Question

write the expression in terms of sine and cosine, and simplify so that no quotients appear in the final expression. (1 + cotθ)(1 - cotθ) - csc²θ

a. 2cot²θ
b. 0
c. 2
d. - 2cot²θ

Explanation:

Step1: Recall trigonometric identities

Recall that $\cot\theta=\frac{\cos\theta}{\sin\theta}$ and $\csc\theta = \frac{1}{\sin\theta}$.

Step2: Expand the product

Expand $(1 + \cot\theta)(1 - \cot\theta)$ using the difference - of - squares formula $(a + b)(a - b)=a^{2}-b^{2}$. Here $a = 1$ and $b=\cot\theta$, so $(1 + \cot\theta)(1 - \cot\theta)=1-\cot^{2}\theta$.

Step3: Substitute identities

The original expression $(1 + \cot\theta)(1 - \cot\theta)-\csc^{2}\theta$ becomes $1-\cot^{2}\theta-\csc^{2}\theta$. Substitute $\cot\theta=\frac{\cos\theta}{\sin\theta}$ and $\csc\theta=\frac{1}{\sin\theta}$:
\[

$$\begin{align*} 1-\frac{\cos^{2}\theta}{\sin^{2}\theta}-\frac{1}{\sin^{2}\theta}&=\frac{\sin^{2}\theta-\cos^{2}\theta - 1}{\sin^{2}\theta}\\ \end{align*}$$

\]

Step4: Use the Pythagorean identity

Recall the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$, or $\sin^{2}\theta=1 - \cos^{2}\theta$. Substitute $\sin^{2}\theta$ in the above expression:
\[

$$\begin{align*} \frac{\sin^{2}\theta-\cos^{2}\theta - 1}{\sin^{2}\theta}&=\frac{(1 - \cos^{2}\theta)-\cos^{2}\theta - 1}{\sin^{2}\theta}\\ &=\frac{1-\cos^{2}\theta-\cos^{2}\theta - 1}{\sin^{2}\theta}\\ &=\frac{- 2\cos^{2}\theta}{\sin^{2}\theta}\\ &=- 2\cot^{2}\theta \end{align*}$$

\]

Answer:

D. $-2\cot^{2}\theta$