Question

Write the first expression in terms of the second if the terminal point determined by $t$is in the given quadrant.

${\sec }^{2}t{\sin }^{2}t,\cos t;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}anyquadrant}$

Short answer

The first expression in terms of the second if the terminal point determined by $t$is in any quadrant is$\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{t}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{t}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{t}}{\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{t}}$.

Step-by-step solution

Step 1. State the Trigonometric identities.

The reciprocal identities are $\csc t=\frac{1}{\sin t}$, $\sec t=\frac{1}{\cos t}$, $\cot t=\frac{1}{\tan t}$, $\tan t=\frac{\sin t}{\cos t}$and $\cot =\frac{\cos t}{\sin t}$.

The Pythagorean identities are:

1.${\sin }^{2}t+{\cos }^{2}t=1$

2. ${\tan }^{2}t+1={\sec }^{2}t$

3.${\cot }^{2}t+1={\csc }^{2}t$

Step 2. Rewrite the expression sint and sect in terms of cost. 

Using Pythagorean Identity ${\sin }^{2}t+{\cos }^{2}t=1$, rewrite the expression ${\sin }^{2}t$into terms of ${\cos }^{2}t$.

$$\begin{gathered} {\sin }^{2}t+{\cos }^{2}t=1 \\ {\sin }^{2}t=1-{\cos }^{2}t\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\cos }^{2}t\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}onbothsides} \end{gathered}$$

Next use reciprocal identity, $\sec t=\frac{1}{\cos t}$, to rewrite the expression ${\sec }^{2}t$into terms of ${\cos }^{2}t$.

$$\begin{gathered} {\sec }^{2}t={\left(\frac{1}{\cos t}\right)}^2 \\ =\frac{1}{{\cos }^{2}t} \end{gathered}$$

Step 3. State the conclusion.

The final expression is:

$$\begin{gathered} {\sec }^{2}t{\sin }^{2}t=\frac{1}{{\cos }^{2}t}\times \left(1-{\cos }^{2}t\right) \\ =\frac{1-{\cos }^{2}t}{{\cos }^{2}t} \end{gathered}$$

Since ${\cos }^{2}t>0$, for any real number $t$, therefore, ${\sec }^{2}t{\sin }^{2}t=\frac{1-{\cos }^{2}t}{{\cos }^{2}t}$ in any quadrant.