Question:

Step 1: Establish the identity for angles summing to $45^\circ$. For any angles $A$ and $B$ such that $A + B = 45^\circ$, we have $B = 45^\circ - A$. Using the tangent subtraction formula, $\tan(45^\circ - A) = \frac{\tan 45^\circ - \tan A}{1 + \tan 45^\circ \tan A} = \frac{1 - \tan A}{1 + \tan A}$. Consider the product $(1 + \tan A)(1 + \tan B)$: $$ (1 + \tan A)(1 + \tan B) = (1 + \tan A)(1 + \tan(45^\circ - A)) $$ Substitute the expression for $\tan(45^\circ - A)$: $$ (1 + \tan A)\left(1 + \frac{1 - \tan A}{1 + \tan A}\right) = (1 + \tan A)\left(\frac{(1 + \tan A) + (1 - \tan A)}{1 + \tan A}\right) $$ $$ = (1 + \tan A)\left(\frac{2}{1 + \tan A}\right) = 2 $$ Thus, if $A + B = 45^\circ$, then $(1 + \tan A)(1 + \tan B) = 2$. Step 2: Apply the identity to the given product. The given product is $P = (1 + \tan 5^\circ)(1 + \tan 10^\circ)\cdots(1 + \tan 45^\circ)$. The angles are $5^\circ, 10^\circ, 15^\circ, 20^\circ, 25^\circ, 30^\circ, 35^\circ, 40^\circ, 45^\circ$. There are 9 terms in total. We can group these terms into pairs where the sum of the angles is $45^\circ$: \begin{align*} (1 + \tan 5^\circ)(1 + \tan 40^\circ) &= 2 \\ (1 + \tan 10^\circ)(1 + \tan 35^\circ) &= 2 \\ (1 + \tan 15^\circ)(1 + \tan 30^\circ) &= 2 \\ (1 + \tan 20^\circ)(1 + \tan 25^\circ) &= 2 \end{align*} These four pairs contribute a factor of $2^4$. The remaining term is $(1 + \tan 45^\circ)$. Since $\tan 45^\circ = 1$, this term evaluates to: $$ (1 + \tan 45^\circ) = (1 + 1) = 2 $$ The total product is the product of these pairs and the remaining term: $$ P = (2 \times 2 \times 2 \times 2) \times 2 = 2^4 \times 2^1 = 2^5 $$ Step 3: Determine the value of $k$. We are given that the product equals $2^{k+1}$. $$ 2^{k+1} = 2^5 $$ Equating the exponents, we get: $$ k+1 = 5 $$ $$ k = 4 $$