Question

If $(1 + \tan1^\circ)(1 + \tan2^\circ)\dots(1 + \tan45^\circ) = 2^n$, then $n =$ ?

• A. 0
• B. 32
• C. 23
• D. 2

Hint:

Use symmetry: $\tan\theta \cdot \tan(45^\circ - \theta)$ type identities for simplification.

Answer 1

The Correct Option is C

Pair terms: $\tan1^\circ \cdot \tan44^\circ$, $\tan2^\circ \cdot \tan43^\circ$, ..., and note: \[ (1 + \tan1^\circ)(1 + \tan44^\circ),\quad (1 + \tan2^\circ)(1 + \tan43^\circ), \dots \] Each pair gives some product; total pairs = 22, plus 1 for $\tan45^\circ = 1$ So the total product becomes: \[ (1 + \tan1^\circ)(1 + \tan2^\circ)\dots(1 + \tan45^\circ) = 2^{23} \Rightarrow n = 23 \]