfrac{1-\sin(u)}{1-\sin(u)} \right) \\
&= \dfrac{\cos(u)(1-\sin(u))}{1-\sin^2(u)} \\
&= \dfrac{\cos(u)(1-\sin(u))}{\cos^2(u)} \\
&= \dfrac{1-\sin(u)}{\cos(u)} \\
&= \dfrac{1}{\cos(u)} - \dfrac{\sin(u)}{\cos(u)} \\
&= \sec(u) - \tan(u),
\end{array}$$
as was to be shown.

**Section 3.4 #47:** Prove the identity
$$\dfrac{1+\sin(\theta)}{1-\sin(\theta)} = \dfrac{\csc(\theta)+1}{\csc(\theta)-1}.$$
*Solution:* The right-hand side seems more complicated than the left hand side, so we will start with the right-hand side and derive the left-hand side. We will begin by using the basic trignometric identity $\csc(\theta) = \dfrac{1}{\sin(\theta)}$. Afterwards the rest follows from basic algebra, where we will use the formula $\dfrac{\frac{a}{b}}{\frac{c}{d}} = \dfrac{a}{b} \dfrac{d}{c}$. Compute
$$\begin{array}{ll}
\dfrac{\csc(\theta)+1}{\csc(\theta)-1} &= \dfrac{\frac{1}{\sin(\theta)}+1}{\frac{1}{\sin(\theta)}-1} \\
&= \dfrac{\frac{1+\sin(\theta)}{\sin(\theta)}}{\frac{1-\sin(\theta)}{\sin(\theta)}} \\
&= \dfrac{1+\sin(\theta)}{\sin(\theta)} \dfrac{\sin(\theta)}{1-\sin(\theta)} \\
&= \dfrac{1+\sin(\theta)}{1-\sin(\theta)},
\end{array}$$
as was to be shown.

**Section 3.4 #58:** Prove the identity
$$\dfrac{\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)} = \dfrac{\tan(\theta)}{1-\tan^2(\theta)}.$$
*Solution:* It appears that the right-hand-side is "more difficult" -- both have numerators and denominators but the right hand side has tangents while the left-hand isde only has sines and cosines. We will rewrite the right-hand side using sines and cosines and algebraically simplify by getting a common denominator and then using the algebraic fact that $\dfrac{\frac{a}{b}}{\frac{c}{d}} = \dfrac{a}{b} \dfrac{d}{c}$:
$$\begin{array}{ll}
\dfrac{\tan(\theta)}{1-\tan^2(\theta)} &= \dfrac{\frac{\sin(\theta)}{\cos(\theta)}}{1-\frac{\sin^2(\theta)}{\cos^2(\theta)}} \\
&= \dfrac{\frac{\sin(\theta)}{\cos(\theta)}}{\frac{\cos^2(\theta)-\sin^2(\theta)}{\cos^2(\theta)}} \\
&= \dfrac{\sin(\theta)}{\cos(\theta)} \dfrac{\cos^2(\theta)}{\cos^2(\theta)-\sin^2(\theta)} \\
&=\dfrac{\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)},
\end{array}$$
as was to be shown.

**Section 3.4 #86:** Prove the identity
$$\dfrac{1-2\cos^2(\theta)}{\sin(\theta)\cos(\theta)} = \tan(\theta) - \cot(\theta).$$
*Solution:* We will start with the right-hand side. We will rewrite it using sines and cosines and then algebrically simplify by finding a common denominator and then using the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$ rearranged to $\sin^2(\theta) = 1-\cos^2(\theta)$:
$$\begin{array}{ll}
\tan(\theta) - \cot(\theta) &= \dfrac{\sin(\theta)}{\cos(\theta)} - \dfrac{\cos(\theta)}{\sin(\theta)} \\
&= \dfrac{\sin^2(\theta)-\cos^2(\theta)}{\cos(\theta)\sin(\theta)} \\
&=\dfrac{1-\cos^2(\theta)-\cos^2(\theta)}{\cos(\theta)\sin(\theta)} \\
&= \dfrac{1-2\cos^2(\theta)}{\cos(\theta)\sin(\theta)},
\end{array}$$
as was to be shown.