Question

If
\[ (1+\tan1^\circ)(1+\tan2^\circ)\cdots(1+\tan45^\circ)=2^n, \] then \(n=\)

• A. \(0\)
• B. \(32\)
• C. \(23\)
• D. \(2\)

Hint:

For products involving \(\tan1^\circ,\tan2^\circ,\ldots,\tan45^\circ\), pair angles whose sum is \(45^\circ\).

Answer 1

The Correct Option is C

Step 1: Use the identity for complementary angles.
We know that
\[ \tan(45^\circ-\theta)=\frac{1-\tan\theta}{1+\tan\theta} \]
Now, consider
\[ (1+\tan\theta)(1+\tan(45^\circ-\theta)) \]
Substituting the identity,
\[ (1+\tan\theta)\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right) \]
\[ =(1+\tan\theta)\left(\frac{1+\tan\theta+1-\tan\theta}{1+\tan\theta}\right) \]
\[ =(1+\tan\theta)\left(\frac{2}{1+\tan\theta}\right) \]
\[ =2 \]

Step 2: Pair the terms.
The terms from \(1^\circ\) to \(44^\circ\) can be paired as
\[ (1^\circ,44^\circ), (2^\circ,43^\circ), \ldots, (22^\circ,23^\circ) \]
There are \(22\) such pairs.
Each pair gives product \(2\).
Therefore, product of first \(44\) terms is
\[ 2^{22} \]

Step 3: Include the remaining term.
The remaining term is
\[ 1+\tan45^\circ \]
Since
\[ \tan45^\circ=1 \]
we get
\[ 1+\tan45^\circ=2 \]
Thus, total product is
\[ 2^{22}\cdot 2=2^{23} \]

Step 4: Compare with \(2^n\).
Given,
\[ (1+\tan1^\circ)(1+\tan2^\circ)\cdots(1+\tan45^\circ)=2^n \]
So,
\[ 2^n=2^{23} \]
Hence,
\[ n=23 \]

Step 5: Final conclusion.
Therefore,
\[ \boxed{23} \]