Find $k$ if
\[(\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = k + \tan^2 \alpha + \cot^2 \alpha.\]
We have that
\begin{align*}
k &= (\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 - \tan^2 \alpha - \cot^2 \alpha \\
&= \left( \sin \alpha + \frac{1}{\sin \alpha} \right)^2 + \left( \cos \alpha + \frac{1}{\cos \alpha} \right)^2 - \frac{\sin^2 \alpha}{\cos^2 \alpha} - \frac{\cos^2 \alpha}{\sin^2 \alpha} \\
&= \sin^2 \alpha + 2 + \frac{1}{\sin^2 \alpha} + \cos^2 \alpha + 2 + \frac{1}{\cos^2 \alpha} - \frac{\sin^2 \alpha}{\cos^2 \alpha} - \frac{\cos^2 \alpha}{\sin^2 \alpha} \\
&= 5 + \frac{1 - \sin^2 \alpha}{\cos^2 \alpha} + \frac{1 - \cos^2 \alpha}{\sin^2 \alpha} \\
&= 5 + \frac{\cos^2 \alpha}{\cos^2 \alpha} + \frac{\sin^2 \alpha}{\sin^2 \alpha} \\
&= \boxed{7}.
\end{align*}
7
Level 3
Precalculus
Problem: Find $k$ if
\[(\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = k + \tan^2 \alpha + \cot^2 \alpha.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for $k$, we start by expanding the left-hand side of the equation. 

First, we expand $(\sin \alpha + \csc \alpha)^2$:
\[
(\sin \alpha + \csc \alpha)^2 = \sin^2 \alpha + 2\sin \alpha \csc \alpha + \csc^2 \alpha.
\]
Since $\csc \alpha = \frac{1}{\sin \alpha}$, we have:
\[
2\sin \alpha \csc \alpha = 2 \quad \text{and} \quad \csc^2 \alpha = \frac{1}{\sin^2 \alpha}.
\]
Thus,
\[
(\sin \alpha + \csc \alpha)^2 = \sin^2 \alpha + 2 + \frac{1}{\sin^2 \alpha}.
\]

Next, we expand $(\cos \alpha + \sec \alpha)^2$:
\[
(\cos \alpha + \sec \alpha)^2 = \cos^2 \alpha + 2\cos \alpha \sec \alpha + \sec^2 \alpha.
\]
Since $\sec \alpha = \frac{1}{\cos \alpha}$, we have:
\[
2\cos \alpha \sec \alpha = 2 \quad \text{and} \quad \sec^2 \alpha = \frac{1}{\cos^2 \alpha}.
\]
Thus,
\[
(\cos \alpha + \sec \alpha)^2 = \cos^2 \alpha + 2 + \frac{1}{\cos^2 \alpha}.
\]

Now, we combine both expansions:
\[
(\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = \left(\sin^2 \alpha + \frac{1}{\sin^2 \alpha}\right) + \left(\cos^2 \alpha + \frac{1}{\cos^2 \alpha}\right) + 4.
\]

Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$, we can rewrite the expression:
\[
= 1 + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} + 4.
\]

Next, we simplify $\frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha}$:
\[
\frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} = \frac{\cos^2 \alpha + \sin^2 \alpha}{\sin^2 \alpha \cos^2 \alpha} = \frac{1}{\sin^2 \alpha \cos^2 \alpha}.
\]

Thus, we have:
\[
(\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = 5 + \frac{1}{\sin^2 \alpha \cos^2 \alpha}.
\]

Now, we relate this to the right-hand side of the original equation:
\[
k + \tan^2 \alpha + \cot^2 \alpha.
\]
Using the identities $\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}$ and $\cot^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha}$, we find:
\[
\tan^2 \alpha + \cot^2 \alpha = \frac{\sin^4 \alpha + \cos^4 \alpha}{\sin^2 \alpha \cos^2 \alpha}.
\]
Using the identity $\sin^4 \alpha + \cos^4 \alpha = (\sin^2 \alpha + \cos^2 \alpha)^2 - 2\sin^2 \alpha \cos^2 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha$, we have:
\[
\tan^2 \alpha + \cot^2 \alpha = \frac{1 - 2\sin^2 \alpha \cos^2 \alpha}{\sin^2 \alpha \cos^2 \alpha}.
\]

Now, substituting this back into the equation gives:
\[
5 + \frac{1}{\sin^2 \alpha \cos^2 \alpha} = k + \frac{1 - 2\sin^2 \alpha \cos^2 \alpha}{\sin^2 \alpha \cos^2 \alpha}.
\]

To isolate $k$, we rearrange:
\[
k = 5 + \frac{1}{\sin^2 \alpha \cos^2 \alpha} - \frac{1 - 2\sin^2 \alpha \cos^2 \alpha}{\sin^2 \alpha \cos^2 \alpha}.
\]
This simplifies to:
\[
k = 5 + \frac{1 + 2\sin^2 \alpha \cos^2 \alpha - 1}{\sin^2 \alpha \cos^2 \alpha} = 5 + \frac{2\sin^2 \alpha \cos^2 \alpha}{\sin^2 \alpha \cos^2 \alpha} = 5 + 2 = 7.
\]

Thus, the final answer is $7$.<|END_OF_LATENT|> The final answer is $7$.